ChemWiki - User contributions [en]

MRD:pc2715

2017-05-19T15:27:33Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

[[File:GgCapture1.PNG]]

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

[[File:GgCapture2.PNG]]

[[File:GgCapture3.PNG]]

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

[[File:GgCapture8.PNG]]
[[File:GgCapture9.PNG]]
[[File:GgCapture10.PNG]]
[[File:GgCapture11.PNG]]
[[File:GgCapture12.PNG]]

===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

MRD:pc2715

2017-05-19T15:26:37Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

[[File:GgCapture1.PNG|200px|thumb|right]]

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

[[File:GgCapture2.PNG|200px|thumb|right]]

[[File:GgCapture3.PNG|200px|thumb|right]]

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

[[File:GgCapture8.PNG|200px|thumb|right]]
[[File:GgCapture9.PNG|200px|thumb|right]]
[[File:GgCapture10.PNG|200px|thumb|right]]
[[File:GgCapture11.PNG|200px|thumb|right]]
[[File:GgCapture12.PNG|200px|thumb|right]]

===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

MRD:pc2715

2017-05-19T15:25:45Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

[[File:GgCapture1.PNG|200px|thumb]]

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

[[File:GgCapture2.PNG|200px|thumb]]

[[File:GgCapture3.PNG|200px|thumb]]

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

[[File:GgCapture8.PNG|200px|thumb]]
[[File:GgCapture9.PNG|200px|thumb]]
[[File:GgCapture10.PNG|200px|thumb]]
[[File:GgCapture11.PNG|200px|thumb]]
[[File:GgCapture12.PNG|200px|thumb]]

===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

MRD:pc2715

2017-05-19T15:23:02Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

[[File:GgCapture1.PNG]]

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

[[File:GgCapture2.PNG]]

[[File:GgCapture3.PNG]]

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

Capture 8-12
===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

File:GgCapture12.PNG

2017-05-19T15:21:41Z

Pc2715:

File:GgCapture11.PNG

2017-05-19T15:21:22Z

Pc2715:

File:GgCapture10.PNG

2017-05-19T15:21:00Z

Pc2715:

File:GgCapture9.PNG

2017-05-19T15:20:40Z

Pc2715:

File:GgCapture8.PNG

2017-05-19T15:20:18Z

Pc2715:

File:GgCapture5.PNG

2017-05-19T15:19:46Z

Pc2715:

File:GgCapture4.PNG

2017-05-19T15:19:25Z

Pc2715:

File:GgCapture3.PNG

2017-05-19T15:19:06Z

Pc2715:

File:GgCapture2.PNG

2017-05-19T15:18:46Z

Pc2715:

MRD:pc2715

2017-05-19T15:18:14Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

[[File:GgCapture1.PNG]]

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

Capture 2
Capture 3

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

Capture 8-12
===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

MRD:pc2715

2017-05-19T15:17:35Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

[[File:GgCapture1.PNG|200px|thumb]]

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

Capture 2
Capture 3

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

Capture 8-12
===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

File:GgCapture1.PNG

2017-05-19T15:16:45Z

Pc2715:

MRD:pc2715

2017-05-19T15:15:47Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

[[File:ggCapture1.png|200px|thumb]]

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

Capture 2
Capture 3

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

Capture 8-12
===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

MRD:pc2715

2017-05-19T15:15:19Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

[[File:ggCapture1.png|200px|thumb|left]]

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

Capture 2
Capture 3

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

Capture 8-12
===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

MRD:pc2715

2017-05-19T15:14:52Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

[[File:ggCapture1.png|200px|thumb|left]]

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

Capture 2
Capture 3

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

Capture 8-12
===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

MRD:pc2715

2017-05-19T15:09:00Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

[[File:Capture 1.png|200px|thumb|left]]

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

Capture 2
Capture 3

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

Capture 8-12
===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

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Pc2715: Pc2715 uploaded a new version of File:Capture12.PNG

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MRD:pc2715

2017-05-19T15:02:18Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

Capture 1

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

Capture 2
Capture 3

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

Capture 8-12
===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

MRD:pc2715

2017-05-19T14:38:38Z

Pc2715:

== EXERCISE 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===

Both gradients of the potential energy at minimum and at the transition structure are both zero. The minimum point and the transition structure can be distinguished by using the curvature curve. If the trajectory starting position is on the transition point, it will be there forever. But a small change in the radius will cause a change in the potential energy, making it roll down the hill on either one side, reactant or the product. However, if it is in the minimum energy surface, it will at either end of the reaction, product or the reactant.

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

R of 1 and 2 is 0.908. The internuclear distances for both radii did not change with time. This means that the molecules are in a position where dV(r)/dr = 0 but it is not the minimium peak. It is where the transition state is as a small change in the position cause the change in energy level at only one way, toward the reactant or the product.

===Comment on how the mep and the trajectory you just calculated differ.===

MEP projection is much shorter and there is no oscillation as it corresponded to the infinitely slow motion. However, with dynamic calculation, oscillation was taken into consideration and the trajectory shown had more oscillation.

===Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border="1"
|+caption
! P1 !! P2 !! Description
|-
| -1.25 || -2.5 || Reactive- The energy is enough to move from the product to the reactant
|-
| -1.5 || -2.0 || Unreactive - The reactant don’t have enough energy to overcome the transition energy barrier
|-
| -1.5 || -2.5 || Reactive - There is enough energy to move from the the reactant to the product.
|-
| -2.5 || -5.0 || Unreactive - Reactant have enough energy to overcome to barrier but it reverse toward the reactant because of barrier recrossing
|-
| -2.5 || -5.2 || Reactive - Overall reaction is reactive even though the reaction went back to the reactant for a short time. The high momentum push the reaction toward the product.
|}

===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?===

The main assumptions of Transition State Theory are; the reactants do not need to be in a correct orientation for the reaction to occur, every collision have enough energy to overcome the energy barrier, and the successful collision rate is 100%.
The prediction rate value will be higher than the experimental value as the experimental do not have a 100% successful rate and so the rate of turning into product will be lower than expected.

== Problem 2, F - H - H system==

===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?
Locate the approximate position of the transition state.===

H + HF is endothermic and H + F2 is exothermic. This is due to the fact that H-F bond is a lot stronger than H-H.
The energy needed to break H-F is more than the energy release from making H-F, making H + HF reaction endothermic.
However, in H + F2, the energy used to break F-F will be less than the energy released when H-F formed, making H + F2 reaction exothermic.
Transition start position:
Distance (H-F/AB) = 1.815
Distance (H-H/BC) = 0.745
The activation energy for H + HF and H + F2 are 0.5 kcal/mol and 30.5 kcal/mol.

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

A set of initial conditions that results in a reactive trajectory is AB(HF) = 2.10 Å, BC(HH) = 0.75 Å, p(AB) = -2.12, and p(BC) = 2.38.
From looking at the Internuclear Momenta vs Time, a transfer of energy between translational and vibrational energy was observed. The released energy was transfer to the vibrational energy in the created H-F bond. This can be confirmed by looking at the oscillation of H-F (A-B) after the transition state.

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state===

Polanyi’s empirical rules tells that the translational energy is more effective in increasing the efficiency of the reaction than the vibrational state for the early transition state. The opposite can be applied to the late transition state.
F + H2
When transitional energy is small at -0.5 and vibrational energy is large at 2, the reaction is unsuccessful. However, if the condition were changed to a large transitional energy, at -0.8, and small vibrational energy at 0.1, the reaction become successful. This means that F + H2 as an early transition state.
H + HF
When transitional energy is small at 1.5 and vibrational energy is small at -0.5, the reaction is unsuccessful. However, if the condition were changed to a smaller transitional energy, at 0.1, and larger vibrational energy at -2.1, the reaction become successful. This means that F + H2 as an late transition state.

MRD:pc2715

2017-05-15T16:29:03Z

Pc2715: Created page with "== EXERCISE 1: H + H2 system == === What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly exp..."

Pc27151

2017-05-09T13:53:56Z

Pc2715: Created page with "yay"

yay

Pc2715

2016-02-26T15:42:42Z

Pc2715:

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

Nitrogen atom is more negative because it is more electronegative atom than the hydrogen.

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|left|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

The carbon atom is more negative because it is more electronegative atom than the hydrogen. However, the molecule is a non polar molecule because the charges canceled out.

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|500px]]

1s bonding orbital

The energy level of this orbital is at -10.16707 which is too low to form bond with other orbital. It doesn't involve in any chemical bonding.

[[File:Palita MO 2.png|500px|]]

2s bonding orbital

2s orbital is larger than the 1s orbital, Therefore, it can overlap with other orbital.

[[File:Palita MO 4.png|500px|]]

2p bonding orbital.

Methane contain 3 of this p orbitals. They all located at the same energy level of -0.38831. The 3 orbitals orientate in 3 different direction; x, y or z. The p orbital of the carbon overlap with the 1s orbital of the hydrogen; the AO that contributed to this orbital is 1s from hydrogen and 2p from carbon giving this MO.

However, the 2s and the three 2p can mixed to form 4 sp3 orbitals. The MO of the 4 sp3 orbitals were not shown as the program did not show the hybridised orbital formed.

[[File:Palita LUMO 6.png|500px|]]

2s antibonding orbital.

This is the antibonding orbital of this orbital. The energy level is higher than the bonding orbital, it is at 0.11824.

[[File:Palita anti 7.png|500px|]]

2p antibonding orbital.

This is the antibonding orbital of this orbital. The energy level of this orbital is at 0.17677.

Pc2715

2016-02-26T15:39:04Z

Pc2715: /* Animating Vibration */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|left|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|500px]]

1s bonding orbital

The energy level of this orbital is at -10.16707 which is too low to form bond with other orbital. It doesn't involve in any chemical bonding.

[[File:Palita MO 2.png|500px|]]

2s bonding orbital

2s orbital is larger than the 1s orbital, Therefore, it can overlap with other orbital.

[[File:Palita MO 4.png|500px|]]

2p bonding orbital.

Methane contain 3 of this p orbitals. They all located at the same energy level of -0.38831. The 3 orbitals orientate in 3 different direction; x, y or z. The p orbital of the carbon overlap with the 1s orbital of the hydrogen; the AO that contributed to this orbital is 1s from hydrogen and 2p from carbon giving this MO.

However, the 2s and the three 2p can mixed to form 4 sp3 orbitals. The MO of the 4 sp3 orbitals were not shown as the program did not show the hybridised orbital formed.

[[File:Palita LUMO 6.png|500px|]]

2s antibonding orbital.

This is the antibonding orbital of this orbital. The energy level is higher than the bonding orbital, it is at 0.11824.

[[File:Palita anti 7.png|500px|]]

2p antibonding orbital.

This is the antibonding orbital of this orbital. The energy level of this orbital is at 0.17677.

Pc2715

2016-02-26T15:38:27Z

Pc2715: /* Animating Vibration */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|left|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|left|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|500px]]

1s bonding orbital

The energy level of this orbital is at -10.16707 which is too low to form bond with other orbital. It doesn't involve in any chemical bonding.

[[File:Palita MO 2.png|500px|]]

2s bonding orbital

2s orbital is larger than the 1s orbital, Therefore, it can overlap with other orbital.

[[File:Palita MO 4.png|500px|]]

2p bonding orbital.

Methane contain 3 of this p orbitals. They all located at the same energy level of -0.38831. The 3 orbitals orientate in 3 different direction; x, y or z. The p orbital of the carbon overlap with the 1s orbital of the hydrogen; the AO that contributed to this orbital is 1s from hydrogen and 2p from carbon giving this MO.

However, the 2s and the three 2p can mixed to form 4 sp3 orbitals. The MO of the 4 sp3 orbitals were not shown as the program did not show the hybridised orbital formed.

[[File:Palita LUMO 6.png|500px|]]

2s antibonding orbital.

This is the antibonding orbital of this orbital. The energy level is higher than the bonding orbital, it is at 0.11824.

[[File:Palita anti 7.png|500px|]]

2p antibonding orbital.

This is the antibonding orbital of this orbital. The energy level of this orbital is at 0.17677.

Pc2715

2016-02-26T15:38:11Z

Pc2715: /* Animating Vibration */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|left|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|left|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|500px]]

1s bonding orbital

The energy level of this orbital is at -10.16707 which is too low to form bond with other orbital. It doesn't involve in any chemical bonding.

[[File:Palita MO 2.png|500px|]]

2s bonding orbital

2s orbital is larger than the 1s orbital, Therefore, it can overlap with other orbital.

[[File:Palita MO 4.png|500px|]]

2p bonding orbital.

Methane contain 3 of this p orbitals. They all located at the same energy level of -0.38831. The 3 orbitals orientate in 3 different direction; x, y or z. The p orbital of the carbon overlap with the 1s orbital of the hydrogen; the AO that contributed to this orbital is 1s from hydrogen and 2p from carbon giving this MO.

However, the 2s and the three 2p can mixed to form 4 sp3 orbitals. The MO of the 4 sp3 orbitals were not shown as the program did not show the hybridised orbital formed.

[[File:Palita LUMO 6.png|500px|]]

2s antibonding orbital.

This is the antibonding orbital of this orbital. The energy level is higher than the bonding orbital, it is at 0.11824.

[[File:Palita anti 7.png|500px|]]

2p antibonding orbital.

This is the antibonding orbital of this orbital. The energy level of this orbital is at 0.17677.

Pc2715

2016-02-26T15:37:20Z

Pc2715: /* Animating Vibration */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|left|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|500px]]

1s bonding orbital

The energy level of this orbital is at -10.16707 which is too low to form bond with other orbital. It doesn't involve in any chemical bonding.

[[File:Palita MO 2.png|500px|]]

2s bonding orbital

2s orbital is larger than the 1s orbital, Therefore, it can overlap with other orbital.

[[File:Palita MO 4.png|500px|]]

2p bonding orbital.

Methane contain 3 of this p orbitals. They all located at the same energy level of -0.38831. The 3 orbitals orientate in 3 different direction; x, y or z. The p orbital of the carbon overlap with the 1s orbital of the hydrogen; the AO that contributed to this orbital is 1s from hydrogen and 2p from carbon giving this MO.

However, the 2s and the three 2p can mixed to form 4 sp3 orbitals. The MO of the 4 sp3 orbitals were not shown as the program did not show the hybridised orbital formed.

[[File:Palita LUMO 6.png|500px|]]

2s antibonding orbital.

This is the antibonding orbital of this orbital. The energy level is higher than the bonding orbital, it is at 0.11824.

[[File:Palita anti 7.png|500px|]]

2p antibonding orbital.

This is the antibonding orbital of this orbital. The energy level of this orbital is at 0.17677.

Pc2715

2016-02-26T15:36:23Z

Pc2715: /* Molecular Orbitals */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|500px]]

1s bonding orbital

The energy level of this orbital is at -10.16707 which is too low to form bond with other orbital. It doesn't involve in any chemical bonding.

[[File:Palita MO 2.png|500px|]]

2s bonding orbital

2s orbital is larger than the 1s orbital, Therefore, it can overlap with other orbital.

[[File:Palita MO 4.png|500px|]]

2p bonding orbital.

Methane contain 3 of this p orbitals. They all located at the same energy level of -0.38831. The 3 orbitals orientate in 3 different direction; x, y or z. The p orbital of the carbon overlap with the 1s orbital of the hydrogen; the AO that contributed to this orbital is 1s from hydrogen and 2p from carbon giving this MO.

However, the 2s and the three 2p can mixed to form 4 sp3 orbitals. The MO of the 4 sp3 orbitals were not shown as the program did not show the hybridised orbital formed.

[[File:Palita LUMO 6.png|500px|]]

2s antibonding orbital.

This is the antibonding orbital of this orbital. The energy level is higher than the bonding orbital, it is at 0.11824.

[[File:Palita anti 7.png|500px|]]

2p antibonding orbital.

This is the antibonding orbital of this orbital. The energy level of this orbital is at 0.17677.

Pc2715

2016-02-26T15:34:52Z

Pc2715: /* Molecular Orbitals */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|500px]]

1s bonding orbital

The energy level of this orbital is at -10.16707 which is too low to form bond with other orbital. It doesn't involve in any chemical bonding.

[[File:Palita MO 2.png|500px|]]

2s bonding orbital

2s orbital is larger than the 1s orbital, Therefore, it can overlap with other orbital.

[[File:Palita MO 4.png|500px|]]

2p bonding orbital.

Methane contain 3 of this p orbitals. They all located at the same energy level of -0.38831. The 3 orbitals orientate in 3 different direction; x, y or z. The p orbital of the carbon overlap with the 1s orbital of the hydrogen; the AO that contributed to this orbital is 1s from hydrogen and 2p from carbon giving this MO.

However, the 2s and the three 2p can mixed to form 4 sp3 orbitals. The MO of the 4 sp3 orbitals were not shown as the program did not show the hybridised orbital formed.

[[File:Palita LUMO 6.png|500px|]]

2s antibonding orbital.

This is the antibonding orbital of this orbital.

[[File:Palita anti 7.png|500px|]]

2p antibonding orbital.

This is the antibonding orbital of this orbital.

Pc2715

2016-02-26T15:26:28Z

Pc2715: /* Molecular Orbitals */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|500px]]

1s bonding orbital The energy level of this orbital is at -10.16707 which is too low to form bond with other orbital. It doesn't involve in any chemical bonding.

[[File:Palita MO 2.png|500px|]]

2s bonding orbital 2s orbital is larger than the 1s orbital, Therefore, it can overlap with other orbital.

[[File:Palita MO 4.png|500px|]]

2p bonding orbital. Methane contain 3 of this p orbitals. They all located at the same energy level of -0.38831. The 3 orbitals orientate in 3 different direction; x, y or z. The p orbital of the carbon overlap with the 1s orbital of the hydrogen; the AO that contributed to this orbital is 1s from hydrogen and 2p from carbon.

However, the 2s and the three 2p can mixed to form 4 sp3 orbitals. The MO of the 4 sp3 orbitals were not shown as the program did not show the hybridised orbital formed.

[[File:Palita LUMO 6.png|500px|]]

2s antibonding orbital.

[[File:Palita anti 7.png|500px|]]

3p orbital of Methane.

Pc2715

2016-02-26T15:25:37Z

Pc2715: /* Molecular Orbitals */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|500px]]
1s bonding orbital The energy level of this orbital is at -10.16707 which is too low to form bond with other orbital. It doesn't involve in any chemical bonding.

[[File:Palita MO 2.png|500px|]]
2s bonding orbital 2s orbital is larger than the 1s orbital, Therefore, it can overlap with other orbital.

[[File:Palita MO 4.png|500px|]]
2p bonding orbital. Methane contain 3 of this p orbitals. They all located at the same energy level of -0.38831. The 3 orbitals orientate in 3 different direction; x, y or z. The p orbital of the carbon overlap with the 1s orbital of the hydrogen; the AO that contributed to this orbital is 1s from hydrogen and 2p from carbon.

However, the 2s and the three 2p can mixed to form 4 sp3 orbitals. The MO of the 4 sp3 orbitals were not shown as the program did not show the hybridised orbital formed.

[[File:Palita LUMO 6.png|500px|]]
2s antibonding orbital.

[[File:Palita anti 7.png|500px|]]
3p orbital of Methane.

Pc2715

2016-02-26T14:35:27Z

Pc2715: /* Molecular Orbitals */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|500px]]
1s orbital of Methane

[[File:Palita MO 2.png|300px|2s orbital of Methane.]]

[[File:Palita MO 4.png|300px|2p bong orbital of Methane.]]

[[File:Palita LUMO 6.png|300px|2s antibonding orbital of Methane.]]

[[File:Palita anti 7.png|300px|2p antibonding orbital of Methane.]]

Pc2715

2016-02-26T14:35:12Z

Pc2715: /* Molecular Orbitals */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|300px]]
1s orbital of Methane

[[File:Palita MO 2.png|300px|2s orbital of Methane.]]

[[File:Palita MO 4.png|300px|2p bong orbital of Methane.]]

[[File:Palita LUMO 6.png|300px|2s antibonding orbital of Methane.]]

[[File:Palita anti 7.png|300px|2p antibonding orbital of Methane.]]

Pc2715

2016-02-26T14:34:13Z

Pc2715: /* Molecular Orbitals */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|300px|thumb|1s orbital of Methane.]]

[[File:Palita MO 2.png|300px|2s orbital of Methane.]]

[[File:Palita MO 4.png|300px|2p bong orbital of Methane.]]

[[File:Palita LUMO 6.png|300px|2s antibonding orbital of Methane.]]

[[File:Palita anti 7.png|300px|2p antibonding orbital of Methane.]]

Pc2715

2016-02-26T14:33:33Z

Pc2715: /* Molecular Orbitals */

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|300px|1s orbital of Methane.]]

[[File:Palita MO 2.png|300px|2s orbital of Methane.]]

[[File:Palita MO 4.png|300px|2p bong orbital of Methane.]]

[[File:Palita LUMO 6.png|300px|2s antibonding orbital of Methane.]]

[[File:Palita anti 7.png|300px|2p antibonding orbital of Methane.]]

Pc2715

2016-02-26T14:31:27Z

Pc2715:

== Ammonia ==
=== Optimized Ammonia Molecule ===

N-H bond distance = 1.01798

H-N-H bond angle = 105.741

=== Summary ===

Molecule = NH3

Calculation Method = RB3LYP

Basic Set = 6-31G(d.p)

Final Energy E(RG3LYP) = -56.55776873 a.u.

RMS Gradient Norn = 0.00000485

Point Group = C3V

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
Predicted change in Energy=-5.986280D-10
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.018 -DE/DX = 0.0 !
! R2 R(1,3) 1.018 -DE/DX = 0.0 !
! R3 R(1,4) 1.018 -DE/DX = 0.0 !
! A1 A(2,1,3) 105.7412 -DE/DX = 0.0 !
! A2 A(2,1,4) 105.7412 -DE/DX = 0.0 !
! A3 A(3,1,4) 105.7412 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -111.8571 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA NH3 OPTF POP.LOG| here]]

<jmol><jmolApplet>
<title>Ammonia</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA NH3 OPTF POP.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Animating Vibration ===

[[File:Palita screenshot vibration.png|700px|thumb|Display Vibrations.]]

How many modes do you expect from the 3N-6 rule?

= 6

Which modes are degenerate (ie have the same energy)?

= 2 and 3, 5 and 6

Which modes are "bending" vibrations and which are "bond stretch" vibrations?

=The bond stretch vibration modes are 4,5 and 6, while the bending vibration modes are 1, 2 and 3.

Which mode is highly symmetric?

= 4

One mode is known as the "umbrella" mode, which one is this?

= 1

How many bands would you expect to see in an experimental spectrum of gaseous ammonia?
2 Because there are only 2 visible peaks in the IR spectrum.

=== Charge Distribution of Ammonia ===
N = -1.125

H = 0.375

=== Haber-Bosch Process ===

N2 + 3H2 -> 2NH3

==== N2 summary ====

What is the molecule? = N2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p0

What is the final energy E(RB3LYP) in atomic units (au)? = -109.52412868 a.u.

What is the RMS gradient? = 0.00000060

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
Predicted change in Energy=-3.401050D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.1055 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad

</pre>

The optimisation file is liked to [[Media:PALITA N2.LOG| here]]

<jmol><jmolApplet>
<title>Nitrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA N2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 2457.33 || 0.0000
|}

==== Charge Distribution of Nitrogen ====

N = 0

Charge Distribution is zero because the molecule is non polar.

==== H2 summary ====

What is the molecule? = H2

What is the calculation method? = RB3LYP

What is the basis set? = 6-31G(d,p)

What is the final energy E(RB3LYP) in atomic units (au)? = -1.17853936 a.u.

What is the RMS gradient? = 0.00000017

What is the point group of your molecule? = Dinf H

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
Predicted change in Energy=-1.164079D-13
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 0.7428 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA H2.LOG| here]]

<jmol><jmolApplet>
<title>Hydrogen</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA H2.LOG</uploadedFileContents>
</jmolApplet></jmol>

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 4465.68 || 0.0000
|}

==== Charge Distribution of Hydrogen====

H = 0

Charge Distribution is zero because the molecule is non polar.

''' Energy '''

E(NH3)= -56.55776873 a.u.

2*E(NH3)= -113.11553746

E(N2)= -109.52412868 a.u.

E(H2)= -1.17853936 a.u.

3*E(H2)= -3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -0.05579070 a.u

Therefore, ΔE is -146.48 kJ/mol. The ammonia product is more stable than gaseous reactants.

== Methane ==

=== Optimized Methane Molecule ===

C-H bond distance = 1.09197

H-C-H bond angle = 109.47122

=== Methane summary ===

molecule name = Methane CH4

calculation method = RB3LYP

basis set = 6-31G(d,p)

final energy E(RB3LYP) in atomic units (au)? = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

the point group of your molecule = TD

<pre>
Item Value Threshold Converged?
Maximum Force 0.000063 0.000450 YES
RMS Force 0.000034 0.000300 YES
Maximum Displacement 0.000179 0.001800 YES
RMS Displacement 0.000095 0.001200 YES
Predicted change in Energy=-2.255986D-08
Optimization completed.
-- Stationary point found.
----------------------------
! Optimized Parameters !
! (Angstroms and Degrees) !
-------------------------- --------------------------
! Name Definition Value Derivative Info. !
--------------------------------------------------------------------------------
! R1 R(1,2) 1.092 -DE/DX = -0.0001 !
! R2 R(1,3) 1.092 -DE/DX = -0.0001 !
! R3 R(1,4) 1.092 -DE/DX = -0.0001 !
! R4 R(1,5) 1.092 -DE/DX = -0.0001 !
! A1 A(2,1,3) 109.4712 -DE/DX = 0.0 !
! A2 A(2,1,4) 109.4712 -DE/DX = 0.0 !
! A3 A(2,1,5) 109.4712 -DE/DX = 0.0 !
! A4 A(3,1,4) 109.4712 -DE/DX = 0.0 !
! A5 A(3,1,5) 109.4712 -DE/DX = 0.0 !
! A6 A(4,1,5) 109.4712 -DE/DX = 0.0 !
! D1 D(2,1,4,3) -120.0 -DE/DX = 0.0 !
! D2 D(2,1,5,3) 120.0 -DE/DX = 0.0 !
! D3 D(2,1,5,4) -120.0 -DE/DX = 0.0 !
! D4 D(3,1,5,4) 120.0 -DE/DX = 0.0 !
--------------------------------------------------------------------------------
GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
</pre>

The optimisation file is liked to [[Media:PALITA CH4.LOG| here]]

 List Of Frequency- Display Vibrations 

{| class="wikitable" border="1"
|+ List Of Frequency
! Mode !! Freq !! Infrared
|-
| 1 || 1356.20 || 14.1008
|-
| 2 || 1356.20 || 14.1008
|-
| 3 || 1356.20 || 14.1008
|-
| 4 || 1578.58 || 0.0000
|-
| 5 || 1578.58 || 0.0000
|-
| 6 || 3046.46 || 0.0000
|-
| 7 || 3162.33 || 25.3343
|-
| 8 || 3162.33 || 25.3343
|-
| 9 || 3162.33 || 25.3343
|}

=== Animating Vibration ===

[[File:Palita ch4 vibration.png|700px|thumb|Display Vibrations.]]

<jmol><jmolApplet>
<title>Methane</title>
<color>black</color>
<size>300</size>
<uploadedFileContents>PALITA CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>

=== Charge distribution of Methane ===

C = -0.930

H = 0.233

=== Molecular Orbitals ===

[[File:Palita MO 1.1.png|thumb|1s orbital of Methane.]]

[[File:Palita MO 2.png|thumb|2s orbital of Methane.]]

[[File:Palita MO 4.png|thumb|2p bong orbital of Methane.]]

[[File:Palita LUMO 6.png|thumb|2s antibonding orbital of Methane.]]

[[File:Palita anti 7.png|thumb|2p antibonding orbital of Methane.]]