ChemWiki - User contributions [en]

Rep:Mod:CMB22301333464

2019-05-16T12:57:11Z

Fiw17: /* B3LYP/6-31G(d.p)LANL2DZ */

=== BH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBBH31.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000161 0.000450 YES
RMS Force 0.000105 0.000300 YES
Maximum Displacement 0.000637 0.001800 YES
RMS Displacement 0.000417 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_BH3_FREQ.LOG| bh3_frequency.log]]

<pre>

Low frequencies --- -0.2458 -0.1130 -0.0053 43.9715 45.1306 45.1313
Low frequencies --- 1163.6034 1213.5913 1213.5940

</pre>

<jmol><jmolApplet>
<title>bh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

====Vibrational spectrum for BH3====
{| class="wikitable"
|+
|-
|wavenumber (cm-1 || Intensity (arbitrary units) || symmetry || IR active? || type
|-
|1164
|92
|A2"
|yes
|out-of-plane bend
|-
|1214
|14
|E'
|very slight
|in plane bend
|-
|1214
|14
|E'
|very slight
|in plane bend
|-
|2580
|0
|A1'
|no
|totally symmetric stretch
|-
|2713
|126
|E'
|yes
|asymmetric stretch
|-
|2713
|126
|E'
|yes
|asymmetric stretch
|}

[[file:CMBIRS1.PNG]]

Here we can see that there are 3 peaks in the spectrum but we have 6 vibrational frequencies, how is this? This can be explained by the fact that there are 2 sets of degenerate vibrations with frequencies 1213 cm-1 and 2713 cm-1 thus reducing the number of expected peaks to 4 then the vibration with frequency 2580 results in no change in dipole moment meaning it is not IR active and not seen.

[[file:CMBMOFINAL.PNG]]

=== NH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBNH3ST.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000014 0.001800 YES
RMS Displacement 0.000009 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NH3_OPT_FREQ.LOG| nh3_frequency.log]]

<pre>

Low frequencies --- -0.0128 -0.0024 0.0007 7.1034 8.1048 8.1051
Low frequencies --- 1089.3834 1693.9368 1693.9368

</pre>

<jmol><jmolApplet>
<title>nh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NH3_OPT_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== NH3BH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBNH3BH3ST.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000122 0.000450 YES
RMS Force 0.000058 0.000300 YES
Maximum Displacement 0.000582 0.001800 YES
RMS Displacement 0.000320 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NH3BH3_OPT_FREQ.LOG| nh3bh3_frequency.log]]

<pre>

Low frequencies --- -0.0007 -0.0006 0.0010 16.8481 17.4133 37.2932
Low frequencies --- 265.8219 632.2116 639.3277

</pre>

<jmol><jmolApplet>
<title>nh3nh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NH3BH3_OPT_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Association Energy Calculation ===

E(NH3)=-56.55776873 a.u

E(BH3)=-26.61532349 a.u

E(NH3BH3)=-83.22468893 a.u

ΔE=E(NH3BH3)-(E(NH3)+E(BH3)) = -0.05160 a.u

1 a.u (Hartree) = 2625.5 kJ mol-1

ΔE = -135 kJ mol-1

MAKE COMMENTS ABOUT THE BOND STRENGTHS AND ASK DEMONSTRATOR ABOUT COMPARISON

=== NI3===

==== B3LYP/6-31G(d.p)LANL2DZ ====

[[file:CMBNHI3.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000068 0.000450 YES
RMS Force 0.000044 0.000300 YES
Maximum Displacement 0.000493 0.001800 YES
RMS Displacement 0.000333 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NI3_FREQ_C3V.LOG| ni3_frequency.log]]

<pre>

Low frequencies --- -62.2503 -62.2469 -61.8931 -0.0117 0.0016 0.0074
Low frequencies --- 134.0272 134.0273 196.1461

</pre>

<jmol><jmolApplet>
<title>ni3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NI3_OPT_C3V.LOG</uploadedFileContents>
<script>frame 1.12</script>
</jmolApplet></jmol>

INCLUDE BOND DISTACNEC

Rep:Mod:CMB22301333464

2019-05-16T12:41:36Z

Fiw17: /* B3LYP/6-31G(d.p) level GEN? */

Rep:Mod:CMB22301333464

2019-05-16T12:40:26Z

Fiw17: /* B3LYP/6-31G(d.p) level GEN? */

=== BH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBBH31.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000161 0.000450 YES
RMS Force 0.000105 0.000300 YES
Maximum Displacement 0.000637 0.001800 YES
RMS Displacement 0.000417 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_BH3_FREQ.LOG| bh3_frequency.log]]

<pre>

Low frequencies --- -0.2458 -0.1130 -0.0053 43.9715 45.1306 45.1313
Low frequencies --- 1163.6034 1213.5913 1213.5940

</pre>

<jmol><jmolApplet>
<title>bh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

====Vibrational spectrum for BH3====
{| class="wikitable"
|+
|-
|wavenumber (cm-1 || Intensity (arbitrary units) || symmetry || IR active? || type
|-
|1164
|92
|A2"
|yes
|out-of-plane bend
|-
|1214
|14
|E'
|very slight
|in plane bend
|-
|1214
|14
|E'
|very slight
|in plane bend
|-
|2580
|0
|A1'
|no
|totally symmetric stretch
|-
|2713
|126
|E'
|yes
|asymmetric stretch
|-
|2713
|126
|E'
|yes
|asymmetric stretch
|}

[[file:CMBIRS1.PNG]]

Here we can see that there are 3 peaks in the spectrum but we have 6 vibrational frequencies, how is this? This can be explained by the fact that there are 2 sets of degenerate vibrations with frequencies 1213 cm-1 and 2713 cm-1 thus reducing the number of expected peaks to 4 then the vibration with frequency 2580 results in no change in dipole moment meaning it is not IR active and not seen.

[[file:CMBMOFINAL.PNG]]

=== NH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBNH3ST.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000014 0.001800 YES
RMS Displacement 0.000009 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NH3_OPT_FREQ.LOG| nh3_frequency.log]]

<pre>

Low frequencies --- -0.0128 -0.0024 0.0007 7.1034 8.1048 8.1051
Low frequencies --- 1089.3834 1693.9368 1693.9368

</pre>

<jmol><jmolApplet>
<title>nh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NH3_OPT_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== NH3BH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBNH3BH3ST.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000122 0.000450 YES
RMS Force 0.000058 0.000300 YES
Maximum Displacement 0.000582 0.001800 YES
RMS Displacement 0.000320 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NH3BH3_OPT_FREQ.LOG| nh3bh3_frequency.log]]

<pre>

Low frequencies --- -0.0007 -0.0006 0.0010 16.8481 17.4133 37.2932
Low frequencies --- 265.8219 632.2116 639.3277

</pre>

<jmol><jmolApplet>
<title>nh3nh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NH3BH3_OPT_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Association Energy Calculation ===

E(NH3)=-56.55776873 a.u

E(BH3)=-26.61532349 a.u

E(NH3BH3)=-83.22468893 a.u

ΔE=E(NH3BH3)-(E(NH3)+E(BH3)) = -0.05160 a.u

1 a.u (Hartree) = 2625.5 kJ mol-1

ΔE = -135 kJ mol-1

MAKE COMMENTS ABOUT THE BOND STRENGTHS AND ASK DEMONSTRATOR ABOUT COMPARISON

=== NI3===

==== B3LYP/6-31G(d.p) level GEN?====

[[file:CMBNHI3.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000068 0.000450 YES
RMS Force 0.000044 0.000300 YES
Maximum Displacement 0.000493 0.001800 YES
RMS Displacement 0.000333 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NI3_FREQ_C3V.LOG| ni3_frequency.log]]

<pre>

Low frequencies --- -62.2503 -62.2469 -61.8931 -0.0117 0.0016 0.0074
Low frequencies --- 134.0272 134.0273 196.1461

</pre>

<jmol><jmolApplet>
<title>ni3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NI3_OPT_C3V.LOG</uploadedFileContents>
<script>frame 1.12</script>
</jmolApplet></jmol>

Rep:Mod:CMB22301333464

2019-05-16T12:39:46Z

Fiw17: /* B3LYP/6-31G(d.p) level GEN? */

=== BH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBBH31.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000161 0.000450 YES
RMS Force 0.000105 0.000300 YES
Maximum Displacement 0.000637 0.001800 YES
RMS Displacement 0.000417 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_BH3_FREQ.LOG| bh3_frequency.log]]

<pre>

Low frequencies --- -0.2458 -0.1130 -0.0053 43.9715 45.1306 45.1313
Low frequencies --- 1163.6034 1213.5913 1213.5940

</pre>

<jmol><jmolApplet>
<title>bh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

====Vibrational spectrum for BH3====
{| class="wikitable"
|+
|-
|wavenumber (cm-1 || Intensity (arbitrary units) || symmetry || IR active? || type
|-
|1164
|92
|A2"
|yes
|out-of-plane bend
|-
|1214
|14
|E'
|very slight
|in plane bend
|-
|1214
|14
|E'
|very slight
|in plane bend
|-
|2580
|0
|A1'
|no
|totally symmetric stretch
|-
|2713
|126
|E'
|yes
|asymmetric stretch
|-
|2713
|126
|E'
|yes
|asymmetric stretch
|}

[[file:CMBIRS1.PNG]]

Here we can see that there are 3 peaks in the spectrum but we have 6 vibrational frequencies, how is this? This can be explained by the fact that there are 2 sets of degenerate vibrations with frequencies 1213 cm-1 and 2713 cm-1 thus reducing the number of expected peaks to 4 then the vibration with frequency 2580 results in no change in dipole moment meaning it is not IR active and not seen.

[[file:CMBMOFINAL.PNG]]

=== NH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBNH3ST.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000014 0.001800 YES
RMS Displacement 0.000009 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NH3_OPT_FREQ.LOG| nh3_frequency.log]]

<pre>

Low frequencies --- -0.0128 -0.0024 0.0007 7.1034 8.1048 8.1051
Low frequencies --- 1089.3834 1693.9368 1693.9368

</pre>

<jmol><jmolApplet>
<title>nh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NH3_OPT_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== NH3BH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBNH3BH3ST.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000122 0.000450 YES
RMS Force 0.000058 0.000300 YES
Maximum Displacement 0.000582 0.001800 YES
RMS Displacement 0.000320 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NH3BH3_OPT_FREQ.LOG| nh3bh3_frequency.log]]

<pre>

Low frequencies --- -0.0007 -0.0006 0.0010 16.8481 17.4133 37.2932
Low frequencies --- 265.8219 632.2116 639.3277

</pre>

<jmol><jmolApplet>
<title>nh3nh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NH3BH3_OPT_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Association Energy Calculation ===

E(NH3)=-56.55776873 a.u

E(BH3)=-26.61532349 a.u

E(NH3BH3)=-83.22468893 a.u

ΔE=E(NH3BH3)-(E(NH3)+E(BH3)) = -0.05160 a.u

1 a.u (Hartree) = 2625.5 kJ mol-1

ΔE = -135 kJ mol-1

MAKE COMMENTS ABOUT THE BOND STRENGTHS AND ASK DEMONSTRATOR ABOUT COMPARISON

=== NI3===

==== B3LYP/6-31G(d.p) level GEN?====

[[file:CMBNHI3.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000068 0.000450 YES
RMS Force 0.000044 0.000300 YES
Maximum Displacement 0.000493 0.001800 YES
RMS Displacement 0.000333 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NI3_FREQ_C3V.LOG| ni3_frequency.log]]

<pre>

Low frequencies --- -62.2503 -62.2469 -61.8931 -0.0117 0.0016 0.0074
Low frequencies --- 134.0272 134.0273 196.1461

</pre>

<jmol><jmolApplet>
<title>ni3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NI3_OPT_C3V.LOG</uploadedFileContents>
<script>frame 1.3</script>
</jmolApplet></jmol>

Rep:Mod:CMB22301333464

2019-05-16T12:38:21Z

Fiw17: /* B3LYP/6-31G(d.p) level GEN? */

=== BH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBBH31.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000161 0.000450 YES
RMS Force 0.000105 0.000300 YES
Maximum Displacement 0.000637 0.001800 YES
RMS Displacement 0.000417 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_BH3_FREQ.LOG| bh3_frequency.log]]

<pre>

Low frequencies --- -0.2458 -0.1130 -0.0053 43.9715 45.1306 45.1313
Low frequencies --- 1163.6034 1213.5913 1213.5940

</pre>

<jmol><jmolApplet>
<title>bh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

====Vibrational spectrum for BH3====
{| class="wikitable"
|+
|-
|wavenumber (cm-1 || Intensity (arbitrary units) || symmetry || IR active? || type
|-
|1164
|92
|A2"
|yes
|out-of-plane bend
|-
|1214
|14
|E'
|very slight
|in plane bend
|-
|1214
|14
|E'
|very slight
|in plane bend
|-
|2580
|0
|A1'
|no
|totally symmetric stretch
|-
|2713
|126
|E'
|yes
|asymmetric stretch
|-
|2713
|126
|E'
|yes
|asymmetric stretch
|}

[[file:CMBIRS1.PNG]]

Here we can see that there are 3 peaks in the spectrum but we have 6 vibrational frequencies, how is this? This can be explained by the fact that there are 2 sets of degenerate vibrations with frequencies 1213 cm-1 and 2713 cm-1 thus reducing the number of expected peaks to 4 then the vibration with frequency 2580 results in no change in dipole moment meaning it is not IR active and not seen.

[[file:CMBMOFINAL.PNG]]

=== NH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBNH3ST.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000014 0.001800 YES
RMS Displacement 0.000009 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NH3_OPT_FREQ.LOG| nh3_frequency.log]]

<pre>

Low frequencies --- -0.0128 -0.0024 0.0007 7.1034 8.1048 8.1051
Low frequencies --- 1089.3834 1693.9368 1693.9368

</pre>

<jmol><jmolApplet>
<title>nh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NH3_OPT_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== NH3BH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBNH3BH3ST.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000122 0.000450 YES
RMS Force 0.000058 0.000300 YES
Maximum Displacement 0.000582 0.001800 YES
RMS Displacement 0.000320 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NH3BH3_OPT_FREQ.LOG| nh3bh3_frequency.log]]

<pre>

Low frequencies --- -0.0007 -0.0006 0.0010 16.8481 17.4133 37.2932
Low frequencies --- 265.8219 632.2116 639.3277

</pre>

<jmol><jmolApplet>
<title>nh3nh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NH3BH3_OPT_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Association Energy Calculation ===

E(NH3)=-56.55776873 a.u

E(BH3)=-26.61532349 a.u

E(NH3BH3)=-83.22468893 a.u

ΔE=E(NH3BH3)-(E(NH3)+E(BH3)) = -0.05160 a.u

1 a.u (Hartree) = 2625.5 kJ mol-1

ΔE = -135 kJ mol-1

MAKE COMMENTS ABOUT THE BOND STRENGTHS AND ASK DEMONSTRATOR ABOUT COMPARISON

=== NI3===

==== B3LYP/6-31G(d.p) level GEN?====

[[file:CMBNHI3.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000068 0.000450 YES
RMS Force 0.000044 0.000300 YES
Maximum Displacement 0.000493 0.001800 YES
RMS Displacement 0.000333 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NI3_FREQ_C3V.LOG| ni3_frequency.log]]

<pre>

Low frequencies --- -62.2503 -62.2469 -61.8931 -0.0117 0.0016 0.0074
Low frequencies --- 134.0272 134.0273 196.1461

</pre>

<jmol><jmolApplet>
<title>ni3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NI3_OPT_C3V.LOG</uploadedFileContents>
</jmolApplet></jmol>

File:CMB NI3 OPT C3V.LOG

2019-05-16T12:38:05Z

Fiw17: Fiw17 uploaded a new version of File:CMB NI3 OPT C3V.LOG

Rep:Mod:CMB22301333464

2019-05-16T12:26:04Z

Fiw17: /* Vibrational spectrum for BH3 */

=== BH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBBH31.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000161 0.000450 YES
RMS Force 0.000105 0.000300 YES
Maximum Displacement 0.000637 0.001800 YES
RMS Displacement 0.000417 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_BH3_FREQ.LOG| bh3_frequency.log]]

<pre>

Low frequencies --- -0.2458 -0.1130 -0.0053 43.9715 45.1306 45.1313
Low frequencies --- 1163.6034 1213.5913 1213.5940

</pre>

<jmol><jmolApplet>
<title>bh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_BH3_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

====Vibrational spectrum for BH3====
{| class="wikitable"
|+
|-
|wavenumber (cm-1 || Intensity (arbitrary units) || symmetry || IR active? || type
|-
|1164
|92
|A2"
|yes
|out-of-plane bend
|-
|1214
|14
|E'
|very slight
|in plane bend
|-
|1214
|14
|E'
|very slight
|in plane bend
|-
|2580
|0
|A1'
|no
|totally symmetric stretch
|-
|2713
|126
|E'
|yes
|asymmetric stretch
|-
|2713
|126
|E'
|yes
|asymmetric stretch
|}

[[file:CMBIRS1.PNG]]

Here we can see that there are 3 peaks in the spectrum but we have 6 vibrational frequencies, how is this? This can be explained by the fact that there are 2 sets of degenerate vibrations with frequencies 1213 cm-1 and 2713 cm-1 thus reducing the number of expected peaks to 4 then the vibration with frequency 2580 results in no change in dipole moment meaning it is not IR active and not seen.

[[file:CMBMOFINAL.PNG]]

=== NH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBNH3ST.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000006 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000014 0.001800 YES
RMS Displacement 0.000009 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NH3_OPT_FREQ.LOG| nh3_frequency.log]]

<pre>

Low frequencies --- -0.0128 -0.0024 0.0007 7.1034 8.1048 8.1051
Low frequencies --- 1089.3834 1693.9368 1693.9368

</pre>

<jmol><jmolApplet>
<title>nh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NH3_OPT_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== NH3BH3 ===

==== B3LYP/6-31G(d.p) level ====

[[file:CMBNH3BH3ST.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000122 0.000450 YES
RMS Force 0.000058 0.000300 YES
Maximum Displacement 0.000582 0.001800 YES
RMS Displacement 0.000320 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NH3BH3_OPT_FREQ.LOG| nh3bh3_frequency.log]]

<pre>

Low frequencies --- -0.0007 -0.0006 0.0010 16.8481 17.4133 37.2932
Low frequencies --- 265.8219 632.2116 639.3277

</pre>

<jmol><jmolApplet>
<title>nh3nh3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NH3BH3_OPT_FREQ.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Association Energy Calculation ===

E(NH3)=-56.55776873 a.u

E(BH3)=-26.61532349 a.u

E(NH3BH3)=-83.22468893 a.u

ΔE=E(NH3BH3)-(E(NH3)+E(BH3)) = -0.05160 a.u

1 a.u (Hartree) = 2625.5 kJ mol-1

ΔE = -135 kJ mol-1

MAKE COMMENTS ABOUT THE BOND STRENGTHS AND ASK DEMONSTRATOR ABOUT COMPARISON

=== NI3===

==== B3LYP/6-31G(d.p) level GEN?====

[[file:CMBNHI3.PNG]]

<pre>

Item Value Threshold Converged?
Maximum Force 0.000068 0.000450 YES
RMS Force 0.000044 0.000300 YES
Maximum Displacement 0.000493 0.001800 YES
RMS Displacement 0.000333 0.001200 YES

</pre>

Frequency analysis log file [[Media:CMB_NI3_FREQ_C3V.LOG| ni3_frequency.log]]

<pre>

Low frequencies --- -62.2503 -62.2469 -61.8931 -0.0117 0.0016 0.0074
Low frequencies --- 134.0272 134.0273 196.1461

</pre>

<jmol><jmolApplet>
<title>ni3</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>CMB_NI3_FREQ_C3V.LOG</uploadedFileContents>
</jmolApplet></jmol>

File:CMBMOFINAL.PNG

2019-05-16T12:25:32Z

Fiw17:

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:57:30Z

Fiw17: /* Question One */

==Exercise one==
===Question One===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat as using a dynamics method means you can get to the exact exact transition point, just very very close to it.

===Question Three===
Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show a minimum energy pathway that a reaction could take which is the minimum energy trajectory along the potential energy surface. Dynamics analysis by contrast shows a reaction trajectory for the entered parameters.

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics!! Contour Plot!!
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
||[[File:Fiw17 Contour 1.PNG]]
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
||[[File:Fiw17 Contour 2.PNG]]
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
||[[File:Fiw17 Contour 3.PNG]]
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
||[[File:Fiw17 Contour 4.PNG]]
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
||[[File:Fiw17 Contour 5.PNG]]
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kcal

Energy of reactants: -103.9 kcal

Activation energy= -103.75--103.90= 0.15 kcal

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kcal

Energy of Reactants= -133.8 kcal

Activation Energy= -103.78--133.85 = 30.07 kcal = 30.1 kcal (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi's Rules===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt whereas changes in translational energy are less effective in changing whether a reaction will occur or not.

==References==

1. Yoder, C. (2019). Common Bond Energies (D. [online] Wiredchemist.com. Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 10 May 2019].

2. Brouard.chem.ox.ac.uk. (2019). Controlling Reagents and Characterising Products. [online] Available at: http://brouard.chem.ox.ac.uk/teaching/dynlectures4to6.pdf [Accessed 10 May 2019].

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:57:13Z

Fiw17: /* Question Three */

==Exercise one==
===Question One===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat as using a dynamics method means you can get to the exact exact transition point, just very very close to it.

===Question Three===
Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show a minimum energy pathway that a reaction could take which is the minimum energy trajectory along the potential energy surface. Dynamics analysis by contrast shows a reaction trajectory for the entered parameters.

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics!! Contour Plot!!
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
||[[File:Fiw17 Contour 1.PNG]]
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
||[[File:Fiw17 Contour 2.PNG]]
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
||[[File:Fiw17 Contour 3.PNG]]
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
||[[File:Fiw17 Contour 4.PNG]]
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
||[[File:Fiw17 Contour 5.PNG]]
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kcal

Energy of reactants: -103.9 kcal

Activation energy= -103.75--103.90= 0.15 kcal

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kcal

Energy of Reactants= -133.8 kcal

Activation Energy= -103.78--133.85 = 30.07 kcal = 30.1 kcal (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi's Rules===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt whereas changes in translational energy are less effective in changing whether a reaction will occur or not.

==References==

1. Yoder, C. (2019). Common Bond Energies (D. [online] Wiredchemist.com. Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 10 May 2019].

2. Brouard.chem.ox.ac.uk. (2019). Controlling Reagents and Characterising Products. [online] Available at: http://brouard.chem.ox.ac.uk/teaching/dynlectures4to6.pdf [Accessed 10 May 2019].

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:54:25Z

Fiw17: /* Question Six: Polanyi */

==Exercise one==
===Question One===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat as using a dynamics method means you can get to the exact exact transition point, just very very close to it.

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show a minimum energy pathway that a reaction could take which is the minimum energy trajectory along the potential energy surface. Dynamics analysis by contrast shows a reaction trajectory for the entered parameters.

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics!! Contour Plot!!
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
||[[File:Fiw17 Contour 1.PNG]]
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
||[[File:Fiw17 Contour 2.PNG]]
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
||[[File:Fiw17 Contour 3.PNG]]
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
||[[File:Fiw17 Contour 4.PNG]]
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
||[[File:Fiw17 Contour 5.PNG]]
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kcal

Energy of reactants: -103.9 kcal

Activation energy= -103.75--103.90= 0.15 kcal

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kcal

Energy of Reactants= -133.8 kcal

Activation Energy= -103.78--133.85 = 30.07 kcal = 30.1 kcal (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi's Rules===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt whereas changes in translational energy are less effective in changing whether a reaction will occur or not.

==References==

1. Yoder, C. (2019). Common Bond Energies (D. [online] Wiredchemist.com. Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 10 May 2019].

2. Brouard.chem.ox.ac.uk. (2019). Controlling Reagents and Characterising Products. [online] Available at: http://brouard.chem.ox.ac.uk/teaching/dynlectures4to6.pdf [Accessed 10 May 2019].

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:53:50Z

Fiw17: /* Question Three */

==Exercise one==
===Question One===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat as using a dynamics method means you can get to the exact exact transition point, just very very close to it.

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show a minimum energy pathway that a reaction could take which is the minimum energy trajectory along the potential energy surface. Dynamics analysis by contrast shows a reaction trajectory for the entered parameters.

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics!! Contour Plot!!
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
||[[File:Fiw17 Contour 1.PNG]]
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
||[[File:Fiw17 Contour 2.PNG]]
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
||[[File:Fiw17 Contour 3.PNG]]
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
||[[File:Fiw17 Contour 4.PNG]]
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
||[[File:Fiw17 Contour 5.PNG]]
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kcal

Energy of reactants: -103.9 kcal

Activation energy= -103.75--103.90= 0.15 kcal

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kcal

Energy of Reactants= -133.8 kcal

Activation Energy= -103.78--133.85 = 30.07 kcal = 30.1 kcal (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt whereas changes in translational energy are less effective in changing whether a reaction will occur or not.

==References==

1. Yoder, C. (2019). Common Bond Energies (D. [online] Wiredchemist.com. Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 10 May 2019].

2. Brouard.chem.ox.ac.uk. (2019). Controlling Reagents and Characterising Products. [online] Available at: http://brouard.chem.ox.ac.uk/teaching/dynlectures4to6.pdf [Accessed 10 May 2019].

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:52:24Z

Fiw17: /* Question Two */

==Exercise one==
===Question One===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat as using a dynamics method means you can get to the exact exact transition point, just very very close to it.

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics!! Contour Plot!!
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
||[[File:Fiw17 Contour 1.PNG]]
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
||[[File:Fiw17 Contour 2.PNG]]
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
||[[File:Fiw17 Contour 3.PNG]]
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
||[[File:Fiw17 Contour 4.PNG]]
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
||[[File:Fiw17 Contour 5.PNG]]
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kcal

Energy of reactants: -103.9 kcal

Activation energy= -103.75--103.90= 0.15 kcal

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kcal

Energy of Reactants= -133.8 kcal

Activation Energy= -103.78--133.85 = 30.07 kcal = 30.1 kcal (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt whereas changes in translational energy are less effective in changing whether a reaction will occur or not.

==References==

1. Yoder, C. (2019). Common Bond Energies (D. [online] Wiredchemist.com. Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 10 May 2019].

2. Brouard.chem.ox.ac.uk. (2019). Controlling Reagents and Characterising Products. [online] Available at: http://brouard.chem.ox.ac.uk/teaching/dynlectures4to6.pdf [Accessed 10 May 2019].

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:51:21Z

Fiw17: /* Question One */

==Exercise one==
===Question One===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics!! Contour Plot!!
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
||[[File:Fiw17 Contour 1.PNG]]
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
||[[File:Fiw17 Contour 2.PNG]]
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
||[[File:Fiw17 Contour 3.PNG]]
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
||[[File:Fiw17 Contour 4.PNG]]
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
||[[File:Fiw17 Contour 5.PNG]]
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kcal

Energy of reactants: -103.9 kcal

Activation energy= -103.75--103.90= 0.15 kcal

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kcal

Energy of Reactants= -133.8 kcal

Activation Energy= -103.78--133.85 = 30.07 kcal = 30.1 kcal (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt whereas changes in translational energy are less effective in changing whether a reaction will occur or not.

==References==

1. Yoder, C. (2019). Common Bond Energies (D. [online] Wiredchemist.com. Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 10 May 2019].

2. Brouard.chem.ox.ac.uk. (2019). Controlling Reagents and Characterising Products. [online] Available at: http://brouard.chem.ox.ac.uk/teaching/dynlectures4to6.pdf [Accessed 10 May 2019].

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:49:06Z

Fiw17: /* Question Four */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics!! Contour Plot!!
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
||[[File:Fiw17 Contour 1.PNG]]
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
||[[File:Fiw17 Contour 2.PNG]]
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
||[[File:Fiw17 Contour 3.PNG]]
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
||[[File:Fiw17 Contour 4.PNG]]
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
||[[File:Fiw17 Contour 5.PNG]]
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kcal

Energy of reactants: -103.9 kcal

Activation energy= -103.75--103.90= 0.15 kcal

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kcal

Energy of Reactants= -133.8 kcal

Activation Energy= -103.78--133.85 = 30.07 kcal = 30.1 kcal (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt whereas changes in translational energy are less effective in changing whether a reaction will occur or not.

==References==

1. Yoder, C. (2019). Common Bond Energies (D. [online] Wiredchemist.com. Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 10 May 2019].

2. Brouard.chem.ox.ac.uk. (2019). Controlling Reagents and Characterising Products. [online] Available at: http://brouard.chem.ox.ac.uk/teaching/dynlectures4to6.pdf [Accessed 10 May 2019].

File:Fiw17 Contour 5.PNG

2019-05-10T16:48:52Z

Fiw17:

File:Fiw17 Contour 4.PNG

2019-05-10T16:47:27Z

Fiw17:

File:Fiw17 Contour 3.PNG

2019-05-10T16:46:45Z

Fiw17:

File:Fiw17 Contour 2.PNG

2019-05-10T16:45:18Z

Fiw17:

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:41:34Z

Fiw17: /* Question Three:Activation Energies */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics!! Contour Plot!!
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
||[[File:Fiw17 Contour 1.PNG]]
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kcal

Energy of reactants: -103.9 kcal

Activation energy= -103.75--103.90= 0.15 kcal

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kcal

Energy of Reactants= -133.8 kcal

Activation Energy= -103.78--133.85 = 30.07 kcal = 30.1 kcal (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt whereas changes in translational energy are less effective in changing whether a reaction will occur or not.

==References==

1. Yoder, C. (2019). Common Bond Energies (D. [online] Wiredchemist.com. Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 10 May 2019].

2. Brouard.chem.ox.ac.uk. (2019). Controlling Reagents and Characterising Products. [online] Available at: http://brouard.chem.ox.ac.uk/teaching/dynlectures4to6.pdf [Accessed 10 May 2019].

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:40:16Z

Fiw17: /* Question Four */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics!! Contour Plot!!
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
||[[File:Fiw17 Contour 1.PNG]]
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt whereas changes in translational energy are less effective in changing whether a reaction will occur or not.

==References==

1. Yoder, C. (2019). Common Bond Energies (D. [online] Wiredchemist.com. Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 10 May 2019].

2. Brouard.chem.ox.ac.uk. (2019). Controlling Reagents and Characterising Products. [online] Available at: http://brouard.chem.ox.ac.uk/teaching/dynlectures4to6.pdf [Accessed 10 May 2019].

File:Fiw17 Contour 1.PNG

2019-05-10T16:39:33Z

Fiw17:

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:36:33Z

Fiw17: /* Question Six: Polanyi */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt whereas changes in translational energy are less effective in changing whether a reaction will occur or not.

==References==

1. Yoder, C. (2019). Common Bond Energies (D. [online] Wiredchemist.com. Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 10 May 2019].

2. Brouard.chem.ox.ac.uk. (2019). Controlling Reagents and Characterising Products. [online] Available at: http://brouard.chem.ox.ac.uk/teaching/dynlectures4to6.pdf [Accessed 10 May 2019].

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:35:34Z

Fiw17: /* Question Six: Polanyi */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening2. Whereas, an increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:27:57Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

The endothermic reaction reaction of HF + H --> H2 + F was carried out with initial conditions of HF distance= 0.91 Å, HH distance= 2.0, HF momentum (represents vibrational energy)= 5.0

{| class="wikitable" border=1
! p(HH) (represents translational energy) !! Reactive?
|-
| -5.0 || Yes
|-
| -4.0 || No
|-
| -3.0 || No
|-
| -2.0 || No
|-
| -1.0 || No
|}

This shows that you need a very large translational energy to cause a significant change whereas when experimenting with the vibrational energy of the HF molecule it was found that a small change in vibrational energy had a larger effect. This is in line with theoretical expectations outlined by Palonyi's Rules.

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:01:24Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:01:09Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:00:59Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:00:43Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T16:00:28Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T15:59:36Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T15:59:05Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below in section 'Question Six').

EXPERIMENT TWO: ENDOTHERMIC REACTION- HF + H--> H2 + F

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T15:57:17Z

Fiw17: /* Question Four: Release of Reaction Energy */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Energy Time Correct1.PNG]] [[File:Fiw17 Momentum energy correct.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. Overall energy is conserved as the change in potential energy from the reaction is converted to kinetic energy of the molecules in the form of vibrational and translational energy. The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

File:Fiw17 Momentum energy correct.PNG

2019-05-10T15:56:46Z

Fiw17:

File:Fiw17 Energy Time Correct1.PNG

2019-05-10T15:55:28Z

Fiw17:

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T15:49:26Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]] [[File:Fiw17 Energy time exothermic 2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H 

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T15:48:57Z

Fiw17: /* Question Four: Release of Reaction Energy */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]] [[File:Fiw17 Energy time exothermic 2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

File:Fiw17 Energy time exothermic 2.PNG

2019-05-10T15:48:41Z

Fiw17:

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T15:35:02Z

Fiw17: /* Question Four: Release of Reaction Energy */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2atom approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T15:34:41Z

Fiw17: /* Question Four: Release of Reaction Energy */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T15:34:02Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

EXPERIMENT ONE: FORWARD EXOTHERMIC REACTION- H2 +F--> HF + H

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T15:25:35Z

Fiw17: /* Question Six: Polanyi */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring while the translational energy has less of an effect.

FORWARD EXOTHERMIC REACTION: H2+ F--> HF + H

The polyani rules theory supports the above experimental results (Fig. A and B) which shows that, for an exthothermic reaction (early transition state), a small change in translational energy (HF momentum) results in a reaction going from not reacting to reacting. (See Figs A vs B). It also supports the experiment (See table in Question Five section) which shows that changes in vibrational energy don't have a significant effect/clear trend effect on the probability of a reaction occuring.

REVERSE ENDOTHERMIC REACTION: HF + H --> H2 + F

The experiment (See Figs C-F in Section 'Question Five') agrees with the theory states in polyani's rules as it shows that a small change in vibrational energy will result in a reaction occuring that previously didnt. The experimental results also show that large variations in translational energy don't necessarily result in a reaction occuring that doesnt occur at lower translational energies

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T15:04:17Z

Fiw17: /* Exercise Two */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state (exothermic), the translational energy has the most significant effect on the probability of the reaction occuring. For a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. Whereas, and increase in vibration energy will have a much less significant effect.

The rules also state that the reverse is true- for a reaction with an late transition state (endothermic), the vibrational energy has the most significiant effect on the probability of the reaction occuring.

According to Polanyi rules, for a reaction with an early transition state, the vibrational energy doesn't have a significant effect on whether a reaction will happen. However, the translational energy will have a significant effect. The rules state that . This explains the results of the above experiments that shows that there is no significant trend with vibrational energy of the reactants and the outcome of the collision. The theory also supports the difference between Figures A and B as the theory agrees with the fact that a small change in translational energy for this exothermic reaction will result in an increase in probability of the reaction occuring.

Polanyi Rules:

Exothermic reaction--> early transition state-->translational energy is the most significant factor

Endothermic reaction-->late transition state--> vibrational energy is the most significant factor

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T14:55:24Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

===Question Five: Trends with Momentum===
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state, the translational energy has the most significant effect on the probability of the reaction occuring. An increase in the translational energy
According to Polanyi rules, for a reaction with an early transition state, the vibrational energy doesn't have a significant effect on whether a reaction will happen. However, the translational energy will have a significant effect. The rules state that for a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. This explains the results of the above experiments that shows that there is no significant trend with vibrational energy of the reactants and the outcome of the collision. The theory also supports the difference between Figures A and B as the theory agrees with the fact that a small change in translational energy for this exothermic reaction will result in an increase in probability of the reaction occuring.

Polanyi Rules:

Exothermic reaction--> early transition state-->translational energy is the most significant factor

Endothermic reaction-->late transition state--> vibrational energy is the most significant factor

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T14:54:41Z

Fiw17: /* Question Six: Polanyi */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

==Question Five: Trends with Momentum==
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Reactants need a certain amount of energy (the activation energy) in order to react when they collide. This energy can be in the form of translational and vibrational energy. The way that the energy is distributed between the two forms and the amount of each form of energy effects the likelihood of a reaction happening when they collide. The amount to which the probability of the reaction occuring varies with changes in the translational and vibrational energy depends on what type of reaction it is and the position of the transition state.

Polyani's Rules state that for a reaction with an early transition state, the translational energy has the most significant effect on the probability of the reaction occuring. An increase in the translational energy
According to Polanyi rules, for a reaction with an early transition state, the vibrational energy doesn't have a significant effect on whether a reaction will happen. However, the translational energy will have a significant effect. The rules state that for a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. This explains the results of the above experiments that shows that there is no significant trend with vibrational energy of the reactants and the outcome of the collision. The theory also supports the difference between Figures A and B as the theory agrees with the fact that a small change in translational energy for this exothermic reaction will result in an increase in probability of the reaction occuring.

Polanyi Rules:

Exothermic reaction--> early transition state-->translational energy is the most significant factor

Endothermic reaction-->late transition state--> vibrational energy is the most significant factor

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T14:48:51Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

==Question Five: Trends with Momentum==
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules (explained below).

=== Question Six: Polanyi===
Variation in the translational and vibrational

According to Polanyi rules, for a reaction with an early transition state, the vibrational energy doesn't have a significant effect on whether a reaction will happen. However, the translational energy will have a significant effect. The rules state that for a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. This explains the results of the above experiments that shows that there is no significant trend with vibrational energy of the reactants and the outcome of the collision. The theory also supports the difference between Figures A and B as the theory agrees with the fact that a small change in translational energy for this exothermic reaction will result in an increase in probability of the reaction occuring.

Polanyi Rules:

Exothermic reaction--> early transition state-->translational energy is the most significant factor

Endothermic reaction-->late transition state--> vibrational energy is the most significant factor

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T14:41:03Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

==Question Five: Trends with Momentum==
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5, HF distance= 2.0, HH distance= 0.74)

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules.

According to Polanyi rules, for a reaction with an early transition state, the vibrational energy doesn't have a significant effect on whether a reaction will happen. However, the translational energy will have a significant effect. The rules state that for a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. This explains the results of the above experiments that shows that there is no significant trend with vibrational energy of the reactants and the outcome of the collision. The theory also supports that

Polanyi Rules:
Exothermic reactiom--> early transition state-->translational energy is the most significant factor
Endothermic reaction-->late transition state--> vibrational energy is the most significant factor

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T14:32:12Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

==Question Five: Trends with Momentum==
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules.

According to Polanyi rules, vibrational energy doesn't have a significant effect on whether a reaction will happen when it has an early transition state whereas translational energy of the reactants has a significant effect. The rules state that for a reaction with an early transition state, a greater amount of translational energy will increase the proabability of a reaction happening. This explains the results of the above experiments that shows that there is no significant trend with vibrational energy of the reactants and the outcome of the collision. The theory also supports that

Exothermic reaction- early transition state, translational energy is important
Endothermic reaction- late transition state, vibrational energy

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T14:25:30Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

==Question Five: Trends with Momentum==
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Figures A and B show that a small difference in translational energy significantly increases the likelihood of a reaction occuring. This can be explained by Polanyi Rules.

Polyani Rules state that

Exothermic reaction- early transition state, translational energy is important
Endothermic reaction- late transition state, vibrational energy

MDR:Physical Comp Lab Francesca Wittmann 01330365

2019-05-10T14:18:41Z

Fiw17: /* Question Five: Trends with Momentum */

==Exercise one==
===Question One===
1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transition state is the point along the reaction trajectory which is a saddle point of potential energy. I.e it is a point where there is simultaneous a maximum and minimum potential energy. At a saddle point the gradient (first derivative) along both axes is zero.

In order to identify where the saddle point is on a potential energy surface diagram, two orthogonal functions are defined. The function Q1 is the potential energy with respect to the line that is orthogonal to the max potential energy point along the minimum energy pathway(see multicoloured squiggly line on diagram). Function Q2 is the potential energy with respect to the line that is tangential to the max potential energy along the minimum energy pathway.

[[File:Fiw17 Contour Map H2.PNG|500 px]]

To find the saddle point, you would differentiate along both Q1 and Q2. There should be a point where Q2 is at a maximum whilst Q1 is simultaneously at a minimum.
At this point:

δV/δQ1=δV/δQ2=0 and δV2/δQ12 >0 and δV2/δQ22 <0

A local minimum will not have the lowest potential energy on a global (i.e. total range) scale. Firstly you would differentiate across the whole range to find anywhere that had a potential gradient of 0 to find all of the minima. You can distinguish a local minimum from a global minimum by comparing the potential energy (i.e. y coordinate) at each minimum. The one with the lowest
potential energy is the global minimum.

===Question Two===
2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

Best estimate of transition state is at A-B distance= B-C distance= 0.907 Å
[[File:Fiw17 Molecule Distances.PNG]]

This was estimated using Hammond's postulate and trial and error dynamics analysis. Hammond's postulate states that the structure of a transition state will resemble either the reactants or products more depending on what the transition state is closer in energy too. In this case, the the reaction is very symmetrical. The reactants are the same as the products and so have the same energy level as the products. Therefore the transition state is exactly in the middle of the 'two' structures, half way between the reactants and products (due to symmetry of the reaction and Hammond's postulate). This means that the three atoms involved will all be the same distance apart at the transition state. I.e. A-B= B-C (See diagram) Therefore, trials were carried out using equal distances for AB and BC. The best estimate for the transition state was the bond distance where the molecule oscillated the least (i.e most flat line inter nuclear distance vs time plot).

[[File:Fiw17 IntervsTime Dynamics.PNG]]

This diagram was calculated through trial and error using dynamics method and so the line isnt perfectly flat

===Question Three===
3.Comment on how the MEP and the trajectory you just calculated differ.

The dynamics analysis method carries over momentum from the previous result so even with the gradient reaches zero, it moves past that point meaning the line doesn't stop at the transition point and the
Could do trial and error but this will take a long time- this is because Dynamics analysis carries over the momentum from the previous iteration so even if the gradient gets to zero, it will move past that and the line wont stop at the transition point. It will oscillate around that point. This is why the graph of inter nuclear distance vs time (using trial and error and dynamics analysis) is still is not completely flat.

MEP fixes this problem, minimum energy pathway, momentum and velocities are reset for each infinitesimally small step. Therefore when the gradient hits zero it will stay at this point which is the transition state saddle point. This is a more accurate and easier way to find the saddle point than using the dynamics method.

The calculated MEP distance was found to be 0.907742 Å

[[File:Fiw17 IntervsTime MEP.PNG]]

MEPs show

===Question Four===

Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99
|| Yes
||AB collides slowly and smoothly with C, BC move away together
|-
| -1.5 || -2.0 || -100
|| No
||Slow moving, slightly toward each other, bounce of each other unreacted,
|-
| -1.5 || -2.5 ||-99|| Yes
||AB collides with C and reacts resulting in vibrating molecule of BC moving away
|-
| -2.5 || -5.0 ||-85||No||AB collides with C. B dissociates from A, bounces of C and reconnects with A to move away as an unreacted oscillating molecule of AB
|-
| -2.5 || -5.2 ||-83 ||Yes|| AB collides with C. B bonds to C then bounces of of A then moves away as a vibrating molecule connected to C
|}

This table shows that just adding more energy to a reaction does not necessarily mean that the molecules will react. The momenta of the particles is important. Similarly, more momentum does not necessarily lead to a reaction as can can be seen from the case with p1= -2.5 and p2= -5.0. This is because the molecules need to have the right amount of energy to react and then move away from each other. If they have too much energy they could just dissociate and bounce off of each other like in p1=-2.4,p2=-5.0. Therefore, not only do the particles need enough energy/momentum to get over the transition state, they can't have too much or they can come back over the transition state to the reactants.

===Question Five===
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is the idea that there is a type of equilibrium (quasi-equilibrium) between the the reactant and products. This equilibrium is at a saddle point of the potential energy surface and is an activated structure which is the transition state. At this transition state there is an activated complex that is high in energy and converts the reactants to the products (i.e. the reactants have to pass through the transition state to get to the products).

The main assumptions of transition state theory are that quantum tunnelling effects are negligible and the Born-Oppenheimer approximation holds true. Also, it relies on the assumption that the reactants are allowed to thermally equilibrate meaning that their energies are Boltzmann distributed and that the reactants are in constant equilibrium with the transition state.

==Exercise Two==
===Question One===
1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

H2 + F --> HF + H

This is an exothermic reaction as can be seen from the surface plot. (See image where A=F, B=H, C=H). This is in agreement with the theory and relative bond strengths as H-F (565 kJ/mol)1 is a much stronger bond than H2 (432 kJ/mol)1

[[File:Fiw17 surfaceplotfor H2F HFH.PNG|500px]]

Hammond's postulate states that a transition state will resemble the reactants or products depending on the activated complex's (transition state's) relative energy compared to the reactants or products. As this reaction is exothermic, the transition state will be early and so nearer the reactants in the potential energy surface.

HF + H --> H2 + F

This is an endothermic reaction, as the potential energy surface plot shows that the reactants are lower in energy than the products.

[[File:Fiw17 Contour Map H2 2.PNG|500px]]

Using Hammond's postulate, it can be assumed that the transition state (activated complex) is closer in energy to the products than the reactants as endothermic reactions tend to have late transition states.

===Question Two: Approximate Transition States===

The approximate transition state position for the reaction For H2 + F --> HF + H was found to be when HF distance (AB)= 1.808 Å and HH distance (BC)= 0.745.

[[File:Fiw17 TransitionState.PNG|400px]] [[File:Fiw17 transitionstate internucleardistance.PNG|400px]] [[File:Fiw17 transitionState contour.PNG|400px]]

The approximate transition state position for the reaction for HF + H -->H2 + F was also found to be 1.808 Å for HF and 0.754 Å for the HH distance.
A=H, B=H, C=F

[[File:Fiw17 Contour Map otherway.PNG|400px]] [[File:Fiw17 Internuclear H2.PNG|400px]] [[File:Fiw17 ContourMap secondequation.PNG|400px]]

===Question Three:Activation Energies===

EXOTHERMIC REACTION: H2 + F--> HF + H
Once the transition state was found, the MEP paramaters were displaced slightly from these values so that the minimum energy pathway falls back to the reactants level (very small drop). The energy time graph was then analysed to see the energy difference which is the activation energy.

[[File:Fiw17 Energy Time Exothermic.PNG]]

Energy at transition state: -103.8 kJ/mol

Energy of reactants: -103.9 kJ/mol

Activation energy= -103.75--103.90= 0.15 kJ/mol

ENDOTHERMIC REACTION: HF + H--> H2 + F
Transition state found, MEP parameters displaced slightly towards the reactants side from saddle point so that the minimum energy pathway 'rolls' back to the reactants side. The energy vs time graph was then analysed to find the activation energy.

[[File:Fiw17 ActivationEnergy reaction1.PNG]]

Energy at transition state= -103.8 kJ/mol

Energy of Reactants= -133.8 kJ/mol

Activation Energy= -103.78--133.85 = 30.07 kJ/mol = 30.1 kJ/mol (1 dp)

===Question Four: Release of Reaction Energy===

A reactive trajectory was found for the reaction of H2 and F with initial parameters:

HF distance= 1.5 Å, HH distance= 0.8 Å, HF momentum= -1.5 HH momentum= -1.5 giving the following energy vs momentum graph. The diagram shows that the H2</atom> approaches the F atom, bounces off of it twice and then joins with the F to make HF.

[[File:Fiw17 Momentum Energy ReactedFandH2.PNG]]

As this reaction is an exothermic reaction energy is released to the surroundings. The potential energy change is conserved and released as kinetic energy in the form of vibration energy of the HF and translational energy of the molecules.

The energy vs momentum graph (see above Diagram) for this reaction shows this release in energy as the vibrational energy of the H2 molecule before the reaction (yellow early vibrations) are much less than the vibrations of the HF molecule after the reaction (blue late vibrations)

The release of potential energy as kinetic energy could also be measured experimentally using calorimetry tests.

==Question Five: Trends with Momentum==
This section investigates the effect of varying the translational and vibrational energy of the reactants on the likelihood of a reaction occuring.

The reaction of H2 + F --> HF + F was analysed with initial conditions HH distance = 0.74 Å, HF distance = 2.0 Å and momentum of FH= -0.5

This tests the effect of changing the vibrational energy of the reactants on the likelihood of a reaction occuring.

{| class="wikitable" border=1
! p(HF) !! Reactive?
|-
| -3.0 || Yes
|-
| -2.5 || Yes
|-
| -2.0 || Yes
|-
| -1.5 || No
|-
| -1.0 || Yes
|-
| -0.5 || Yes
|-
| -0 || Yes
|-
| 0.5 || Yes
|-
| 1.0 || No
|-
| 1.5 || Yes
|-
| 2.0 || Yes
|-
| 2.5 || No
|-
| 3.0 || Yes
|}

These results don't show a clear trend. This implies that the vibration energy doesn't have a very significant effect on this specific reaction. This can be explained by the exothermic nature of the reaction and Polanyi Rules which are explained below.

The effect of translational energy on the likelihood of a reaction occuring by varying the HF momentum very slightly from -0.5 to -0.8. The energy vs momentum graphs for both situations are shown and compared below to analyse the effect of varying translational energy. (Other initial conditions: HH momentum= -1.5

[[File:Fiw17_momentum_energy_reacted_translationallow.PNG|thumb|left|Fig.A- HF momentum= -0.5, No reaction|500px]] [[File:Fiw17 momentum energy reacted translationalhigh.PNG|thumb|left|Fig.B- HF momentum= -0.8, Reaction|500px]]

Experiment: HF distance= 2.0 Å, HH distance= 0.74 Å, HF momentum (translational)= -0.8, HH momentum (vibrational)=

Exothermic reaction- early transition state, translational energy is important
Endothermic reaction- late transition state, vibrational energy