ChemWiki - User contributions [en]

Lammps and plumed

2021-06-24T10:02:55Z

Fdp18:

This wiki serves as an overview of how to use the PLUMED<ref>The Plumed Consortium. (2019). Promoting transparency and reproducibility in enhanced molecular simulations. Nature Methods, 16(8), 670–673. https://doi.org/10.1038/s41592-019-0506-8</ref> <ref>Tribello, G. A., Bonomi, M., Branduardi, D., Camilloni, C., & Bussi, G. (2014). PLUMED 2: New feathers for an old bird. Computer Physics Communications, 185(2), 604–613. https://doi.org/10.1016/j.cpc.2013.09.018</ref> software package to perform biased molecular dynamics simulations in LAMMPS.

=Using LAMMPS=
The very first point of call for everything involving LAMMPS is the official [https://lammps.sandia.gov/doc/Manual.html LAMMPS manual]. One thing is crucial to know: LAMMPS does '''not''' create/contain force fields! You need to define the topology and the force field, LAMMPS only does the simulation itself. For starters, you can find a few example input files here: [[:File:Simple_example.zip]]. Here I use the classical (and classic) CL&P force field for ionic liquids.<ref>Canongia Lopes, J. N., & Padua, A. A. H. (2012). CL&P: A generic and systematic force field for ionic liquids modeling. Theoretical Chemistry Accounts, 131(3), 1129. https://doi.org/10.1007/s00214-012-1129-7</ref><ref>Canongia Lopes, J. N., Deschamps, J., & Padua, A. A. H. (2004). Modeling Ionic Liquids Using a Systematic All-Atom Force Field. The Journal of Physical Chemistry B, 108(6), 2038–2047. https://doi.org/10.1021/jp0362133</ref><ref>Canongia Lopes, J. N., & Padua, A. A. H. (2004). Molecular Force Field for Ionic Liquids Composed of Triflate or Bistriflylimide Anions. The Journal of Physical Chemistry B, 108(43), 16893–16898. https://doi.org/10.1021/jp0476545</ref> The content of the files and what you should watch out for is described in the following.
; run_2D.pbs : This is the runscript. It runs the [[https://github.com/z-gong/lammps.git ttdamp branch of LAMMPS]], compiled with PLUMED 2.6.0. If you want to reproduce my simulations with Drude particles further down: The required thermostat should now be implemented in the USER-DRUDE package.<ref>Goloviznina, K., Gong, Z., Costa Gomes, M. F., & Pádua, A. A. H. (2021). Extension of the CL&Pol Polarizable Force Field to Electrolytes, Protic Ionic Liquids, and Deep Eutectic Solvents. Journal of Chemical Theory and Computation, 17(3), 1606–1617. https://doi.org/10.1021/acs.jctc.0c01002</ref> Have a look beforehand what packages you need etc, this is very different depending on the application.
; METAD.inp : The LAMMPS input file. Line 37 invokes the fix plumed, with a separate input file named "plumedfile.dat". it runs for 10 ns which is ok, since there is only one ion. It needs the topology given as a data file.
; start.dat : This is the LAMMPS data file, containing information about the simulation box, connectivity, bonded force field parameters, cartesian coordinates and velocities.
; plumedfile.dat : Contains the instructions for PLUMED. This one is an example for 2D well-tempered Metadynamics, the two "collective" variables are the two C-S-N-S dihedral angles in the [NTf2]- anion.
After you run the simulation, you can process the results with the following tools:
; sumhills.pbs : You can use this to sum the gaussians from the Metadynamics simulation to produce a two dimensional free energy surface (fes.dat)
; plothills.py : Shows the hill height (of deposited gaussians) as a function of time, as well as the evolution of one of the collective variables.
; plotcolvar_2D.py : Plots the collective variable values as dots on a 2d map.

If you only want to test LAMMPS without PLUMED, you can change the runscript to:
<pre>
module load gcc
module load intel-suite mpi
module load fftw
module load lammps/11Aug17
cd $PBS_O_WORKDIR
mpiexec lammps -in METAD.inp
</pre>
However, if you do so, you also need to comment out the "fix plumed" in METAD.inp, otherwise LAMMPS terminates with the following error:
<pre>
ERROR: Unknown fix style plumed (../modify.cpp:854)
Last command: fix 1 all plumed plumedfile plumedfile.dat outfile plumed.out
</pre>

=Using PLUMED (2.6.0)=
PLUMED is a powerful software package, and I strongly recommend having a look at the [[http://www.plumed.org/doc-v2.5/user-doc/html/_colvar.html PLUMED documentation]]. There are a lot of tutorials on there, and it is worth having a look at the built-in colvars (and multicolvars). Similarly, there are a few good articles describing how nucleation and phase transitions can be modelled with PLUMED.<ref>Giberti, F., Salvalaglio, M., Mazzotti, M., & Parrinello, M. (2015). Insight into the nucleation of urea crystals from the melt. Chemical Engineering Science, 121, 51–59. https://doi.org/10.1016/j.ces.2014.08.032</ref><ref>Bussi, G., & Tribello, G. A. (2019). Analyzing and Biasing Simulations with PLUMED. In Methods in Molecular Biology (Vol. 2022, pp. 529–578). https://doi.org/10.1007/978-1-4939-9608-7_21</ref><ref>Tribello, G. A., Giberti, F., Sosso, G. C., Salvalaglio, M., & Parrinello, M. (2017). Analyzing and Driving Cluster Formation in Atomistic Simulations. Journal of Chemical Theory and Computation, 13(3), 1317–1327. https://doi.org/10.1021/acs.jctc.6b01073</ref> Here I will only show an example which is typical for me, and you can find a stripped-down version in [[:File:BMIMOTf.zip]].

The system in the example is [C4C1Im][OTf], an ionic liquid with the triflate anion. The force field is the CL&Pol force field developed by Goloviznina ''et al.''.<ref>Goloviznina, K., Canongia Lopes, J. N., Costa Gomes, M., & Pádua, A. A. H. (2019). Transferable, Polarizable Force Field for Ionic Liquids. Journal of Chemical Theory and Computation, 15(11), 5858–5871. https://doi.org/10.1021/acs.jctc.9b00689</ref> This anion, F3C-SO3, has an inherent directionality along its C3 axis. Thus we could hypothesise that there might be a liquid-liquid phase transition to an ordered state where all the OTf anions are aligned.<ref>Anaredy, R. S., & Shaw, S. K. (2018). Developing Distinct Chemical Environments in Ionic Liquid Films. The Journal of Physical Chemistry C, 122(34), 19731–19737. https://doi.org/10.1021/acs.jpcc.8b06608</ref> Now have a look at the '''plumedfile.dat''' in the [[:File:BMIMOTf.zip]] example pack. It has the following sections:
* A header, defining units and file i/o flush, also [[http://www.plumed.org/doc-v2.5/user-doc/html/_r_e_s_t_a_r_t.html "RESTART"]] if the simulation is to be restarted - important not to overwrite existing files but rather to continue!
* A section where groups are defined. Here I define all the CF3 groups as their centre of mass as c1,c2,...,c512 (because I have 512 ion pairs in the simulation). Similarly, the centre of mass of the SO3 groups is defined as s1-s512. Note that this simulation contains drude particles, if you wanted to be perfect you would include those in the definition. However, the Drude particles react pretty much instantaneously, so if you need to include them in your definition then your biasing is probably unphysical anyway.
<pre>
c1: COM ATOMS=12801-12804
(...all the other ones...)
c512: COM ATOMS=16889-16892
s1: COM ATOMS=12805-12808
(...all the other ones...)
s512: COM ATOMS=16893-16896
</pre>
* Then we define [[http://www.plumed.org/doc-v2.5/user-doc/html/_m_o_l_e_c_u_l_e_s.html "molecules"]], here we use the centre of mass as the position of the anion. Note that this is a multicolvar that contains all the vectors from the CF3 groups to the SO3 groups!
<pre>
MOLECULES ...
MOL1=c1,s1
(...all the other ones...)
MOL512=c512,s512
LABEL=CSvectors
... MOLECULES
</pre>
* We condense the multicolvar to the actual collective variable which we later want to bias. This collective variable is the [[http://www.plumed.org/doc-v2.5/user-doc/html/_s_m_a_c.html SMAC variable]]. The switching function SWITCH selects molecules whose centres of mass are within a certain distance. The kernel KERNEL1 selects pairs of anions whose CS vectors are parallel. If we wanted to include antiparallel orientations as well, then we would need to add another kernel, etc. So far, SMAC is another multicolvar; each anion gets assigned a real number which is zero if there are no ordered anions nearby, and 1 if all nearby anions are aligned. Now we want to count those anions which are "ordered enough", which is done by another switching function in MORE_THAN1. It is possible to of course define more than 1 MORE_THAN, and to print them all - this is useful to see which values are reasonable for the switching function.
<pre>
SMAC ...
SPECIES=CSvectors # this is the "DATA" from the manual
SWITCH={RATIONAL R_0=9.0 NN=12 D_MAX=13.0} #This should be alright for BMIMOTf
SWITCH_COORD={RATIONAL R_0=0.001} #That should be the "psi". we don't want to exclude neighbour-less molecules at this stage.
KERNEL1={GAUSSIAN CENTER=0 SIGMA=0.480} # the kernelfunction for 0° angle / parallel
MORE_THAN1={RATIONAL R_0=0.5 NN=6 D_MAX=1.0}
LABEL=smac
... SMAC
</pre>
* Now we have a collective variable, smac.morethan-1, which takes values between 0 (no alignment whatsoever) and 512 (all anions are perfectly aligned). We can now bias this collective variable with the [[http://www.plumed.org/doc-v2.5/user-doc/html/_m_e_t_a_d.html METAD keyword]]:
<pre>
METAD ...
LABEL=metad
ARG=smac.morethan-1
HEIGHT=5.0
BIASFACTOR=5000
TEMP=298.15
PACE=200
GRID_MIN=0.0
GRID_MAX=512.0
GRID_BIN=2048
SIGMA=1.0
FILE=HILLS
WALKERS_N=25
WALKERS_ID=XXXX
WALKERS_DIR=../HILLS
... METAD
</pre>
HEIGHT is the maximum height of the deposited gaussians (decreases over time because we're using well-tempered metadynamics, see also the BIASFACTOR).<ref>Barducci, A., Bussi, G., & Parrinello, M. (2008). Well-Tempered Metadynamics: A Smoothly Converging and Tunable Free-Energy Method. Physical Review Letters, 100(2), 020603. https://doi.org/10.1103/PhysRevLett.100.020603</ref> TEMP is the temperature in K of the simulation, and PACE defines how often gaussians are deposited. Here we deposit on a grid which makes the calculation more efficient, since PLUMED doesn't need to keep track of thousands of gaussians, but rather only the cumulative grid file. SIGMA defines the width of the gaussians. What is interesting here are the WALKERS keywords. I am using 25 walkers here, which means that 25 simulations are running in parallel, each of them depositing gaussians on a common grid. The Walkers ID, XXXX, goes from 0 to 24 accordingly. Each of those should be run in their own directory to avoid a mess, and the common hills file directory should also be a clean separate directory.

If you look at the LAMMPS input file, producerestart.inp, you will see that it is slightly different than before. We are here reading a restart file restart1.lmp rather than the data file, and during the simulation restart files are written as well. Hence we can just resubmit the whole array job with the 25 walkers when it is finished but we need some more sampling. The pair coefficients take up quite a bit of space here, which is why they are in their own LAMMPS input file, pair-p-sc.lmp. The 25 restart files should not be the same even to begin with, otherwise that region gets oversamples. I personally like driving the simulation from the minimum colvar to the maximum colvar with a [[http://www.plumed.org/doc-v2.5/user-doc/html/_m_o_v_i_n_g_r_e_s_t_r_a_i_n_t.html movingrestraint]], meanwhile dumping restart files with LAMMPS so I start roughly evenly spaced in collective variable space.
=Using PLUMED (hack-the-tree)=
Some people are also interested in more complex anions, such as the [NTf2]- anion, which we have looked at in isolation in the first simple example.<ref>Anaredy, R. S., & Shaw, S. K. (2016). Long-Range Ordering of Ionic Liquid Fluid Films. Langmuir, 32(20), 5147–5154. https://doi.org/10.1021/acs.langmuir.6b00304</ref> This anion can exist in two states, cis and trans.<ref>Philippi, F., Pugh, D., Rauber, D., Welton, T., & Hunt, P. A. (2020). Conformational design concepts for anions in ionic liquids. Chemical Science, 11(25), 6405–6422. https://doi.org/10.1039/D0SC01379J</ref> The cis conformer has a larger charge arm (the "dipole moment" when the centre of mass is chosen as the origin of the coordinate system). Thus, the collective variable needs to be large only when the anion is in cis, when other ions in the neighbourhood are in cis as well, if the sulphur-sulphur vectors are aligned, and if the carbon-sulphur vectors are aligned. This is not possible with the current official version of PLUMED, thus the following refers to the experimental [[https://github.com/plumed/plumed2/tree/hack-the-tree hack-the-tree]] version. See also the files in [[:File:BMIMTFSI.zip]].

The definition of groups is similar to [C4C1Im][OTf], only that more vectors are required since the molecular anion is more complex. We use the SS vectors (from one SO2 group to the other, each defined as centre of mass) and the CC-SS vectors (from the centre of mass of both CF3 groups to the centre of mass of both SO2 groups).
What is interesting is the end of the plumedfile:

<pre>
osl: MORE_THAN ARG1=ccssdistance SWITCH={RATIONAL R_0=0.8 NN=6 D_MAX=1.7}
smac_cmap: CONTACT_MATRIX GROUP=ssvectors SWITCH={RATIONAL R_0=9.0 D_0=2.0 NN=12 D_MAX=13.0}
smac_dmat: DOTPRODUCT_MATRIX GROUP1=osl
smac_tor1: TORSIONS_MATRIX GROUP1=ssvectors.x GROUP2=ssvectors.y GROUP3=ssvectors.z POSITIONS=ssvectors
smac_ktor1: KERNEL ARG1=smac_tor1 CENTER=0 SIGMA=0.480 TYPE=gaussian
smac_ktor2: KERNEL ARG1=smac_tor1 CENTER=pi SIGMA=0.480 TYPE=gaussian
smac_ksum1: COMBINE ARG1=smac_ktor1 ARG2=smac_ktor2 PERIODIC=NO
smac_tor2: TORSIONS_MATRIX GROUP1=ccssvectors.x GROUP2=ccssvectors.y GROUP3=ccssvectors.z POSITIONS=ssvectors
smac_ktor3: KERNEL ARG1=smac_tor2 CENTER=0 SIGMA=0.480 TYPE=gaussian
smac_prod: MATHEVAL ARG1=smac_cmap.w ARG2=smac_dmat ARG3=smac_ksum1 ARG4=smac_ktor3 VAR=a,b,c,d FUNC=a*b*c*d PERIODIC=NO
smac: COORDINATIONNUMBER WEIGHT=smac_prod
smac_denom: COORDINATIONNUMBER WEIGHT=smac_cmap.w
smac_avg: MATHEVAL ARG1=smac ARG2=smac_denom FUNC=x/y PERIODIC=NO
smac_mt: MORE_THAN ARG=smac_avg SWITCH={RATIONAL R_0=0.5 NN=6 D_MAX=1.0}
final: COMBINE ARG=smac_mt PERIODIC=NO
</pre>

I will explain the commands in reverse order, since I believe that this is easier to understand. the "final" colvar is the sum of all "smac_mt", which in turn is obtained from "smac_avg" by applying a switching function. This switching function essentially amplifies changes in "smac_avg". Both "smac_avg" and "smac_mt" are already multicolvars, i.e. one value per anion. "smac_avg" smoothly varies between 0 (disordered surroundings) and 1 (perfectly ordered surroundings) for every anion, whereas "smac_mt" should be 0 if the anion surroundings are too disordered (even if "smac_avg" is not 0) and 1 if they are ordered enough (even if "smac_avg" is not 1). "smac_avg" Itself is obtained by dividing "smac" by "smac_denom", which has the purpose of normalising smac to be between 0 and 1. Correspondingly, "smac_denom" is the number of neighbours. Thus, "smac" is the variable (more precisely, multicolvar) that counts how many other anions in the neighbourhood of each anion fulfill the above criteria. To this end, for each pair of anions, the product of "smac_cmap.w", "smac_dmat", "smac_ksum1" and "smac_ktor3" is calculated, and summed up per anion (done by "COORDINATIONNUMBER WEIGHT=smac_prod"). Here, "smac_cmap" decides whether a pair of anions is close enough by means of a switching function on the distance, "smac_dmat" selects only anions which are cis by means of a switching function on the distance between the centre of mass of the CF3 groups and the centre of mass of the SO2 groups, "smac_ksum1" selects anions whose SS vectors are either parallel ("smac_ktor1") or antiparallel ("smac_ktor2"), and finally "smac_ktor3" selects anions whose CC-SS vectors are aligned.

=References=
<References>

Lammps and plumed

2021-05-26T12:53:53Z

Fdp18: /* Using PLUMED (hack-the-tree) */

Lammps and plumed

2021-05-26T12:30:26Z

Fdp18:

File:BMIMTFSI.zip

2021-05-26T12:29:46Z

Fdp18: BMIMTFSI example, PLUMED special version / hack-the-tree branch.

BMIMTFSI example, PLUMED special version / hack-the-tree branch.

Lammps and plumed

2021-05-26T12:15:29Z

Fdp18: /* Using PLUMED (2.6.0) */

Lammps and plumed

2021-05-26T12:14:07Z

Fdp18: /* Using LAMMPS */

This wiki serves as an overview of how to use the PLUMED<ref>The Plumed Consortium. (2019). Promoting transparency and reproducibility in enhanced molecular simulations. Nature Methods, 16(8), 670–673. https://doi.org/10.1038/s41592-019-0506-8</ref> <ref>Tribello, G. A., Bonomi, M., Branduardi, D., Camilloni, C., & Bussi, G. (2014). PLUMED 2: New feathers for an old bird. Computer Physics Communications, 185(2), 604–613. https://doi.org/10.1016/j.cpc.2013.09.018</ref> software package to perform biased molecular dynamics simulations in LAMMPS.

=Using LAMMPS=
The very first point of call for everything involving LAMMPS is the official [https://lammps.sandia.gov/doc/Manual.html LAMMPS manual]. One thing is crucial to know: LAMMPS does '''not''' create/contain force fields! You need to define the topology and the force field, LAMMPS only does the simulation itself. For starters, you can find a few example input files here: [[:File:Simple_example.zip]]. Here I use the classical (and classic) CL&P force field for ionic liquids.<ref>Canongia Lopes, J. N., & Padua, A. A. H. (2012). CL&P: A generic and systematic force field for ionic liquids modeling. Theoretical Chemistry Accounts, 131(3), 1129. https://doi.org/10.1007/s00214-012-1129-7</ref><ref>Canongia Lopes, J. N., Deschamps, J., & Padua, A. A. H. (2004). Modeling Ionic Liquids Using a Systematic All-Atom Force Field. The Journal of Physical Chemistry B, 108(6), 2038–2047. https://doi.org/10.1021/jp0362133</ref><ref>Canongia Lopes, J. N., & Padua, A. A. H. (2004). Molecular Force Field for Ionic Liquids Composed of Triflate or Bistriflylimide Anions. The Journal of Physical Chemistry B, 108(43), 16893–16898. https://doi.org/10.1021/jp0476545</ref> The content of the files and what you should watch out for is described in the following.
; run_2D.pbs : This is the runscript. It runs the [[https://github.com/z-gong/lammps.git ttdamp branch of LAMMPS]], compiled with PLUMED 2.6.0. If you want to reproduce my simulations with Drude particles further down: The required thermostat should now be implemented in the USER-DRUDE package.<ref>Goloviznina, K., Gong, Z., Costa Gomes, M. F., & Pádua, A. A. H. (2021). Extension of the CL&Pol Polarizable Force Field to Electrolytes, Protic Ionic Liquids, and Deep Eutectic Solvents. Journal of Chemical Theory and Computation, 17(3), 1606–1617. https://doi.org/10.1021/acs.jctc.0c01002</ref> Have a look beforehand what packages you need etc, this is very different depending on the application.
; METAD.inp : The LAMMPS input file. Line 37 invokes the fix plumed, with a separate input file named "plumedfile.dat". it runs for 10 ns which is ok, since there is only one ion. It needs the topology given as a data file.
; start.dat : This is the LAMMPS data file, containing information about the simulation box, connectivity, bonded force field parameters, cartesian coordinates and velocities.
; plumedfile.dat : Contains the instructions for PLUMED. This one is an example for 2D well-tempered Metadynamics, the two "collective" variables are the two C-S-N-S dihedral angles in the [NTf2]- anion.
After you run the simulation, you can process the results with the following tools:
; sumhills.pbs : You can use this to sum the gaussians from the Metadynamics simulation to produce a two dimensional free energy surface (fes.dat)
; plothills.py : Shows the hill height (of deposited gaussians) as a function of time, as well as the evolution of one of the collective variables.
; plotcolvar_2D.py : Plots the collective variable values as dots on a 2d map.

If you only want to test LAMMPS without PLUMED, you can change the runscript to:
<pre>
module load gcc
module load intel-suite mpi
module load fftw
module load lammps/11Aug17
cd $PBS_O_WORKDIR
mpiexec lammps -in METAD.inp
</pre>
However, if you do so, you also need to comment out the "fix plumed" in METAD.inp, otherwise LAMMPS terminates with the following error:
<pre>
ERROR: Unknown fix style plumed (../modify.cpp:854)
Last command: fix 1 all plumed plumedfile plumedfile.dat outfile plumed.out
</pre>

=Using PLUMED (2.6.0)=
PLUMED is a powerful software package, and I strongly recommend having a look at the [[http://www.plumed.org/doc-v2.5/user-doc/html/_colvar.html PLUMED documentation]]. There are a lot of tutorials on there, and it is worth having a look at the built-in colvars (and multicolvars). Similarly, there are a few good articles describing how nucleation and phase transitions can be modelled with PLUMED.<ref>Giberti, F., Salvalaglio, M., Mazzotti, M., & Parrinello, M. (2015). Insight into the nucleation of urea crystals from the melt. Chemical Engineering Science, 121, 51–59. https://doi.org/10.1016/j.ces.2014.08.032</ref><ref>Bussi, G., & Tribello, G. A. (2019). Analyzing and Biasing Simulations with PLUMED. In Methods in Molecular Biology (Vol. 2022, pp. 529–578). https://doi.org/10.1007/978-1-4939-9608-7_21</ref><ref>Tribello, G. A., Giberti, F., Sosso, G. C., Salvalaglio, M., & Parrinello, M. (2017). Analyzing and Driving Cluster Formation in Atomistic Simulations. Journal of Chemical Theory and Computation, 13(3), 1317–1327. https://doi.org/10.1021/acs.jctc.6b01073</ref> Here I will only show an example which is typical for me, and you can find a stripped-down version in [[:File:BMIMOTf.zip]].

The system in the example is [C4C1Im][OTf], an ionic liquid with the triflate anion. This anion, F3C-SO3, has an inherent directionality along its C3 axis. Thus we could hypothesise that there might be a liquid-liquid phase transition to an ordered state where all the OTf anions are aligned.<ref>Anaredy, R. S., & Shaw, S. K. (2018). Developing Distinct Chemical Environments in Ionic Liquid Films. The Journal of Physical Chemistry C, 122(34), 19731–19737. https://doi.org/10.1021/acs.jpcc.8b06608</ref> Now have a look at the '''plumedfile.dat''' in the [[:File:BMIMOTf.zip]] example pack. It has the following sections:
* A header, defining units and file i/o flush, also [[http://www.plumed.org/doc-v2.5/user-doc/html/_r_e_s_t_a_r_t.html "RESTART"]] if the simulation is to be restarted - important not to overwrite existing files but rather to continue!
* A section where groups are defined. Here I define all the CF3 groups as their centre of mass as c1,c2,...,c512 (because I have 512 ion pairs in the simulation). Similarly, the centre of mass of the SO3 groups is defined as s1-s512. Note that this simulation contains drude particles, if you wanted to be perfect you would include those in the definition. However, the Drude particles react pretty much instantaneously, so if you need to include them in your definition then your biasing is probably unphysical anyway.
<pre>
c1: COM ATOMS=12801-12804
(...all the other ones...)
c512: COM ATOMS=16889-16892
s1: COM ATOMS=12805-12808
(...all the other ones...)
s512: COM ATOMS=16893-16896
</pre>
* Then we define [[http://www.plumed.org/doc-v2.5/user-doc/html/_m_o_l_e_c_u_l_e_s.html "molecules"]], here we use the centre of mass as the position of the anion. Note that this is a multicolvar that contains all the vectors from the CF3 groups to the SO3 groups!
<pre>
MOLECULES ...
MOL1=c1,s1
(...all the other ones...)
MOL512=c512,s512
LABEL=CSvectors
... MOLECULES
</pre>
* We condense the multicolvar to the actual collective variable which we later want to bias. This collective variable is the [[http://www.plumed.org/doc-v2.5/user-doc/html/_s_m_a_c.html SMAC variable]]. The switching function SWITCH selects molecules whose centres of mass are within a certain distance. The kernel KERNEL1 selects pairs of anions whose CS vectors are parallel. If we wanted to include antiparallel orientations as well, then we would need to add another kernel, etc. So far, SMAC is another multicolvar; each anion gets assigned a real number which is zero if there are no ordered anions nearby, and 1 if all nearby anions are aligned. Now we want to count those anions which are "ordered enough", which is done by another switching function in MORE_THAN1. It is possible to of course define more than 1 MORE_THAN, and to print them all - this is useful to see which values are reasonable for the switching function.
<pre>
SMAC ...
SPECIES=CSvectors # this is the "DATA" from the manual
SWITCH={RATIONAL R_0=9.0 NN=12 D_MAX=13.0} #This should be alright for BMIMOTf
SWITCH_COORD={RATIONAL R_0=0.001} #That should be the "psi". we don't want to exclude neighbour-less molecules at this stage.
KERNEL1={GAUSSIAN CENTER=0 SIGMA=0.480} # the kernelfunction for 0° angle / parallel
MORE_THAN1={RATIONAL R_0=0.5 NN=6 D_MAX=1.0}
LABEL=smac
... SMAC
</pre>
* Now we have a collective variable, smac.morethan-1, which takes values between 0 (no alignment whatsoever) and 512 (all anions are perfectly aligned). We can now bias this collective variable with the [[http://www.plumed.org/doc-v2.5/user-doc/html/_m_e_t_a_d.html METAD keyword]]:
<pre>
METAD ...
LABEL=metad
ARG=smac.morethan-1
HEIGHT=5.0
BIASFACTOR=5000
TEMP=298.15
PACE=200
GRID_MIN=0.0
GRID_MAX=512.0
GRID_BIN=2048
SIGMA=1.0
FILE=HILLS
WALKERS_N=25
WALKERS_ID=XXXX
WALKERS_DIR=../HILLS
... METAD
</pre>
HEIGHT is the maximum height of the deposited gaussians (decreases over time because we're using well-tempered metadynamics, see also the BIASFACTOR).<ref>Barducci, A., Bussi, G., & Parrinello, M. (2008). Well-Tempered Metadynamics: A Smoothly Converging and Tunable Free-Energy Method. Physical Review Letters, 100(2), 020603. https://doi.org/10.1103/PhysRevLett.100.020603</ref> TEMP is the temperature in K of the simulation, and PACE defines how often gaussians are deposited. Here we deposit on a grid which makes the calculation more efficient, since PLUMED doesn't need to keep track of thousands of gaussians, but rather only the cumulative grid file. SIGMA defines the width of the gaussians. What is interesting here are the WALKERS keywords. I am using 25 walkers here, which means that 25 simulations are running in parallel, each of them depositing gaussians on a common grid. The Walkers ID, XXXX, goes from 0 to 24 accordingly. Each of those should be run in their own directory to avoid a mess, and the common hills file directory should also be a clean separate directory.

If you look at the LAMMPS input file, producerestart.inp, you will see that it is slightly different than before. We are here reading a restart file restart1.lmp rather than the data file, and during the simulation restart files are written as well. Hence we can just resubmit the whole array job with the 25 walkers when it is finished but we need some more sampling. The pair coefficients take up quite a bit of space here, which is why they are in their own LAMMPS input file, pair-p-sc.lmp. The 25 restart files should not be the same even to begin with, otherwise that region gets oversamples. I personally like driving the simulation from the minimum colvar to the maximum colvar with a [[http://www.plumed.org/doc-v2.5/user-doc/html/_m_o_v_i_n_g_r_e_s_t_r_a_i_n_t.html movingrestraint]], meanwhile dumping restart files with LAMMPS so I start roughly evenly spaced in collective variable space.
==References==
<References>

Lammps and plumed

2021-05-26T12:07:10Z

Fdp18: /* Using PLUMED (2.6.0) */

Lammps and plumed

2021-05-26T12:04:21Z

Fdp18:

Lammps and plumed

2021-05-26T12:01:57Z

Fdp18: /* Using PLUMED (2.6.0) */

Lammps and plumed

2021-05-26T11:52:35Z

Fdp18:

File:BMIMOTf.zip

2021-05-26T11:29:48Z

Fdp18: BMIMOTf lammps example with plumed

BMIMOTf lammps example with plumed

Lammps and plumed

2021-05-26T11:18:47Z

Fdp18:

Lammps and plumed

2021-05-26T10:04:54Z

Fdp18: /* Using LAMMPS */

File:Simple example.zip

2021-05-26T09:48:12Z

Fdp18: single TFSI anion for lammps and also plumed

single TFSI anion for lammps and also plumed

Lammps and plumed

2021-05-26T09:44:41Z

Fdp18: Created page with "This wiki serves as an overview of how to use the PLUMED<ref>The Plumed Consortium. (2019). Promoting transparency and reproducibility in enhanced molecular simulations. Natur..."

Third year simulation experiment

2020-10-26T14:51:21Z

Fdp18: /* Assessment */

'''<big>This is the optional experiment which may be chosen by any third year student. If you are looking for the Programming-Ising experiment, you will find it [[Third_year_CMP_compulsory_experiment|here]].</big>'''

 '''To run this experiment, you will need access to the Imperial College supercomputers. If you have VPN installed, you should be able to access the computers directly.'''

'''If you have not yet installed VPN, you may 1) use Remote Desktop connection (VPN not needed) or 2) install VPN.'''

Please follow the instructions to use one of these two options:

A) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/remotely-access-my-college-computer/remote-desktop-access-for-students/ Remote Desktop connection]

B) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/virtual-private-network-vpn/ using VPN]

==Introduction==

Computer simulation is widely used to study a huge variety of chemical phenomena, from the behaviour of exotic materials under extreme conditions to protein folding and the properties of biological systems such as lipid membranes. In this experiment, we hope to give you a gentle introduction to one of the most powerful methods for the simulation of chemical systems, '''molecular dynamics simulation'''.

This course closely follows some of the ideas introduced in Professor Bresme's statistical thermodynamics lecture course. Do not worry if you are doing this experiment before the lectures begin, everything that you need to know to complete the experiment will be explained in these instructions. We will begin with a brief overview of the fundamental theory behind the method, before you start running your own simulations of a simple liquid using the college's [https://dx.doi.org/10.14469/hpc/2232 high performance computing] facilities.

At the end of this experiment, you will have performed your own simulations using state of the art software packages used by researchers all around the world, and used those simulations to calculate both structural and dynamic properties of a simple liquid. You will have seen how the concepts of statistical physics introduced by Professor Bresme are needed to calculate thermodynamic quantities such as temperature and pressure in computer simulations, and you will see how computers can be used to validate those concepts.

All of the information that you need to complete the experiment is provided in these wiki pages. We have also tried to provide links to external resources and relevant textbooks where possible — unless explicitly stated, reading these resources '''is not required'''; they are provided only as further information for those interested in the subject.

==Assessment==

At the end of this experiment you must submit a report (pdf format via turnitin) and a zip file with inputs and outputs. The report should be structured:

* Introduction Questions (20% of the total mark)
* Results and Discussion (60% of total mark)
* Conclusion Questions (20% of total mark)
* References (see note below)

Relevant supplementary material can be added at the end of the report so long as it supports your discussion. Please limit the introduction and conclusion sections (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font). It is certainly possible to answer the introduction and conclusion questions with under this length limit, and as with most scientific writing, shorter concise but well thought out answers are preferable. We have not imposed a hard word limit because, equally, you should spend the majority of this lab learning, not editing and rewriting answers to just satisfy the word limit.

Please note that five "floating" marks have been reserved in the results and discussion section. These can be used to reward particularly insightful comments and/or explanations, on the basis of good/bad presentation, use of scientific language, or for other reasons.

It is important to provide appropriate references. 'Appropriate' refers to both quality (books and peer-reviewed papers are preferred over a link to wikipedia) and quantity (make sure to include references where you used them, but there is no point quoting 10 papers for a statement like 'macroscopic systems consist of a large number of atoms'). Please note that presenting external information as your own counts as plagiarism! As a bare minimum, you will need to reference the information you present in your introduction.

'''Copy&Paste will not be accepted, even if referenced! Every report is automatically checked for plagiarism.'''

'''The final task of the lab (task 10 - MSD diffusion simulations) is NOT part of the lab this year.''' The section has been left on the wiki for any interested student to give it a go. However, it should NOT be submitted as part of your lab report, and will not be marked.

==Getting Help==

Please feel free to ask any demonstrators for help during the lab sessions - that is what we are here for. Questions can also be asked via email.

The member of academic staff responsible for this exercise is Professor Fernando Bresme (f.bresme@imperial.ac.uk).

==Structure of this Experiment==

This experimental manual has been broken up into a number of subsections. To help you plan your time it is suggested you complete the following at these times:

Monday (morning session): Theory - Introduction to molecular dynamics simulations

Monday (afternoon session): Theory + Equilibration (submit your files for running)

Tuesday (morning session): Equilibration (analyse your files)

Tuesday (afternoon session): Running simulations under specific conditions (submit and read)

Thursday (morning session): Checkpoint for progress. Submit your input files for the radial distribution function and analyse your equation of state from the previous section

Thursday (afternoon session).

Friday (morning session): Analyse MSD diffusion simulations

Friday (afternoon session): Analyse MSD diffusion simulations. Report write-up.

Direct links to each of them may be found below. You should attempt them in order, and you should complete all of them to finish the experiment.
# [[Third year simulation experiment/Files to download|Downloading the Files]]
# [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]]
# [[Third_year_simulation_experiment/Equilibration|Equilibration]]
# [[Third_year_simulation_experiment/Running_simulations_under_specific_conditions|Running simulations under specific conditions]]
# [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]]
# [[Third year simulation experiment/Dynamical properties and the diffusion coefficient|Dynamical properties and the diffusion coefficient]]

== Introduction Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# Give a brief account on what molecular dynamics is, and (broadly speaking) what it aims to achieve. [5]
# Give two examples of instances where molecular dynamics simulations have/can be used (in a helpful way) instead of or in synergy with experimental techniques. In each case explain why. [10] (''Please consider only chemically relevant examples. E.g. To study geochemical processes which occur under extreme temperature/pressure conditions, and therefore cannot be studied experimentally.'')
# Explain what is meant by a thermodynamic ensemble, and the term "conserved quantity". [2]
# What is the Ergodic hypothesis? Explain its relevance to molecular dynamics simulations. [3]

== Conclusion Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# In this lab, you have used the Lennard-Jones potential exclusively. Give one example of a system that can be described well using only LJ potentials, and one that cannot. In each case explain why. [4]
# What cut-off have you been using in your simulations? What are the advantages and disadvantages for using a shorter/longer cut-off? [3]
# What are finite size effects? Do you think they are significant in the simulations you have performed? Why? [3]
# Algorithms such as SHAKE and RATTLE (holonomic constraints) allow MD simulations to be performed while fixing bond lengths. Why is this desirable? [3] ''(Hint: think about the timestep.)''
# In the simulations you have performed ergodicity has not been an issue. Describe a system in which "brute force" MD struggles to achieve ergodic sampling. Describe one "enhanced sampling" technique that can be used to overcome this. [7]

Third year simulation experiment

2020-10-26T09:17:18Z

Fdp18: /* Assessment */

'''<big>This is the optional experiment which may be chosen by any third year student. If you are looking for the Programming-Ising experiment, you will find it [[Third_year_CMP_compulsory_experiment|here]].</big>'''

 '''To run this experiment, you will need access to the Imperial College supercomputers. If you have VPN installed, you should be able to access the computers directly.'''

'''If you have not yet installed VPN, you may 1) use Remote Desktop connection (VPN not needed) or 2) install VPN.'''

Please follow the instructions to use one of these two options:

A) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/remotely-access-my-college-computer/remote-desktop-access-for-students/ Remote Desktop connection]

B) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/virtual-private-network-vpn/ using VPN]

==Introduction==

Computer simulation is widely used to study a huge variety of chemical phenomena, from the behaviour of exotic materials under extreme conditions to protein folding and the properties of biological systems such as lipid membranes. In this experiment, we hope to give you a gentle introduction to one of the most powerful methods for the simulation of chemical systems, '''molecular dynamics simulation'''.

This course closely follows some of the ideas introduced in Professor Bresme's statistical thermodynamics lecture course. Do not worry if you are doing this experiment before the lectures begin, everything that you need to know to complete the experiment will be explained in these instructions. We will begin with a brief overview of the fundamental theory behind the method, before you start running your own simulations of a simple liquid using the college's [https://dx.doi.org/10.14469/hpc/2232 high performance computing] facilities.

At the end of this experiment, you will have performed your own simulations using state of the art software packages used by researchers all around the world, and used those simulations to calculate both structural and dynamic properties of a simple liquid. You will have seen how the concepts of statistical physics introduced by Professor Bresme are needed to calculate thermodynamic quantities such as temperature and pressure in computer simulations, and you will see how computers can be used to validate those concepts.

All of the information that you need to complete the experiment is provided in these wiki pages. We have also tried to provide links to external resources and relevant textbooks where possible — unless explicitly stated, reading these resources '''is not required'''; they are provided only as further information for those interested in the subject.

==Assessment==

At the end of this experiment you must submit a report (pdf format via turnitin) and a zip file with inputs and outputs. The report should be structured:

* Introduction Questions (20% of the total mark)
* Results and Discussion (60% of total mark)
* Conclusion Questions (20% of total mark)
* References (see note below)

Relevant supplementary material can be added at the end of the report so long as it supports your discussion. Please limit the introduction and conclusion sections (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font). It is certainly possible to answer the introduction and conclusion questions with under this length limit, and as with most scientific writing, shorter concise but well thought out answers are preferable. We have not imposed a hard word limit because, equally, you should spend the majority of this lab learning, not editing and rewriting answers to just satisfy the word limit.

Please note that five "floating" marks have been reserved in the results and discussion section. These can be used to reward particularly insightful comments and/or explanations, on the basis of good/bad presentation, use of scientific language, or for other reasons.

It is important to provide appropriate references. 'Appropriate' refers to both quality (books and peer-reviewed papers are preferred over a link to wikipedia) and quantity (make sure to include references where you used them, but there is no point quoting 10 papers for a statement like 'macroscopic systems consist of a large number of atoms'). Please note that presenting external information as your own counts as plagiarism! As a bare minimum, you will need to reference the information you present in your introduction. '''Copy&Paste will not be accepted, even if referenced! Every report is automatically checked for plagiarism.'''

'''The final task of the lab (task 10 - MSD diffusion simulations) is NOT part of the lab this year.''' The section has been left on the wiki for any interested student to give it a go. However, it should NOT be submitted as part of your lab report, and will not be marked.

==Getting Help==

Please feel free to ask any demonstrators for help during the lab sessions - that is what we are here for. Questions can also be asked via email.

The member of academic staff responsible for this exercise is Professor Fernando Bresme (f.bresme@imperial.ac.uk).

==Structure of this Experiment==

This experimental manual has been broken up into a number of subsections. To help you plan your time it is suggested you complete the following at these times:

Monday (morning session): Theory - Introduction to molecular dynamics simulations

Monday (afternoon session): Theory + Equilibration (submit your files for running)

Tuesday (morning session): Equilibration (analyse your files)

Tuesday (afternoon session): Running simulations under specific conditions (submit and read)

Thursday (morning session): Checkpoint for progress. Submit your input files for the radial distribution function and analyse your equation of state from the previous section

Thursday (afternoon session).

Friday (morning session): Analyse MSD diffusion simulations

Friday (afternoon session): Analyse MSD diffusion simulations. Report write-up.

Direct links to each of them may be found below. You should attempt them in order, and you should complete all of them to finish the experiment.
# [[Third year simulation experiment/Files to download|Downloading the Files]]
# [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]]
# [[Third_year_simulation_experiment/Equilibration|Equilibration]]
# [[Third_year_simulation_experiment/Running_simulations_under_specific_conditions|Running simulations under specific conditions]]
# [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]]
# [[Third year simulation experiment/Dynamical properties and the diffusion coefficient|Dynamical properties and the diffusion coefficient]]

== Introduction Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# Give a brief account on what molecular dynamics is, and (broadly speaking) what it aims to achieve. [5]
# Give two examples of instances where molecular dynamics simulations have/can be used (in a helpful way) instead of or in synergy with experimental techniques. In each case explain why. [10] (''Please consider only chemically relevant examples. E.g. To study geochemical processes which occur under extreme temperature/pressure conditions, and therefore cannot be studied experimentally.'')
# Explain what is meant by a thermodynamic ensemble, and the term "conserved quantity". [2]
# What is the Ergodic hypothesis? Explain its relevance to molecular dynamics simulations. [3]

== Conclusion Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# In this lab, you have used the Lennard-Jones potential exclusively. Give one example of a system that can be described well using only LJ potentials, and one that cannot. In each case explain why. [4]
# What cut-off have you been using in your simulations? What are the advantages and disadvantages for using a shorter/longer cut-off? [3]
# What are finite size effects? Do you think they are significant in the simulations you have performed? Why? [3]
# Algorithms such as SHAKE and RATTLE (holonomic constraints) allow MD simulations to be performed while fixing bond lengths. Why is this desirable? [3] ''(Hint: think about the timestep.)''
# In the simulations you have performed ergodicity has not been an issue. Describe a system in which "brute force" MD struggles to achieve ergodic sampling. Describe one "enhanced sampling" technique that can be used to overcome this. [7]

Third year simulation experiment

2020-10-26T09:15:48Z

Fdp18: /* Assessment */

'''<big>This is the optional experiment which may be chosen by any third year student. If you are looking for the Programming-Ising experiment, you will find it [[Third_year_CMP_compulsory_experiment|here]].</big>'''

 '''To run this experiment, you will need access to the Imperial College supercomputers. If you have VPN installed, you should be able to access the computers directly.'''

'''If you have not yet installed VPN, you may 1) use Remote Desktop connection (VPN not needed) or 2) install VPN.'''

Please follow the instructions to use one of these two options:

A) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/remotely-access-my-college-computer/remote-desktop-access-for-students/ Remote Desktop connection]

B) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/virtual-private-network-vpn/ using VPN]

==Introduction==

Computer simulation is widely used to study a huge variety of chemical phenomena, from the behaviour of exotic materials under extreme conditions to protein folding and the properties of biological systems such as lipid membranes. In this experiment, we hope to give you a gentle introduction to one of the most powerful methods for the simulation of chemical systems, '''molecular dynamics simulation'''.

This course closely follows some of the ideas introduced in Professor Bresme's statistical thermodynamics lecture course. Do not worry if you are doing this experiment before the lectures begin, everything that you need to know to complete the experiment will be explained in these instructions. We will begin with a brief overview of the fundamental theory behind the method, before you start running your own simulations of a simple liquid using the college's [https://dx.doi.org/10.14469/hpc/2232 high performance computing] facilities.

At the end of this experiment, you will have performed your own simulations using state of the art software packages used by researchers all around the world, and used those simulations to calculate both structural and dynamic properties of a simple liquid. You will have seen how the concepts of statistical physics introduced by Professor Bresme are needed to calculate thermodynamic quantities such as temperature and pressure in computer simulations, and you will see how computers can be used to validate those concepts.

All of the information that you need to complete the experiment is provided in these wiki pages. We have also tried to provide links to external resources and relevant textbooks where possible — unless explicitly stated, reading these resources '''is not required'''; they are provided only as further information for those interested in the subject.

==Assessment==

At the end of this experiment you must submit a report (pdf format via turnitin) and a zip file with inputs and outputs. The report should be structured:

* Introduction Questions (20% of the total mark)
* Results and Discussion (60% of total mark)
* Conclusion Questions (20% of total mark)
* References (see note below)

Relevant supplementary material can be added at the end of the report so long as it supports your discussion. Please limit the introduction and conclusion sections (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font). It is certainly possible to answer the introduction and conclusion questions with under this length limit, and as with most scientific writing, shorter concise but well thought out answers are preferable. We have not imposed a hard word limit because, equally, you should spend the majority of this lab learning, not editing and rewriting answers to just satisfy the word limit.

Please note that five "floating" marks have been reserved in the results and discussion section. These can be used to reward particularly insightful comments and/or explanations, on the basis of good/bad presentation, use of scientific language, or for other reasons.

It is important to provide appropriate references. 'Appropriate' refers to both quality (books and peer-reviewed papers are preferred over a link to wikipedia) and quantity (make sure to include references where you used them, but there is no point quoting 10 papers for a statement like 'macroscopic systems consist of a large number of atoms'). Please note that presenting external information as your own counts as plagiarism! As a bare minimum, you will need to reference the information you present in your introduction. '''Copy&Paste will not be accepted, even if referenced!'''

'''The final task of the lab (task 10 - MSD diffusion simulations) is NOT part of the lab this year.''' The section has been left on the wiki for any interested student to give it a go. However, it should NOT be submitted as part of your lab report, and will not be marked.

==Getting Help==

Please feel free to ask any demonstrators for help during the lab sessions - that is what we are here for. Questions can also be asked via email.

The member of academic staff responsible for this exercise is Professor Fernando Bresme (f.bresme@imperial.ac.uk).

==Structure of this Experiment==

This experimental manual has been broken up into a number of subsections. To help you plan your time it is suggested you complete the following at these times:

Monday (morning session): Theory - Introduction to molecular dynamics simulations

Monday (afternoon session): Theory + Equilibration (submit your files for running)

Tuesday (morning session): Equilibration (analyse your files)

Tuesday (afternoon session): Running simulations under specific conditions (submit and read)

Thursday (morning session): Checkpoint for progress. Submit your input files for the radial distribution function and analyse your equation of state from the previous section

Thursday (afternoon session).

Friday (morning session): Analyse MSD diffusion simulations

Friday (afternoon session): Analyse MSD diffusion simulations. Report write-up.

Direct links to each of them may be found below. You should attempt them in order, and you should complete all of them to finish the experiment.
# [[Third year simulation experiment/Files to download|Downloading the Files]]
# [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]]
# [[Third_year_simulation_experiment/Equilibration|Equilibration]]
# [[Third_year_simulation_experiment/Running_simulations_under_specific_conditions|Running simulations under specific conditions]]
# [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]]
# [[Third year simulation experiment/Dynamical properties and the diffusion coefficient|Dynamical properties and the diffusion coefficient]]

== Introduction Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# Give a brief account on what molecular dynamics is, and (broadly speaking) what it aims to achieve. [5]
# Give two examples of instances where molecular dynamics simulations have/can be used (in a helpful way) instead of or in synergy with experimental techniques. In each case explain why. [10] (''Please consider only chemically relevant examples. E.g. To study geochemical processes which occur under extreme temperature/pressure conditions, and therefore cannot be studied experimentally.'')
# Explain what is meant by a thermodynamic ensemble, and the term "conserved quantity". [2]
# What is the Ergodic hypothesis? Explain its relevance to molecular dynamics simulations. [3]

== Conclusion Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# In this lab, you have used the Lennard-Jones potential exclusively. Give one example of a system that can be described well using only LJ potentials, and one that cannot. In each case explain why. [4]
# What cut-off have you been using in your simulations? What are the advantages and disadvantages for using a shorter/longer cut-off? [3]
# What are finite size effects? Do you think they are significant in the simulations you have performed? Why? [3]
# Algorithms such as SHAKE and RATTLE (holonomic constraints) allow MD simulations to be performed while fixing bond lengths. Why is this desirable? [3] ''(Hint: think about the timestep.)''
# In the simulations you have performed ergodicity has not been an issue. Describe a system in which "brute force" MD struggles to achieve ergodic sampling. Describe one "enhanced sampling" technique that can be used to overcome this. [7]

Third year simulation experiment

2020-10-12T15:33:00Z

Fdp18: /* Assessment */

'''<big>This is the optional experiment which may be chosen by any third year student. If you are looking for the compulsory simulation experiment for students studying chemistry with molecular physics, you will find it [[Third_year_CMP_compulsory_experiment|here]].</big>'''

 '''To run this experiment, you will need access to the Imperial College supercomputers. If you have VPN installed, you should be able to access the computers directly.'''

'''If you have not yet installed VPN, you may 1) use Remote Desktop connection (VPN not needed) or 2) install VPN.'''

Please follow the instructions to use one of these two options:

A) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/remotely-access-my-college-computer/remote-desktop-access-for-students/ Remote Desktop connection]

B) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/virtual-private-network-vpn/ using VPN]

==Introduction==

Computer simulation is widely used to study a huge variety of chemical phenomena, from the behaviour of exotic materials under extreme conditions to protein folding and the properties of biological systems such as lipid membranes. In this experiment, we hope to give you a gentle introduction to one of the most powerful methods for the simulation of chemical systems, '''molecular dynamics simulation'''.

This course closely follows some of the ideas introduced in Professor Bresme's statistical thermodynamics lecture course. Do not worry if you are doing this experiment before the lectures begin, everything that you need to know to complete the experiment will be explained in these instructions. We will begin with a brief overview of the fundamental theory behind the method, before you start running your own simulations of a simple liquid using the college's [https://dx.doi.org/10.14469/hpc/2232 high performance computing] facilities.

At the end of this experiment, you will have performed your own simulations using state of the art software packages used by researchers all around the world, and used those simulations to calculate both structural and dynamic properties of a simple liquid. You will have seen how the concepts of statistical physics introduced by Professor Bresme are needed to calculate thermodynamic quantities such as temperature and pressure in computer simulations, and you will see how computers can be used to validate those concepts.

All of the information that you need to complete the experiment is provided in these wiki pages. We have also tried to provide links to external resources and relevant textbooks where possible — unless explicitly stated, reading these resources '''is not required'''; they are provided only as further information for those interested in the subject.

==Assessment==

At the end of this experiment you must submit a report (pdf format via turnitin) and a zip file with inputs and outputs. The report should be structured:

* Introduction Questions (20% of the total mark)
* Results and Discussion (60% of total mark)
* Conclusion Questions (20% of total mark)
* References (see note below)

Relevant supplementary material can be added at the end of the report so long as it supports your discussion. Please limit the introduction and conclusion sections (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font). It is certainly possible to answer the introduction and conclusion questions with under this length limit, and as with most scientific writing, shorter concise but well thought out answers are preferable. We have not imposed a hard word limit because, equally, you should spend the majority of this lab learning, not editing and rewriting answers to just satisfy the word limit.

Please note that five "floating" marks have been reserved in the results and discussion section. These can be used to reward particularly insightful comments and/or explanations, on the basis of good/bad presentation, use of scientific language, or for other reasons.

It is important to provide appropriate references. 'Appropriate' refers to both quality (books and peer-reviewed papers are preferred over a link to wikipedia) and quantity (make sure to include references where you used them, but there is no point quoting 10 papers for a statement like 'macroscopic systems consist of a large number of atoms'). Please note that presenting external information as your own counts as plagiarism! As a bare minimum, you will need to reference the information you present in your introduction.

'''The final task of the lab (task 10 - MSD diffusion simulations) is NOT part of the lab this year.''' The section has been left on the wiki for any interested student to give it a go. However, it should NOT be submitted as part of your lab report, and will not be marked.

==Getting Help==

Please feel free to ask any demonstrators for help during the lab sessions - that is what we are here for. Questions can also be asked via email.

The member of academic staff responsible for this exercise is Professor Fernando Bresme (f.bresme@imperial.ac.uk).

==Structure of this Experiment==

This experimental manual has been broken up into a number of subsections. To help you plan your time it is suggested you complete the following at these times:

Monday (morning session): Theory - Introduction to molecular dynamics simulations

Monday (afternoon session): Theory + Equilibration (submit your files for running)

Tuesday (morning session): Equilibration (analyse your files)

Tuesday (afternoon session): Running simulations under specific conditions (submit and read)

Thursday (morning session): Checkpoint for progress. Submit your input files for the radial distribution function and analyse your equation of state from the previous section

Thursday (afternoon session).

Friday (morning session): Analyse MSD diffusion simulations

Friday (afternoon session): Analyse MSD diffusion simulations. Report write-up.

Direct links to each of them may be found below. You should attempt them in order, and you should complete all of them to finish the experiment.
# [[Third year simulation experiment/Files to download|Downloading the Files]]
# [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]]
# [[Third_year_simulation_experiment/Equilibration|Equilibration]]
# [[Third_year_simulation_experiment/Running_simulations_under_specific_conditions|Running simulations under specific conditions]]
# [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]]
# [[Third year simulation experiment/Dynamical properties and the diffusion coefficient|Dynamical properties and the diffusion coefficient]]

== Introduction Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# Give a brief account on what molecular dynamics is, and (broadly speaking) what it aims to achieve. [5]
# Give two examples of instances where molecular dynamics simulations have/can be used (in a helpful way) instead of or in synergy with experimental techniques. In each case explain why. [10] (''Please consider only chemically relevant examples. E.g. To study geochemical processes which occur under extreme temperature/pressure conditions, and therefore cannot be studied experimentally.'')
# Explain what is meant by a thermodynamic ensemble, and the term "conserved quantity". [2]
# What is the Ergodic hypothesis? Explain its relevance to molecular dynamics simulations. [3]

== Conclusion Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# In this lab, you have used the Lennard-Jones potential exclusively. Give one example of a system that can be described well using only LJ potentials, and one that cannot. In each case explain why. [4]
# What cut-off have you been using in your simulations? What are the advantages and disadvantages for using a shorter/longer cut-off? [3]
# What are finite size effects? Do you think they are significant in the simulations you have performed? Why? [3]
# Algorithms such as SHAKE and RATTLE (holonomic constraints) allow MD simulations to be performed while fixing bond lengths. Why is this desirable? [3] ''(Hint: think about the timestep.)''
# In the simulations you have performed ergodicity has not been an issue. Describe a system in which "brute force" MD struggles to achieve ergodic sampling. Describe one "enhanced sampling" technique that can be used to overcome this. [7]

Third year simulation experiment/Dynamical properties and the diffusion coefficient

2020-10-12T15:15:29Z

Fdp18: /* Simulations in this Section */

'''<big>This is the sixth (and final) section of the third year simulation experiment. You can return to the previous page, [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]], or go back to the [[Third_year_simulation_experiment|Introduction]].</big>'''

In this final section, we are going to make measurements to get some idea of how much the atoms in our simulation move around. We can characterise this by the diffusion coefficient, which we will calculate by two different approaches.

==Simulations in this Section==

'''<big>TASK 10: In the D subfolder, there is a file ''liq.in'' that will run a simulation at specified density and temperature to calculate the mean squared displacement and velocity autocorrelation function of your system. Run one of these simulations for a vapour, liquid, and solid. You have also been given some simulated data from much larger systems (approximately one million atoms). You will need these files later. make a plot for each of your simulations (solid, liquid, and gas), showing the mean squared displacement (the "total" MSD) as a function of timestep. Are these as you would expect? Estimate <math>D</math> in each case. Be careful with the units! Repeat this procedure for the MSD data that you were given from the one million atom simulations. Compare your data to the million atoms. [0 marks - voluntary] </big>'''

==Mean Squared Displacement==
The easiest way to measure <math>D</math> is by exploiting its connection to the [http://en.wikipedia.org/wiki/Mean_squared_displacement mean squared displacement].

<math>D = \frac{1}{6}\frac{\partial\left\langle r^2\left(t\right)\right\rangle}{\partial t}</math>

Note that in general, we expect the simulation to take a little time to establish this linear behaviour!

The simulations that you have already performed have recorded the MSD for you. Download the "optional output-2" file for each of the simulations, and give it a suitable name. These are text files which contain 5 columns: the first is the number of timesteps since the start of the simulation (*not* the elapsed time in reduced units), the next three contain the mean squared displacement for each of the Cartesian directions (x,y,z), and the final column contains the "total" mean squared displacement.

'''<big>This is the seventh (and final) section of the third year simulation experiment. You can return to the previous page, [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]], or go back to the [[Third_year_simulation_experiment|Introduction]].</big>'''

Third year simulation experiment

2020-10-03T10:02:46Z

Fdp18:

'''<big>This is the optional experiment which may be chosen by any third year student. If you are looking for the compulsory simulation experiment for students studying chemistry with molecular physics, you will find it [[Third_year_CMP_compulsory_experiment|here]].</big>'''

 '''To run this experiment, you will need access to the Imperial College supercomputers. If you have VPN installed, you should be able to access the computers directly.'''

'''If you have not yet installed VPN, you may 1) use Remote Desktop connection (VPN not needed) or 2) install VPN.'''

Please follow the instructions to use one of these two options:

A) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/remotely-access-my-college-computer/remote-desktop-access-for-students/ Remote Desktop connection]

B) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/virtual-private-network-vpn/ using VPN]

==Introduction==

Computer simulation is widely used to study a huge variety of chemical phenomena, from the behaviour of exotic materials under extreme conditions to protein folding and the properties of biological systems such as lipid membranes. In this experiment, we hope to give you a gentle introduction to one of the most powerful methods for the simulation of chemical systems, '''molecular dynamics simulation'''.

This course closely follows some of the ideas introduced in Professor Bresme's statistical thermodynamics lecture course. Do not worry if you are doing this experiment before the lectures begin, everything that you need to know to complete the experiment will be explained in these instructions. We will begin with a brief overview of the fundamental theory behind the method, before you start running your own simulations of a simple liquid using the college's [https://dx.doi.org/10.14469/hpc/2232 high performance computing] facilities.

At the end of this experiment, you will have performed your own simulations using state of the art software packages used by researchers all around the world, and used those simulations to calculate both structural and dynamic properties of a simple liquid. You will have seen how the concepts of statistical physics introduced by Professor Bresme are needed to calculate thermodynamic quantities such as temperature and pressure in computer simulations, and you will see how computers can be used to validate those concepts.

All of the information that you need to complete the experiment is provided in these wiki pages. We have also tried to provide links to external resources and relevant textbooks where possible — unless explicitly stated, reading these resources '''is not required'''; they are provided only as further information for those interested in the subject.

==Assessment==

At the end of this experiment you must submit a report (pdf format via turnitin) and a zip file with inputs and outputs. The report should be structured:

* Introduction Questions (20% of the total mark)
* Results and Discussion (60% of total mark)
* Conclusion Questions (20% of total mark)
* References

Relevant supplementary material can be added at the end of the report so long as it supports your discussion. Please limit the introduction and conclusion sections (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font). It is certainly possible to answer the introduction and conclusion questions with under this length limit, and as with most scientific writing, shorter concise but well thought out answers are preferable. We have not imposed a hard word limit because, equally, you should spend the majority of this lab learning, not editing and rewriting answers to just satisfy the word limit.

Please note that five "floating" marks have been reserved in the results and discussion section. These can be used to reward particularly insightful comments and/or explanations, on the basis of good/bad presentation, use of scientific language, or for other reasons.

The final task of the lab (task 10 - MSD diffusion simulations) is NOT part of the lab this year. The section has been left on the wiki for any interested student to give it a go. However, it should NOT be submitted as part of your lab report, and will not be marked.

==Getting Help==

Please feel free to ask any demonstrators for help during the lab sessions - that is what we are here for. Questions can also be asked via email.

The member of academic staff responsible for this exercise is Professor Fernando Bresme (f.bresme@imperial.ac.uk).

==Structure of this Experiment==

This experimental manual has been broken up into a number of subsections. To help you plan your time it is suggested you complete the following at these times:

Monday (morning session): Theory - Introduction to molecular dynamics simulations

Monday (afternoon session): Theory + Equilibration (submit your files for running)

Tuesday (morning session): Equilibration (analyse your files)

Tuesday (afternoon session): Running simulations under specific conditions (submit and read)

Thursday (morning session): Checkpoint for progress. Submit your input files for the radial distribution function and analyse your equation of state from the previous section

Thursday (afternoon session).

Friday (morning session): Analyse MSD diffusion simulations

Friday (afternoon session): Analyse MSD diffusion simulations. Report write-up.

Direct links to each of them may be found below. You should attempt them in order, and you should complete all of them to finish the experiment.
# [[Third year simulation experiment/Files to download|Downloading the Files]]
# [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]]
# [[Third_year_simulation_experiment/Equilibration|Equilibration]]
# [[Third_year_simulation_experiment/Running_simulations_under_specific_conditions|Running simulations under specific conditions]]
# [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]]
# [[Third year simulation experiment/Dynamical properties and the diffusion coefficient|Dynamical properties and the diffusion coefficient]]

== Introduction Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# Give a brief account on what molecular dynamics is, and (broadly speaking) what it aims to achieve. [5]
# Give two examples of instances where molecular dynamics simulations have/can be used (in a helpful way) instead of or in synergy with experimental techniques. In each case explain why. [10] (''Please consider only chemically relevant examples. E.g. To study geochemical processes which occur under extreme temperature/pressure conditions, and therefore cannot be studied experimentally.'')
# Explain what is meant by a thermodynamic ensemble, and the term "conserved quantity". [2]
# What is the Ergodic hypothesis? Explain its relevance to molecular dynamics simulations. [3]

== Conclusion Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# In this lab, you have used the Lennard-Jones potential exclusively. Give one example of a system that can be described well using only LJ potentials, and one that cannot. In each case explain why. [4]
# What cut-off have you been using in your simulations? What are the advantages and disadvantages for using a shorter/longer cut-off? [3]
# What are finite size effects? Do you think they are significant in the simulations you have performed? Why? [3]
# Algorithms such as SHAKE and RATTLE (holonomic constraints) allow MD simulations to be performed while fixing bond lengths. Why is this desirable? [3] ''(Hint: think about the timestep.)''
# In the simulations you have performed ergodicity has not been an issue. Describe a system in which "brute force" MD struggles to achieve ergodic sampling. Describe one "enhanced sampling" technique that can be used to overcome this. [7]

Third year simulation experiment

2020-10-03T10:01:23Z

Fdp18:

'''<big>This is the optional experiment which may be chosen by any third year student. If you are looking for the compulsory simulation experiment for students studying chemistry with molecular physics, you will find it [[Third_year_CMP_compulsory_experiment|here]].</big>'''

 To run this experiment, you will need access to the Imperial College supercomputers. If you have VPN installed, you should be able to access the computers directly.

If you have not VPN installed, you may 1) use Remote Desktop connection (VPN not needed) or 2) install VPN.

Please follow the instructions to use one of these two options:

A) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/remotely-access-my-college-computer/remote-desktop-access-for-students/ Remote Desktop connection]

B) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/virtual-private-network-vpn/ using VPN]

==Introduction==

Computer simulation is widely used to study a huge variety of chemical phenomena, from the behaviour of exotic materials under extreme conditions to protein folding and the properties of biological systems such as lipid membranes. In this experiment, we hope to give you a gentle introduction to one of the most powerful methods for the simulation of chemical systems, '''molecular dynamics simulation'''.

This course closely follows some of the ideas introduced in Professor Bresme's statistical thermodynamics lecture course. Do not worry if you are doing this experiment before the lectures begin, everything that you need to know to complete the experiment will be explained in these instructions. We will begin with a brief overview of the fundamental theory behind the method, before you start running your own simulations of a simple liquid using the college's [https://dx.doi.org/10.14469/hpc/2232 high performance computing] facilities.

At the end of this experiment, you will have performed your own simulations using state of the art software packages used by researchers all around the world, and used those simulations to calculate both structural and dynamic properties of a simple liquid. You will have seen how the concepts of statistical physics introduced by Professor Bresme are needed to calculate thermodynamic quantities such as temperature and pressure in computer simulations, and you will see how computers can be used to validate those concepts.

All of the information that you need to complete the experiment is provided in these wiki pages. We have also tried to provide links to external resources and relevant textbooks where possible — unless explicitly stated, reading these resources '''is not required'''; they are provided only as further information for those interested in the subject.

==Assessment==

At the end of this experiment you must submit a report (pdf format via turnitin) and a zip file with inputs and outputs. The report should be structured:

* Introduction Questions (20% of the total mark)
* Results and Discussion (60% of total mark)
* Conclusion Questions (20% of total mark)
* References

Relevant supplementary material can be added at the end of the report so long as it supports your discussion. Please limit the introduction and conclusion sections (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font). It is certainly possible to answer the introduction and conclusion questions with under this length limit, and as with most scientific writing, shorter concise but well thought out answers are preferable. We have not imposed a hard word limit because, equally, you should spend the majority of this lab learning, not editing and rewriting answers to just satisfy the word limit.

Please note that five "floating" marks have been reserved in the results and discussion section. These can be used to reward particularly insightful comments and/or explanations, on the basis of good/bad presentation, use of scientific language, or for other reasons.

The final task of the lab (task 10 - MSD diffusion simulations) is NOT part of the lab this year. The section has been left on the wiki for any interested student to give it a go. However, it should NOT be submitted as part of your lab report, and will not be marked.

==Getting Help==

Please feel free to ask any demonstrators for help during the lab sessions - that is what we are here for. Questions can also be asked via email.

The member of academic staff responsible for this exercise is Professor Fernando Bresme (f.bresme@imperial.ac.uk).

==Structure of this Experiment==

This experimental manual has been broken up into a number of subsections. To help you plan your time it is suggested you complete the following at these times:

Monday (morning session): Theory - Introduction to molecular dynamics simulations

Monday (afternoon session): Theory + Equilibration (submit your files for running)

Tuesday (morning session): Equilibration (analyse your files)

Tuesday (afternoon session): Running simulations under specific conditions (submit and read)

Thursday (morning session): Checkpoint for progress. Submit your input files for the radial distribution function and analyse your equation of state from the previous section

Thursday (afternoon session).

Friday (morning session): Analyse MSD diffusion simulations

Friday (afternoon session): Analyse MSD diffusion simulations. Report write-up.

Direct links to each of them may be found below. You should attempt them in order, and you should complete all of them to finish the experiment.
# [[Third year simulation experiment/Files to download|Downloading the Files]]
# [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]]
# [[Third_year_simulation_experiment/Equilibration|Equilibration]]
# [[Third_year_simulation_experiment/Running_simulations_under_specific_conditions|Running simulations under specific conditions]]
# [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]]
# [[Third year simulation experiment/Dynamical properties and the diffusion coefficient|Dynamical properties and the diffusion coefficient]]

== Introduction Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# Give a brief account on what molecular dynamics is, and (broadly speaking) what it aims to achieve. [5]
# Give two examples of instances where molecular dynamics simulations have/can be used (in a helpful way) instead of or in synergy with experimental techniques. In each case explain why. [10] (''Please consider only chemically relevant examples. E.g. To study geochemical processes which occur under extreme temperature/pressure conditions, and therefore cannot be studied experimentally.'')
# Explain what is meant by a thermodynamic ensemble, and the term "conserved quantity". [2]
# What is the Ergodic hypothesis? Explain its relevance to molecular dynamics simulations. [3]

== Conclusion Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# In this lab, you have used the Lennard-Jones potential exclusively. Give one example of a system that can be described well using only LJ potentials, and one that cannot. In each case explain why. [4]
# What cut-off have you been using in your simulations? What are the advantages and disadvantages for using a shorter/longer cut-off? [3]
# What are finite size effects? Do you think they are significant in the simulations you have performed? Why? [3]
# Algorithms such as SHAKE and RATTLE (holonomic constraints) allow MD simulations to be performed while fixing bond lengths. Why is this desirable? [3] ''(Hint: think about the timestep.)''
# In the simulations you have performed ergodicity has not been an issue. Describe a system in which "brute force" MD struggles to achieve ergodic sampling. Describe one "enhanced sampling" technique that can be used to overcome this. [7]

Third year simulation experiment/Equilibration

2020-10-03T09:58:06Z

Fdp18: /* Creating the simulation box */

'''<big>This is the third section of the third year simulation experiment. You can return to the previous page, [[Third_year_simulation_experiment/Introduction_to_molecular_dynamics_simulation|Introduction to molecular dynamics simulation]], or jump ahead to the next section, [[Third year simulation experiment/Running simulations under specific conditions|Running simulations under specific conditions]].</big>'''
 
We will be using the LAMMPS program to carry out our molecular dynamics simulations.
'''In several places in this section, we will ask you to consult the LAMMPS manual to find out things about how the software works. You can find the manual [https://lammps.sandia.gov/doc/Manual.html here].''' We appreciate that the format of this document can make it a little hard to navigate, but it is the definitive resource on how different commands in LAMMPS work, and is therefore invaluable. The files you will need for this section can be found in the intro folder downloaded previously.

===Creating the simulation box===
In the previous section, it was pointed out that before we can start a simulation, we need to know the initial states of all of the atoms in the system. Exactly what information we need about each atom depends on which method of numerical integration we need, but at the very least we need to specify the starting position of each atom. If we wanted to simulate a crystal, this information would be quite easy to come by — we could just look up the crystal structure, and use that to generate coordinates for however many unit cells we wanted. For this purpose, LAMMPS includes a command which generates crystal lattice structures.

Generating coordinates for atoms in a liquid is more difficult. There is no long range order, so we can't use a single point of reference to work out the positions of every other atom like we can in a solid. We could generate a random position for each atom. This would certainly create a disordered structure, but causes larger problems when we try to run the simulation.

Instead, we are going to place the atoms on the lattice points of a simple cubic lattice. This, of course, is not a situation in which the system is likely to be found physically. It turns out, though, that if we simulate for enough time we will find that the atoms rearrange themselves into more realistic configurations. We will discuss towards the end of this section exactly what is meant by "enough time"!

Consider the line in the input file <pre>lattice sc 0.8</pre> This command [https://lammps.sandia.gov/doc/lattice.html (further info)] creates a grid of points forming a simple cubic lattice (one lattice point per unit cell). The parameter <math>0.8</math> specifies the number density (number of lattice points per unit volume). In a corresponding output file, you will see the line <pre>Lattice spacing in x,y,z = 1.07722 1.07722 1.07722</pre> This indicates that the distance between the points of this lattice is <math>1.07722</math> (in reduced units, remember!).

The next lines in the input file are
<pre>
region box block 0 10 0 10 0 10
create_box 1 box
</pre>
The corresponding log file output is
<pre>
Created orthogonal box = (0 0 0) to (10.7722 10.7722 10.7722)
</pre>

The region command [https://lammps.sandia.gov/doc/region.html (further info)] simply defines a geometrical region in space, which we call "box". In this case, "box" is a cube extending ten lattice spacings from the origin in all three dimensions. The subsequent create_box command [https://lammps.sandia.gov/doc/create_box.html (further info)] tells LAMMPS to use the geometrical region called "box" as a template for the simulation box. The number 1 between "create_box" and "box" indicates that our simulation will contain only one type (species) of atom.

So far we have defined a simulation box which is based around a virtual simple cubic lattice. Our box contains 1000 (10x10x10) unit cells of this lattice, and so contains 1000 lattice points. We now need to fill our simulation box with atoms. The input command is <pre>create_atoms 1 box</pre> [https://lammps.sandia.gov/doc/create_atoms.html (further info)] while the log file simply contains an acknowledgement of this <pre>Created 1000 atoms</pre>

The create_atoms command has two arguments; the first tells LAMMPS that all of the atoms that we create will be of type 1. Every atom in the simulation has a type — because we will be simulating a pure fluid, containing only one chemical species, every atom will have the same type. The actual type that we assign to each atom is arbitrary — type 1 does not, for example, need to correspond to the element with atomic number 1 (hydrogen). If we wanted to simulate water, we might make the hydrogen atoms type 1 and the oxygen atoms type 2. We will specify the physical and chemical properties of each atom type later in the input script.

The remaining data in the log file isn't very instructive as it stands — it simply contains a list of the thermodynamic properties of the simulation at certain intervals. In a few sections time, we will plot this data, but for now you can close the log file. Keep the input script open.

===Setting the properties of the atoms===

In addition to their positions, we also need the physical properties of the atoms to be able to perform the simulation. We set these properties on a 'per-type' basis, so that every atom of the same type has the same mass and the same interactions.

So far we have created 1000 atoms, and we know the starting (<math>t = 0</math>) position for each of them. We have also set their masses, and told LAMMPS what sort of forces to calculate between them. The final thing we need to specify to completely specify the initial conditions is the velocity of each atom.

Choosing initial velocities for the atoms is a little easier than choosing initial positions. From the statistical thermodynamics lectures, you should know that, at equilibrium, the velocities of atoms in any system must be distributed according to the [http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution Maxwell-Boltzmann (MB) distribution]. If we know the masses of the atoms, and we know what temperature we want to simulate, then we can determine the relevant MB distribution function. LAMMPS is able to give every atom a random velocity whilst ensuring that overall the MB distribution is followed. This is the purpose of the line

<pre>
velocity all create 1.5 12345 dist gaussian rot yes mom yes
</pre>

You can see the manual page for this command [http://lammps.sandia.gov/doc/velocity.html here], but the key sections are:

* '''all''': the ''group'' of atoms on which the command acts. '''all''' simply specifies that we want every atom to have a velocity assigned to it.
* '''1.5''': the temperature, <math>T</math>, needed to calculate the MB distribution(in reduced units, as always)

===Monitoring thermodynamic properties===

We need to be sure that our simulation is correctly modelling whatever physical system we want to study. It is relatively easy to set up simulations, but how can we be sure that the "results" we get make sense? One of the best ways is to calculate from the simulation things that we can measure in experiment, and see if they agree. For example, we might want to simulate our system at a particular temperature and pressure, and measure the resulting density. If we repeat this over a range of temperatures at the same pressure, we will be able to plot an ''equation of state'', which we could compare to experimental measurements.

LAMMPS is able to calculate a great deal of thermodynamic information for us (you can see a full list of the properties it is able to calculate [http://lammps.sandia.gov/doc/thermo_style.html here]), but in these first simulations we are only interested in those properties specified in these commands:

<pre>
thermo_style custom time etotal temp press
thermo 10
</pre>

The first controls which properties will be printed out in the log file. In this case, we print how much time we have simulated so far (which is ''not'' the same as how long it has taken us to simulate it!), the total energy of the atoms, their temperature, and their pressure. The second line tells LAMMPS to print this information on every 10th timestep.

===Running the simulation===

'''Look at the lines below.'''
<pre>
### SPECIFY TIMESTEP ###
variable timestep equal 0.001
variable n_steps equal floor(100/${timestep})
timestep ${timestep}

### RUN SIMULATION ###
run ${n_steps}
</pre>
'''The second line (starting "variable timestep...") tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001.'''
 

'''<big> It is now time to run your first simulation, submit the input script with the data file in the intro folder of the files you have downloaded Try changing the timestep - what happens when you make the timestep larger?. </big>'''

===Visualising the trajectory===

The '''trajectory files''' contain the positions of all the atoms in the simulation, recorded at a set interval (for all of these simulations, this was every ten timesteps — this is controlled by the '''dump''' command in the input scripts). We use a programme called [http://www.ks.uiuc.edu/Research/vmd/ '''VMD'''] to view these trajectories, which you should find is already installed on both the desktop and laptop computers. You can run VMD from the start menu with '''Start''' -> '''All Programs''' -> '''University of Illinois''' -> '''VMD'''.

====Loading a Trajectory====

We'll start by looking at the output of the 0.02 timestep simulation. In the '''VMD Main''' window, select the menu option '''File''' -> '''New Molecule'''. Click the '''Browse''' button, then select the relevant trajectory file. In the '''Determine file type''' dropdown, select '''LAMMPS Trajectory'''. Then click '''Load'''.

You will see that the '''VMD 1.9.1 OpenGL Display''' window now shows a horrible mess. VMD's default behaviour is to draw lines between atoms which it thinks might be chemically bonded. Our system doesn't model chemical bonds, so we want to turn this off. In the '''VMD Main''' window, select the menu option '''Graphics''' -> '''Representations'''. This shows a list of "representations" of our atoms. You will see that at the moment, there is a single representation listed, and it is selected. It will have the ''Lines'' style, the ''Name'' colour, and the selection ''all''. "Selection" simply tells VMD which atoms we want it to draw. We want to show every atom, so the current selection is fine. The ''name'' colouring method just makes VMD give atoms colours according to their specified type. The colour isn't important to us, so we can leave this be too. The "style" tells VMD what we want it to display for each atom. Change the '''Drawing Method''' from ''Lines'' to ''VDW''. You will see that the mess of lines is replaced by a mess of low resolution, overlapping spheres. Change the '''Sphere Scale''' to 0.3, and the '''Sphere Resolution''' to 17. The result should look a little smoother. Close the '''Graphical Representations''' window. You will notice that in the bottom right of the '''VMD Main''' window, there is a small play button. Click this, and you will see the animated version of your simulation trajectory.

By clicking and dragging with the mouse, you can rotate the simulation box (though this may be sluggish). At any time, you can reset the view by pressing the equals key.

====Tracking a Single Particle====
To illustrate the periodic boundary conditions that we are using, we are going to draw almost all of the atoms as points, but we will pick a single atom at random to draw as a sphere. This will make it easy to see how a single atom moves through the box. Reset the display using the equals key, then use the '''VMD Main''' window controls to pause the trajectory and reset it to the first trajectory (play with the different buttons until you find the one that does this). You should see the perfect cubic lattice. Use the option '''Display''' -> '''Orthographic''' to change the drawing mode, then rotate the displayed crystal so that you are looking at one vertex (looking down the 111 direction, in crystallographic terms).

Open the '''Graphical Representations''' window again. Change the representation style from '''VDW''' to points, then click the '''Create Rep''' button. This creates a second representation, allowing a subset of the atoms to be drawn in a different way. The '''Selected Atoms''' box allows us to choose which atoms this representation applies to. We just want to pick two of them at random — VMD assigns every atom an index, from 0 to N-1. In our case, there are 1000 atoms, so choose two numbers between 0 and 999. Changed the '''Selected Atoms''' field to <pre>index i or index j</pre> where i and j are your chosen numbers, press return, then change the '''Drawing Method''' to '''VDW'''. You should now see only two atoms represented by spheres, with the rest shown as small points. In the '''VMD Main''' window, click play. Try rotating the box, and changing the playback speed.

You will see that sometimes one of the spheres seems to change position across the box very rapidly — this occurs when it reaches one periodic boundary, and is reflected back across the other face. Try playing with some of the other representation types in VMD — it is a very powerful package, which is often used to render images of simulated proteins, so many of its options aren't relevant to our simple system!

===Checking equilibration===

When we first set up a simulation, it is very important to make sure that our system reaches an equilibrium state. We characterise equilibrium by the average values of thermodynamic quantities becoming constant (due to the approximations that we have made, there will always be fluctuations, but the average values will become constant).

In this section, we are going to plot the thermodynamic output of the simulation to see how long it takes to reach the equilibrium state (and indeed, whether this happens at all). Instructions are given below to import data from the LAMMPS log file into Microsoft Excel. Once you have the data in a spreadsheet, you can plot it. If you know how to use some of the other plotting software available on the chemistry computers (like Origin), you are welcome to use it.

# Open a blank Excel workbook
# Copy the data in the textfile into the first cell
# With these data highlighted, click the Data tab and "Text to Columns"
# Click "Delimited", continue and let it be space delimited
# Click finish

'''<big>TASK 7: </big>'''

=== What does it mean for a simulation to "reach equilibrium"? Why is this important in terms of sampling from an ensemble using molecular dynamics? [2] ===

=== Plot the energy (potential, kinetic and total), temperature and pressure, against time for the 0,001 timestep experiment [2] ===

=== Does the simulation reach equilibrium? How can you tell? [2] ===

=== Make a single plot which shows the energy vs. time for the timesteps you have simulated [2]. ===

=== Of the timesteps that you used, which timestep will you use for subsequent simulations and why? [6] ===
''(Think about what is happening "physically" as you increase/decrease the timestep. Also, what features of each timeseries are indicative of the simulation's "health"?)''

'''<big>This is the third section of the third year simulation experiment. You can return to the previous page, [[Third_year_simulation_experiment/Introduction_to_molecular_dynamics_simulation|Introduction to molecular dynamics simulation]], or jump ahead to the next section, [[Third year simulation experiment/Running simulations under specific conditions|Running simulations under specific conditions]].</big>'''

Third year simulation experiment/Files to download

2020-10-03T09:54:20Z

Fdp18: /* Getting the files for the experiment */

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

All of the simulations that you run in this experiment are going to be performed on the Imperial [https://dx.doi.org/10.14469/hpc/2232 High Performance Computing] (HPC) systems. To do this, you must login to one of the HPC machines, add your simulation task to a queue of jobs waiting to run, and then wait for it to finish. For the small-scale simulations that we will perform in this experiment, the queue time should not be too long. However, you should expect to have to wait up to several hours for results to be available, particularly in the later stages!

In each section of the exercise, we have tried to provide a number of mathematical and/or research exercises that you should attempt while you are waiting for the simulations in that section to be completed. You can also use this time to write your report on the previous sections!

In this first section, we will teach you how to login to the HPC computers and submit an example simulation. While you wait for that example to complete, you can move on to the next section and read about the theory of molecular dynamics simulations

==Logging into the HPC Portal==

The simulations that you will perform in this experiment can in principle be performed on a desktop computer. However, these machines can only do a single simulation at a time, and are rather slow. Instead, we are going to run simulations on the college's supercomputer resources. This offers two large advantages. Firstly, even the longest simulations we will perform should take only a few hours. More importantly, you will be able to perform several simulations at the same time. In fact, to use the HPC service you must add your simulations to a queue of "jobs" that the supercomputer will perform. You can add as many jobs as you like to this queue, and simple come back to collect the results some time later.

To start, go to [http://portal.hpc.imperial.ac.uk the HPC portal login page]. You can login with your normal college credentials. We first need to create a "project" with which your simulations will be associated. Click '''Projects''' in the menu on the left hand side, then type a name in the "Add Project" box and click '''Add'''. You are free to choose whichever name you like, but a sensible choice might be something such as "ThirdYearSimulationExpt".

Leave the HPC Portal webpage open for now — we will return to it shortly.

==Getting the files for the experiment==

You can download all of the files that you will need for this experiment [https://github.com/niallj/ImperialChem-Year3SimExpt1415/archive/master.zip from this address]. This .zip archive contains a folder called '''ImperialChem-Year3SimExpt1415-master''', which you should extract to a location of your choice. It contains a number of subfolders — one for each section of the experiment. Every subsequent page of this lab manual will begin with a line telling you which folder contains the necessary files, like this one:

<big>'''THE FILES THAT YOU NEED FOR THIS SECTION ARE FOUND IN THE "Intro" SUBFOLDER'''.</big>

Have a look in the '''"Intro"''' folder now. It contains a file called '''melt_crystal.in''', which you should open with a text editor (like Notepad++ which you can find in Software Hub on your Desktop). This file is called an "input script", and it controls how the simulation software operates. We will perform all of our simulations with a software package called [http://lammps.sandia.gov LAMMPS]. Over the course of the experiment, you will learn what all of the commands in this file mean. The [https://lammps.sandia.gov/doc/Commands.html LAMMPS manual] contains a lot of valuable information about each of the commands, if you don't understand one of them or want to look up what some parameter means you can look it up there. To make life easier, we put "further info" links in the wiki. For now, we are going to use this file to run a few trial simulations.

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

Third year simulation experiment/Files to download

2020-10-03T09:53:41Z

Fdp18: /* Getting the files for the experiment */

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

All of the simulations that you run in this experiment are going to be performed on the Imperial [https://dx.doi.org/10.14469/hpc/2232 High Performance Computing] (HPC) systems. To do this, you must login to one of the HPC machines, add your simulation task to a queue of jobs waiting to run, and then wait for it to finish. For the small-scale simulations that we will perform in this experiment, the queue time should not be too long. However, you should expect to have to wait up to several hours for results to be available, particularly in the later stages!

In each section of the exercise, we have tried to provide a number of mathematical and/or research exercises that you should attempt while you are waiting for the simulations in that section to be completed. You can also use this time to write your report on the previous sections!

In this first section, we will teach you how to login to the HPC computers and submit an example simulation. While you wait for that example to complete, you can move on to the next section and read about the theory of molecular dynamics simulations

==Logging into the HPC Portal==

The simulations that you will perform in this experiment can in principle be performed on a desktop computer. However, these machines can only do a single simulation at a time, and are rather slow. Instead, we are going to run simulations on the college's supercomputer resources. This offers two large advantages. Firstly, even the longest simulations we will perform should take only a few hours. More importantly, you will be able to perform several simulations at the same time. In fact, to use the HPC service you must add your simulations to a queue of "jobs" that the supercomputer will perform. You can add as many jobs as you like to this queue, and simple come back to collect the results some time later.

To start, go to [http://portal.hpc.imperial.ac.uk the HPC portal login page]. You can login with your normal college credentials. We first need to create a "project" with which your simulations will be associated. Click '''Projects''' in the menu on the left hand side, then type a name in the "Add Project" box and click '''Add'''. You are free to choose whichever name you like, but a sensible choice might be something such as "ThirdYearSimulationExpt".

Leave the HPC Portal webpage open for now — we will return to it shortly.

==Getting the files for the experiment==

You can download all of the files that you will need for this experiment [https://github.com/niallj/ImperialChem-Year3SimExpt1415/archive/master.zip from this address]. This .zip archive contains a folder called '''ImperialChem-Year3SimExpt1415-master''', which you should extract to a location of your choice. It contains a number of subfolders — one for each section of the experiment. Every subsequent page of this lab manual will begin with a line telling you which folder contains the necessary files, like this one:

<big>'''THE FILES THAT YOU NEED FOR THIS SECTION ARE FOUND IN THE "Intro" SUBFOLDER'''.</big>

Have a look in the '''"Intro"''' folder now. It contains a file called '''melt_crystal.in''', which you should open with a text editor (like Notepad++ which you can find in Software Hub on your Desktop). This file is called an "input script", and it controls how the simulation software operates. We will perform all of our simulations with a software package called [http://lammps.sandia.gov LAMMPS]. Over the course of the experiment, you will learn what all of the commands in this file mean. For now, we are going to use this file to run a few trial simulations. The [https://lammps.sandia.gov/doc/Commands.html LAMMPS manual] contains a lot of valuable information about each of the commands, if you don't understand one of them or want to look up what some parameter means you can look it up there. To make life easier, we put "further info" links in the wiki.

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

Third year simulation experiment/Files to download

2020-10-03T09:49:11Z

Fdp18: /* Logging into the HPC Portal */

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

All of the simulations that you run in this experiment are going to be performed on the Imperial [https://dx.doi.org/10.14469/hpc/2232 High Performance Computing] (HPC) systems. To do this, you must login to one of the HPC machines, add your simulation task to a queue of jobs waiting to run, and then wait for it to finish. For the small-scale simulations that we will perform in this experiment, the queue time should not be too long. However, you should expect to have to wait up to several hours for results to be available, particularly in the later stages!

In each section of the exercise, we have tried to provide a number of mathematical and/or research exercises that you should attempt while you are waiting for the simulations in that section to be completed. You can also use this time to write your report on the previous sections!

In this first section, we will teach you how to login to the HPC computers and submit an example simulation. While you wait for that example to complete, you can move on to the next section and read about the theory of molecular dynamics simulations

==Logging into the HPC Portal==

The simulations that you will perform in this experiment can in principle be performed on a desktop computer. However, these machines can only do a single simulation at a time, and are rather slow. Instead, we are going to run simulations on the college's supercomputer resources. This offers two large advantages. Firstly, even the longest simulations we will perform should take only a few hours. More importantly, you will be able to perform several simulations at the same time. In fact, to use the HPC service you must add your simulations to a queue of "jobs" that the supercomputer will perform. You can add as many jobs as you like to this queue, and simple come back to collect the results some time later.

To start, go to [http://portal.hpc.imperial.ac.uk the HPC portal login page]. You can login with your normal college credentials. We first need to create a "project" with which your simulations will be associated. Click '''Projects''' in the menu on the left hand side, then type a name in the "Add Project" box and click '''Add'''. You are free to choose whichever name you like, but a sensible choice might be something such as "ThirdYearSimulationExpt".

Leave the HPC Portal webpage open for now — we will return to it shortly.

==Getting the files for the experiment==

You can download all of the files that you will need for this experiment [https://github.com/niallj/ImperialChem-Year3SimExpt1415/archive/master.zip from this address]. This .zip archive contains a folder called '''ImperialChem-Year3SimExpt1415-master''', which you should extract to a location of your choice. It contains a number of subfolders — one for each section of the experiment. Every subsequent page of this lab manual will begin with a line telling you which folder contains the necessary files, like this one:

<big>'''THE FILES THAT YOU NEED FOR THIS SECTION ARE FOUND IN THE "Intro" SUBFOLDER'''.</big>

Have a look in the '''"Intro"''' folder now. It contains a file called '''melt_crystal.in''', which you should open with a text editor (like Notepad++ which you can find in Software Hub on your Desktop). This file is called an "input script", and it controls how the simulation software operates. We will perform all of our simulations with a software package called [http://lammps.sandia.gov LAMMPS]. Over the course of the experiment, you will learn what all of the commands in this file mean. For now, we are going to use this file to run a few trial simulations.

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

Third year simulation experiment

2020-10-03T09:47:04Z

Fdp18:

'''<big>This is the optional experiment which may be chosen by any third year student. If you are looking for the compulsory simulation experiment for students studying chemistry with molecular physics, you will find it [[Third_year_CMP_compulsory_experiment|here]].</big>'''

To run this experiment, you will need access to the Imperial College supercomputers. If you have VPN installed, you should be able to access the computers directly.

If you have not VPN installed, you may 1) use Remote Desktop connection (VPN not needed) or 2) install VPN.

Please follow the instructions to use one of these two options:

A) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/remotely-access-my-college-computer/remote-desktop-access-for-students/ Remote Desktop connection]

B) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/virtual-private-network-vpn/ using VPN]

==Introduction==

Computer simulation is widely used to study a huge variety of chemical phenomena, from the behaviour of exotic materials under extreme conditions to protein folding and the properties of biological systems such as lipid membranes. In this experiment, we hope to give you a gentle introduction to one of the most powerful methods for the simulation of chemical systems, '''molecular dynamics simulation'''.

This course closely follows some of the ideas introduced in Professor Bresme's statistical thermodynamics lecture course. Do not worry if you are doing this experiment before the lectures begin, everything that you need to know to complete the experiment will be explained in these instructions. We will begin with a brief overview of the fundamental theory behind the method, before you start running your own simulations of a simple liquid using the college's [https://dx.doi.org/10.14469/hpc/2232 high performance computing] facilities.

At the end of this experiment, you will have performed your own simulations using state of the art software packages used by researchers all around the world, and used those simulations to calculate both structural and dynamic properties of a simple liquid. You will have seen how the concepts of statistical physics introduced by Professor Bresme are needed to calculate thermodynamic quantities such as temperature and pressure in computer simulations, and you will see how computers can be used to validate those concepts.

All of the information that you need to complete the experiment is provided in these wiki pages. We have also tried to provide links to external resources and relevant textbooks where possible — unless explicitly stated, reading these resources '''is not required'''; they are provided only as further information for those interested in the subject.

==Assessment==

At the end of this experiment you must submit a report (pdf format via turnitin) and a zip file with inputs and outputs. The report should be structured:

* Introduction Questions (20% of the total mark)
* Results and Discussion (60% of total mark)
* Conclusion Questions (20% of total mark)
* References

Relevant supplementary material can be added at the end of the report so long as it supports your discussion. Please limit the introduction and conclusion sections (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font). It is certainly possible to answer the introduction and conclusion questions with under this length limit, and as with most scientific writing, shorter concise but well thought out answers are preferable. We have not imposed a hard word limit because, equally, you should spend the majority of this lab learning, not editing and rewriting answers to just satisfy the word limit.

Please note that five "floating" marks have been reserved in the results and discussion section. These can be used to reward particularly insightful comments and/or explanations, on the basis of good/bad presentation, use of scientific language, or for other reasons.

The final task of the lab (task 10 - MSD diffusion simulations) is NOT part of the lab this year. The section has been left on the wiki for any interested student to give it a go. However, it should NOT be submitted as part of your lab report, and will not be marked.

==Getting Help==

Please feel free to ask any demonstrators for help during the lab sessions - that is what we are here for. Questions can also be asked via email.

The member of academic staff responsible for this exercise is Professor Fernando Bresme (f.bresme@imperial.ac.uk).

==Structure of this Experiment==

This experimental manual has been broken up into a number of subsections. To help you plan your time it is suggested you complete the following at these times:

Monday (morning session): Theory - Introduction to molecular dynamics simulations

Monday (afternoon session): Theory + Equilibration (submit your files for running)

Tuesday (morning session): Equilibration (analyse your files)

Tuesday (afternoon session): Running simulations under specific conditions (submit and read)

Thursday (morning session): Checkpoint for progress. Submit your input files for the radial distribution function and analyse your equation of state from the previous section

Thursday (afternoon session).

Friday (morning session): Analyse MSD diffusion simulations

Friday (afternoon session): Analyse MSD diffusion simulations. Report write-up.

Direct links to each of them may be found below. You should attempt them in order, and you should complete all of them to finish the experiment.
# [[Third year simulation experiment/Files to download|Downloading the Files]]
# [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]]
# [[Third_year_simulation_experiment/Equilibration|Equilibration]]
# [[Third_year_simulation_experiment/Running_simulations_under_specific_conditions|Running simulations under specific conditions]]
# [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]]
# [[Third year simulation experiment/Dynamical properties and the diffusion coefficient|Dynamical properties and the diffusion coefficient]]

== Introduction Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# Give a brief account on what molecular dynamics is, and (broadly speaking) what it aims to achieve. [5]
# Give two examples of instances where molecular dynamics simulations have/can be used (in a helpful way) instead of or in synergy with experimental techniques. In each case explain why. [10] (''Please consider only chemically relevant examples. E.g. To study geochemical processes which occur under extreme temperature/pressure conditions, and therefore cannot be studied experimentally.'')
# Explain what is meant by a thermodynamic ensemble, and the term "conserved quantity". [2]
# What is the Ergodic hypothesis? Explain its relevance to molecular dynamics simulations. [3]

== Conclusion Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# In this lab, you have used the Lennard-Jones potential exclusively. Give one example of a system that can be described well using only LJ potentials, and one that cannot. In each case explain why. [4]
# What cut-off have you been using in your simulations? What are the advantages and disadvantages for using a shorter/longer cut-off? [3]
# What are finite size effects? Do you think they are significant in the simulations you have performed? Why? [3]
# Algorithms such as SHAKE and RATTLE (holonomic constraints) allow MD simulations to be performed while fixing bond lengths. Why is this desirable? [3] ''(Hint: think about the timestep.)''
# In the simulations you have performed ergodicity has not been an issue. Describe a system in which "brute force" MD struggles to achieve ergodic sampling. Describe one "enhanced sampling" technique that can be used to overcome this. [7]

Third year simulation experiment

2020-10-03T09:46:51Z

Fdp18:

'''<big>This is the optional experiment which may be chosen by any third year student. If you are looking for the compulsory simulation experiment for students studying chemistry with molecular physics, you will find it [[Third_year_CMP_compulsory_experiment|here]].</big>'''

To run this experiment, you will need access to the Imperial College supercomputers. If you have VPN installed, you should be able to access the computers directly.

If you have not VPN installed, you may 1) use Remote Desktop connection (VPN not needed) or 2) install VPN.

Please follow the instructions to use one of these two options:

A) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/remotely-access-my-college-computer/remote-desktop-access-for-students/ Remote Desktop connection]

B) [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/virtual-private-network-vpn/
using VPN]

==Introduction==

Computer simulation is widely used to study a huge variety of chemical phenomena, from the behaviour of exotic materials under extreme conditions to protein folding and the properties of biological systems such as lipid membranes. In this experiment, we hope to give you a gentle introduction to one of the most powerful methods for the simulation of chemical systems, '''molecular dynamics simulation'''.

This course closely follows some of the ideas introduced in Professor Bresme's statistical thermodynamics lecture course. Do not worry if you are doing this experiment before the lectures begin, everything that you need to know to complete the experiment will be explained in these instructions. We will begin with a brief overview of the fundamental theory behind the method, before you start running your own simulations of a simple liquid using the college's [https://dx.doi.org/10.14469/hpc/2232 high performance computing] facilities.

At the end of this experiment, you will have performed your own simulations using state of the art software packages used by researchers all around the world, and used those simulations to calculate both structural and dynamic properties of a simple liquid. You will have seen how the concepts of statistical physics introduced by Professor Bresme are needed to calculate thermodynamic quantities such as temperature and pressure in computer simulations, and you will see how computers can be used to validate those concepts.

All of the information that you need to complete the experiment is provided in these wiki pages. We have also tried to provide links to external resources and relevant textbooks where possible — unless explicitly stated, reading these resources '''is not required'''; they are provided only as further information for those interested in the subject.

==Assessment==

At the end of this experiment you must submit a report (pdf format via turnitin) and a zip file with inputs and outputs. The report should be structured:

* Introduction Questions (20% of the total mark)
* Results and Discussion (60% of total mark)
* Conclusion Questions (20% of total mark)
* References

Relevant supplementary material can be added at the end of the report so long as it supports your discussion. Please limit the introduction and conclusion sections (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font). It is certainly possible to answer the introduction and conclusion questions with under this length limit, and as with most scientific writing, shorter concise but well thought out answers are preferable. We have not imposed a hard word limit because, equally, you should spend the majority of this lab learning, not editing and rewriting answers to just satisfy the word limit.

Please note that five "floating" marks have been reserved in the results and discussion section. These can be used to reward particularly insightful comments and/or explanations, on the basis of good/bad presentation, use of scientific language, or for other reasons.

The final task of the lab (task 10 - MSD diffusion simulations) is NOT part of the lab this year. The section has been left on the wiki for any interested student to give it a go. However, it should NOT be submitted as part of your lab report, and will not be marked.

==Getting Help==

Please feel free to ask any demonstrators for help during the lab sessions - that is what we are here for. Questions can also be asked via email.

The member of academic staff responsible for this exercise is Professor Fernando Bresme (f.bresme@imperial.ac.uk).

==Structure of this Experiment==

This experimental manual has been broken up into a number of subsections. To help you plan your time it is suggested you complete the following at these times:

Monday (morning session): Theory - Introduction to molecular dynamics simulations

Monday (afternoon session): Theory + Equilibration (submit your files for running)

Tuesday (morning session): Equilibration (analyse your files)

Tuesday (afternoon session): Running simulations under specific conditions (submit and read)

Thursday (morning session): Checkpoint for progress. Submit your input files for the radial distribution function and analyse your equation of state from the previous section

Thursday (afternoon session).

Friday (morning session): Analyse MSD diffusion simulations

Friday (afternoon session): Analyse MSD diffusion simulations. Report write-up.

Direct links to each of them may be found below. You should attempt them in order, and you should complete all of them to finish the experiment.
# [[Third year simulation experiment/Files to download|Downloading the Files]]
# [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]]
# [[Third_year_simulation_experiment/Equilibration|Equilibration]]
# [[Third_year_simulation_experiment/Running_simulations_under_specific_conditions|Running simulations under specific conditions]]
# [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]]
# [[Third year simulation experiment/Dynamical properties and the diffusion coefficient|Dynamical properties and the diffusion coefficient]]

== Introduction Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# Give a brief account on what molecular dynamics is, and (broadly speaking) what it aims to achieve. [5]
# Give two examples of instances where molecular dynamics simulations have/can be used (in a helpful way) instead of or in synergy with experimental techniques. In each case explain why. [10] (''Please consider only chemically relevant examples. E.g. To study geochemical processes which occur under extreme temperature/pressure conditions, and therefore cannot be studied experimentally.'')
# Explain what is meant by a thermodynamic ensemble, and the term "conserved quantity". [2]
# What is the Ergodic hypothesis? Explain its relevance to molecular dynamics simulations. [3]

== Conclusion Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# In this lab, you have used the Lennard-Jones potential exclusively. Give one example of a system that can be described well using only LJ potentials, and one that cannot. In each case explain why. [4]
# What cut-off have you been using in your simulations? What are the advantages and disadvantages for using a shorter/longer cut-off? [3]
# What are finite size effects? Do you think they are significant in the simulations you have performed? Why? [3]
# Algorithms such as SHAKE and RATTLE (holonomic constraints) allow MD simulations to be performed while fixing bond lengths. Why is this desirable? [3] ''(Hint: think about the timestep.)''
# In the simulations you have performed ergodicity has not been an issue. Describe a system in which "brute force" MD struggles to achieve ergodic sampling. Describe one "enhanced sampling" technique that can be used to overcome this. [7]

Third year simulation experiment/Equilibration

2020-10-03T08:08:01Z

Fdp18: /* What does it mean for a simulation to "reach equilibrium"? Why is this important in terms of sampling form an ensemble using molecular dynamics? [2] */

'''<big>This is the third section of the third year simulation experiment. You can return to the previous page, [[Third_year_simulation_experiment/Introduction_to_molecular_dynamics_simulation|Introduction to molecular dynamics simulation]], or jump ahead to the next section, [[Third year simulation experiment/Running simulations under specific conditions|Running simulations under specific conditions]].</big>'''
 
We will be using the LAMMPS program to carry out our molecular dynamics simulations.
'''In several places in this section, we will ask you to consult the LAMMPS manual to find out things about how the software works. You can find the manual [https://lammps.sandia.gov/doc/Manual.html here].''' We appreciate that the format of this document can make it a little hard to navigate, but it is the definitive resource on how different commands in LAMMPS work, and is therefore invaluable. The files you will need for this section can be found in the intro folder downloaded previously.

===Creating the simulation box===
In the previous section, it was pointed out that before we can start a simulation, we need to know the initial states of all of the atoms in the system. Exactly what information we need about each atom depends on which method of numerical integration we need, but at the very least we need to specify the starting position of each atom. If we wanted to simulate a crystal, this information would be quite easy to come by — we could just look up the crystal structure, and use that to generate coordinates for however many unit cells we wanted. For this purpose, LAMMPS includes a command which generates crystal lattice structures.

Generating coordinates for atoms in a liquid is more difficult. There is no long range order, so we can't use a single point of reference to work out the positions of every other atom like we can in a solid. We could generate a random position for each atom. This would certainly create a disordered structure, but causes larger problems when we try to run the simulation.

Instead, we are going to place the atoms on the lattice points of a simple cubic lattice. This, of course, is not a situation in which the system is likely to be found physically. It turns out, though, that if we simulate for enough time we will find that the atoms rearrange themselves into more realistic configurations. We will discuss towards the end of this section exactly what is meant by "enough time"!

Consider the line in the input file <pre>lattice sc 0.8</pre> This command creates a grid of points forming a simple cubic lattice (one lattice point per unit cell). The parameter <math>0.8</math> specifies the number density (number of lattice points per unit volume). In a corresponding output file, you will see the line <pre>Lattice spacing in x,y,z = 1.07722 1.07722 1.07722</pre> This indicates that the distance between the points of this lattice is <math>1.07722</math> (in reduced units, remember!).

The next lines in the input file are
<pre>
region box block 0 10 0 10 0 10
create_box 1 box
</pre>
The corresponding log file output is
<pre>
Created orthogonal box = (0 0 0) to (10.7722 10.7722 10.7722)
</pre>

The region command simply defines a geometrical region in space, which we call "box". In this case, "box" is a cube extending ten lattice spacings from the origin in all three dimensions. The subsequent create_box command tells LAMMPS to use the geometrical region called "box" as a template for the simulation box. The number 1 between "create_box" and "box" indicates that our simulation will contain only one type (species) of atom.

So far we have defined a simulation box which is based around a virtual simple cubic lattice. Our box contains 1000 (10x10x10) unit cells of this lattice, and so contains 1000 lattice points. We now need to fill our simulation box with atoms. The input command is <pre>create_atoms 1 box</pre> while the log file simply contains an acknowledgement of this <pre>Created 1000 atoms</pre>

The create_atoms command has two arguments; the first tells LAMMPS that all of the atoms that we create will be of type 1. Every atom in the simulation has a type — because we will be simulating a pure fluid, containing only one chemical species, every atom will have the same type. The actual type that we assign to each atom is arbitrary — type 1 does not, for example, need to correspond to the element with atomic number 1 (hydrogen). If we wanted to simulate water, we might make the hydrogen atoms type 1 and the oxygen atoms type 2. We will specify the physical and chemical properties of each atom type later in the input script.

The remaining data in the log file isn't very instructive as it stands — it simply contains a list of the thermodynamic properties of the simulation at certain intervals. In a few sections time, we will plot this data, but for now you can close the log file. Keep the input script open.

===Setting the properties of the atoms===

In addition to their positions, we also need the physical properties of the atoms to be able to perform the simulation. We set these properties on a 'per-type' basis, so that every atom of the same type has the same mass and the same interactions.

So far we have created 1000 atoms, and we know the starting (<math>t = 0</math>) position for each of them. We have also set their masses, and told LAMMPS what sort of forces to calculate between them. The final thing we need to specify to completely specify the initial conditions is the velocity of each atom.

Choosing initial velocities for the atoms is a little easier than choosing initial positions. From the statistical thermodynamics lectures, you should know that, at equilibrium, the velocities of atoms in any system must be distributed according to the [http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution Maxwell-Boltzmann (MB) distribution]. If we know the masses of the atoms, and we know what temperature we want to simulate, then we can determine the relevant MB distribution function. LAMMPS is able to give every atom a random velocity whilst ensuring that overall the MB distribution is followed. This is the purpose of the line

<pre>
velocity all create 1.5 12345 dist gaussian rot yes mom yes
</pre>

You can see the manual page for this command [http://lammps.sandia.gov/doc/velocity.html here], but the key sections are:

* '''all''': the ''group'' of atoms on which the command acts. '''all''' simply specifies that we want every atom to have a velocity assigned to it.
* '''1.5''': the temperature, <math>T</math>, needed to calculate the MB distribution(in reduced units, as always)

===Monitoring thermodynamic properties===

We need to be sure that our simulation is correctly modelling whatever physical system we want to study. It is relatively easy to set up simulations, but how can we be sure that the "results" we get make sense? One of the best ways is to calculate from the simulation things that we can measure in experiment, and see if they agree. For example, we might want to simulate our system at a particular temperature and pressure, and measure the resulting density. If we repeat this over a range of temperatures at the same pressure, we will be able to plot an ''equation of state'', which we could compare to experimental measurements.

LAMMPS is able to calculate a great deal of thermodynamic information for us (you can see a full list of the properties it is able to calculate [http://lammps.sandia.gov/doc/thermo_style.html here]), but in these first simulations we are only interested in those properties specified in these commands:

<pre>
thermo_style custom time etotal temp press
thermo 10
</pre>

The first controls which properties will be printed out in the log file. In this case, we print how much time we have simulated so far (which is ''not'' the same as how long it has taken us to simulate it!), the total energy of the atoms, their temperature, and their pressure. The second line tells LAMMPS to print this information on every 10th timestep.

===Running the simulation===

'''Look at the lines below.'''
<pre>
### SPECIFY TIMESTEP ###
variable timestep equal 0.001
variable n_steps equal floor(100/${timestep})
timestep ${timestep}

### RUN SIMULATION ###
run ${n_steps}
</pre>
'''The second line (starting "variable timestep...") tells LAMMPS that if it encounters the text ${timestep} on a subsequent line, it should replace it by the value given. In this case, the value ${timestep} is always replaced by 0.001.'''
 

'''<big> It is now time to run your first simulation, submit the input script with the data file in the intro folder of the files you have downloaded Try changing the timestep - what happens when you make the timestep larger?. </big>'''

===Visualising the trajectory===

The '''trajectory files''' contain the positions of all the atoms in the simulation, recorded at a set interval (for all of these simulations, this was every ten timesteps — this is controlled by the '''dump''' command in the input scripts). We use a programme called [http://www.ks.uiuc.edu/Research/vmd/ '''VMD'''] to view these trajectories, which you should find is already installed on both the desktop and laptop computers. You can run VMD from the start menu with '''Start''' -> '''All Programs''' -> '''University of Illinois''' -> '''VMD'''.

====Loading a Trajectory====

We'll start by looking at the output of the 0.02 timestep simulation. In the '''VMD Main''' window, select the menu option '''File''' -> '''New Molecule'''. Click the '''Browse''' button, then select the relevant trajectory file. In the '''Determine file type''' dropdown, select '''LAMMPS Trajectory'''. Then click '''Load'''.

You will see that the '''VMD 1.9.1 OpenGL Display''' window now shows a horrible mess. VMD's default behaviour is to draw lines between atoms which it thinks might be chemically bonded. Our system doesn't model chemical bonds, so we want to turn this off. In the '''VMD Main''' window, select the menu option '''Graphics''' -> '''Representations'''. This shows a list of "representations" of our atoms. You will see that at the moment, there is a single representation listed, and it is selected. It will have the ''Lines'' style, the ''Name'' colour, and the selection ''all''. "Selection" simply tells VMD which atoms we want it to draw. We want to show every atom, so the current selection is fine. The ''name'' colouring method just makes VMD give atoms colours according to their specified type. The colour isn't important to us, so we can leave this be too. The "style" tells VMD what we want it to display for each atom. Change the '''Drawing Method''' from ''Lines'' to ''VDW''. You will see that the mess of lines is replaced by a mess of low resolution, overlapping spheres. Change the '''Sphere Scale''' to 0.3, and the '''Sphere Resolution''' to 17. The result should look a little smoother. Close the '''Graphical Representations''' window. You will notice that in the bottom right of the '''VMD Main''' window, there is a small play button. Click this, and you will see the animated version of your simulation trajectory.

By clicking and dragging with the mouse, you can rotate the simulation box (though this may be sluggish). At any time, you can reset the view by pressing the equals key.

====Tracking a Single Particle====
To illustrate the periodic boundary conditions that we are using, we are going to draw almost all of the atoms as points, but we will pick a single atom at random to draw as a sphere. This will make it easy to see how a single atom moves through the box. Reset the display using the equals key, then use the '''VMD Main''' window controls to pause the trajectory and reset it to the first trajectory (play with the different buttons until you find the one that does this). You should see the perfect cubic lattice. Use the option '''Display''' -> '''Orthographic''' to change the drawing mode, then rotate the displayed crystal so that you are looking at one vertex (looking down the 111 direction, in crystallographic terms).

Open the '''Graphical Representations''' window again. Change the representation style from '''VDW''' to points, then click the '''Create Rep''' button. This creates a second representation, allowing a subset of the atoms to be drawn in a different way. The '''Selected Atoms''' box allows us to choose which atoms this representation applies to. We just want to pick two of them at random — VMD assigns every atom an index, from 0 to N-1. In our case, there are 1000 atoms, so choose two numbers between 0 and 999. Changed the '''Selected Atoms''' field to <pre>index i or index j</pre> where i and j are your chosen numbers, press return, then change the '''Drawing Method''' to '''VDW'''. You should now see only two atoms represented by spheres, with the rest shown as small points. In the '''VMD Main''' window, click play. Try rotating the box, and changing the playback speed.

You will see that sometimes one of the spheres seems to change position across the box very rapidly — this occurs when it reaches one periodic boundary, and is reflected back across the other face. Try playing with some of the other representation types in VMD — it is a very powerful package, which is often used to render images of simulated proteins, so many of its options aren't relevant to our simple system!

===Checking equilibration===

When we first set up a simulation, it is very important to make sure that our system reaches an equilibrium state. We characterise equilibrium by the average values of thermodynamic quantities becoming constant (due to the approximations that we have made, there will always be fluctuations, but the average values will become constant).

In this section, we are going to plot the thermodynamic output of the simulation to see how long it takes to reach the equilibrium state (and indeed, whether this happens at all). Instructions are given below to import data from the LAMMPS log file into Microsoft Excel. Once you have the data in a spreadsheet, you can plot it. If you know how to use some of the other plotting software available on the chemistry computers (like Origin), you are welcome to use it.

# Open a blank Excel workbook
# Copy the data in the textfile into the first cell
# With these data highlighted, click the Data tab and "Text to Columns"
# Click "Delimited", continue and let it be space delimited
# Click finish

'''<big>TASK 7: </big>'''

=== What does it mean for a simulation to "reach equilibrium"? Why is this important in terms of sampling from an ensemble using molecular dynamics? [2] ===

=== Plot the energy (potential, kinetic and total), temperature and pressure, against time for the 0,001 timestep experiment [2] ===

=== Does the simulation reach equilibrium? How can you tell? [2] ===

=== Make a single plot which shows the energy vs. time for the timesteps you have simulated [2]. ===

=== Of the timesteps that you used, which timestep will you use for subsequent simulations and why? [6] ===
''(Think about what is happening "physically" as you increase/decrease the timestep. Also, what features of each timeseries are indicative of the simulation's "health"?)''

'''<big>This is the third section of the third year simulation experiment. You can return to the previous page, [[Third_year_simulation_experiment/Introduction_to_molecular_dynamics_simulation|Introduction to molecular dynamics simulation]], or jump ahead to the next section, [[Third year simulation experiment/Running simulations under specific conditions|Running simulations under specific conditions]].</big>'''

Third year simulation experiment/Files to download

2020-10-03T08:00:38Z

Fdp18: /* Logging into the HPC Portal */

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

All of the simulations that you run in this experiment are going to be performed on the Imperial [https://dx.doi.org/10.14469/hpc/2232 High Performance Computing] (HPC) systems. To do this, you must login to one of the HPC machines, add your simulation task to a queue of jobs waiting to run, and then wait for it to finish. For the small-scale simulations that we will perform in this experiment, the queue time should not be too long. However, you should expect to have to wait up to several hours for results to be available, particularly in the later stages!

In each section of the exercise, we have tried to provide a number of mathematical and/or research exercises that you should attempt while you are waiting for the simulations in that section to be completed. You can also use this time to write your report on the previous sections!

In this first section, we will teach you how to login to the HPC computers and submit an example simulation. While you wait for that example to complete, you can move on to the next section and read about the theory of molecular dynamics simulations

==Logging into the HPC Portal==

The simulations that you will perform in this experiment can in principle be performed on a desktop computer. However, these machines can only do a single simulation at a time, and are rather slow. Instead, we are going to run simulations on the college's supercomputer resources. This offers two large advantages. Firstly, even the longest simulations we will perform should take only a few hours. More importantly, you will be able to perform several simulations at the same time. In fact, to use the HPC service you must add your simulations to a queue of "jobs" that the supercomputer will perform. You can add as many jobs as you like to this queue, and simple come back to collect the results some time later.

To start, go to [http://portal.hpc.imperial.ac.uk the HPC portal login page]. You can login with your normal college credentials. We first need to create a "project" with which your simulations will be associated. Click '''Projects''' in the menu on the left hand side, then type a name in the "Add Project" box and click '''Add'''. You are free to choose whichever name you like, but a sensible choice might be something such as "ThirdYearSimulationExpt".

For the HPC portal to work, you need to access it from the college network. The easiest way of doing this from home is [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/virtual-private-network-vpn/ to use a VPN]. From some locations the direct setup of VPN doesn't work, in which case it is easiest to use openVPN, as described under 'Alternative method':

Download and install [https://openvpn.net/client-connect-vpn-for-windows/ the latest OpenVPN client].
Download the [https://openvpn.ic.ac.uk/ic.ovpn OpenVPN configuration file].
Click the start button, type OpenVPN and click on the OpenVPN icon to open the client
Under Import Profile click File, browse to the location of the OpenVPN configuration file that was downloaded in step 2 and open it.
Enter your College username.
Tick Save Password and enter your College password
Click Add
Click on the Open VPN Profile to connect to the VPN

Leave the HPC Portal webpage open for now — we will return to it shortly.

==Getting the files for the experiment==

You can download all of the files that you will need for this experiment [https://github.com/niallj/ImperialChem-Year3SimExpt1415/archive/master.zip from this address]. This .zip archive contains a folder called '''ImperialChem-Year3SimExpt1415-master''', which you should extract to a location of your choice. It contains a number of subfolders — one for each section of the experiment. Every subsequent page of this lab manual will begin with a line telling you which folder contains the necessary files, like this one:

<big>'''THE FILES THAT YOU NEED FOR THIS SECTION ARE FOUND IN THE "Intro" SUBFOLDER'''.</big>

Have a look in the '''"Intro"''' folder now. It contains a file called '''melt_crystal.in''', which you should open with a text editor (like Notepad++ which you can find in Software Hub on your Desktop). This file is called an "input script", and it controls how the simulation software operates. We will perform all of our simulations with a software package called [http://lammps.sandia.gov LAMMPS]. Over the course of the experiment, you will learn what all of the commands in this file mean. For now, we are going to use this file to run a few trial simulations.

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

Third year simulation experiment/Files to download

2020-10-03T07:57:00Z

Fdp18: /* Logging into the HPC Portal */

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

All of the simulations that you run in this experiment are going to be performed on the Imperial [https://dx.doi.org/10.14469/hpc/2232 High Performance Computing] (HPC) systems. To do this, you must login to one of the HPC machines, add your simulation task to a queue of jobs waiting to run, and then wait for it to finish. For the small-scale simulations that we will perform in this experiment, the queue time should not be too long. However, you should expect to have to wait up to several hours for results to be available, particularly in the later stages!

In each section of the exercise, we have tried to provide a number of mathematical and/or research exercises that you should attempt while you are waiting for the simulations in that section to be completed. You can also use this time to write your report on the previous sections!

In this first section, we will teach you how to login to the HPC computers and submit an example simulation. While you wait for that example to complete, you can move on to the next section and read about the theory of molecular dynamics simulations

==Logging into the HPC Portal==

The simulations that you will perform in this experiment can in principle be performed on a desktop computer. However, these machines can only do a single simulation at a time, and are rather slow. Instead, we are going to run simulations on the college's supercomputer resources. This offers two large advantages. Firstly, even the longest simulations we will perform should take only a few hours. More importantly, you will be able to perform several simulations at the same time. In fact, to use the HPC service you must add your simulations to a queue of "jobs" that the supercomputer will perform. You can add as many jobs as you like to this queue, and simple come back to collect the results some time later.

To start, go to [http://portal.hpc.imperial.ac.uk the HPC portal login page]. You can login with your normal college credentials. We first need to create a "project" with which your simulations will be associated. Click '''Projects''' in the menu on the left hand side, then type a name in the "Add Project" box and click '''Add'''. You are free to choose whichever name you like, but a sensible choice might be something such as "ThirdYearSimulationExpt".

For the HPC portal to work, you need to access it from the college network. The easiest way of doing this from home is [https://www.imperial.ac.uk/admin-services/ict/self-service/connect-communicate/remote-access/virtual-private-network-vpn/ to use a VPN]. From some locations the direct setup of VPN doesn't work, in which case it is easiest to use openVPN.

Leave the HPC Portal webpage open for now — we will return to it shortly.

==Getting the files for the experiment==

You can download all of the files that you will need for this experiment [https://github.com/niallj/ImperialChem-Year3SimExpt1415/archive/master.zip from this address]. This .zip archive contains a folder called '''ImperialChem-Year3SimExpt1415-master''', which you should extract to a location of your choice. It contains a number of subfolders — one for each section of the experiment. Every subsequent page of this lab manual will begin with a line telling you which folder contains the necessary files, like this one:

<big>'''THE FILES THAT YOU NEED FOR THIS SECTION ARE FOUND IN THE "Intro" SUBFOLDER'''.</big>

Have a look in the '''"Intro"''' folder now. It contains a file called '''melt_crystal.in''', which you should open with a text editor (like Notepad++ which you can find in Software Hub on your Desktop). This file is called an "input script", and it controls how the simulation software operates. We will perform all of our simulations with a software package called [http://lammps.sandia.gov LAMMPS]. Over the course of the experiment, you will learn what all of the commands in this file mean. For now, we are going to use this file to run a few trial simulations.

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

Third year simulation experiment/Files to download

2020-10-03T07:53:59Z

Fdp18:

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

All of the simulations that you run in this experiment are going to be performed on the Imperial [https://dx.doi.org/10.14469/hpc/2232 High Performance Computing] (HPC) systems. To do this, you must login to one of the HPC machines, add your simulation task to a queue of jobs waiting to run, and then wait for it to finish. For the small-scale simulations that we will perform in this experiment, the queue time should not be too long. However, you should expect to have to wait up to several hours for results to be available, particularly in the later stages!

In each section of the exercise, we have tried to provide a number of mathematical and/or research exercises that you should attempt while you are waiting for the simulations in that section to be completed. You can also use this time to write your report on the previous sections!

In this first section, we will teach you how to login to the HPC computers and submit an example simulation. While you wait for that example to complete, you can move on to the next section and read about the theory of molecular dynamics simulations

==Logging into the HPC Portal==

The simulations that you will perform in this experiment can in principle be performed on a desktop computer. However, these machines can only do a single simulation at a time, and are rather slow. Instead, we are going to run simulations on the college's supercomputer resources. This offers two large advantages. Firstly, even the longest simulations we will perform should take only a few hours. More importantly, you will be able to perform several simulations at the same time. In fact, to use the HPC service you must add your simulations to a queue of "jobs" that the supercomputer will perform. You can add as many jobs as you like to this queue, and simple come back to collect the results some time later.

To start, go to [http://portal.hpc.imperial.ac.uk the HPC portal login page]. You can login with your normal college credentials. We first need to create a "project" with which your simulations will be associated. Click '''Projects''' in the menu on the left hand side, then type a name in the "Add Project" box and click '''Add'''. You are free to choose whichever name you like, but a sensible choice might be something such as "ThirdYearSimulationExpt".

Leave the HPC Portal webpage open for now — we will return to it shortly.

==Getting the files for the experiment==

You can download all of the files that you will need for this experiment [https://github.com/niallj/ImperialChem-Year3SimExpt1415/archive/master.zip from this address]. This .zip archive contains a folder called '''ImperialChem-Year3SimExpt1415-master''', which you should extract to a location of your choice. It contains a number of subfolders — one for each section of the experiment. Every subsequent page of this lab manual will begin with a line telling you which folder contains the necessary files, like this one:

<big>'''THE FILES THAT YOU NEED FOR THIS SECTION ARE FOUND IN THE "Intro" SUBFOLDER'''.</big>

Have a look in the '''"Intro"''' folder now. It contains a file called '''melt_crystal.in''', which you should open with a text editor (like Notepad++ which you can find in Software Hub on your Desktop). This file is called an "input script", and it controls how the simulation software operates. We will perform all of our simulations with a software package called [http://lammps.sandia.gov LAMMPS]. Over the course of the experiment, you will learn what all of the commands in this file mean. For now, we are going to use this file to run a few trial simulations.

'''<big>This is the first section of the third year simulation experiment. You can return to the introduction page, [[Third year simulation experiment]], or jump ahead to the next section, [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]].</big>'''

Third year simulation experiment

2020-10-03T07:53:32Z

Fdp18: /* Introduction */

'''<big>This is the optional experiment which may be chosen by any third year student. If you are looking for the compulsory simulation experiment for students studying chemistry with molecular physics, you will find it [[Third_year_CMP_compulsory_experiment|here]].</big>'''

==Introduction==

Computer simulation is widely used to study a huge variety of chemical phenomena, from the behaviour of exotic materials under extreme conditions to protein folding and the properties of biological systems such as lipid membranes. In this experiment, we hope to give you a gentle introduction to one of the most powerful methods for the simulation of chemical systems, '''molecular dynamics simulation'''.

This course closely follows some of the ideas introduced in Professor Bresme's statistical thermodynamics lecture course. Do not worry if you are doing this experiment before the lectures begin, everything that you need to know to complete the experiment will be explained in these instructions. We will begin with a brief overview of the fundamental theory behind the method, before you start running your own simulations of a simple liquid using the college's [https://dx.doi.org/10.14469/hpc/2232 high performance computing] facilities.

At the end of this experiment, you will have performed your own simulations using state of the art software packages used by researchers all around the world, and used those simulations to calculate both structural and dynamic properties of a simple liquid. You will have seen how the concepts of statistical physics introduced by Professor Bresme are needed to calculate thermodynamic quantities such as temperature and pressure in computer simulations, and you will see how computers can be used to validate those concepts.

All of the information that you need to complete the experiment is provided in these wiki pages. We have also tried to provide links to external resources and relevant textbooks where possible — unless explicitly stated, reading these resources '''is not required'''; they are provided only as further information for those interested in the subject.

==Assessment==

At the end of this experiment you must submit a report (pdf format via turnitin) and a zip file with inputs and outputs. The report should be structured:

* Introduction Questions (20% of the total mark)
* Results and Discussion (60% of total mark)
* Conclusion Questions (20% of total mark)
* References

Relevant supplementary material can be added at the end of the report so long as it supports your discussion. Please limit the introduction and conclusion sections (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font). It is certainly possible to answer the introduction and conclusion questions with under this length limit, and as with most scientific writing, shorter concise but well thought out answers are preferable. We have not imposed a hard word limit because, equally, you should spend the majority of this lab learning, not editing and rewriting answers to just satisfy the word limit.

Please note that five "floating" marks have been reserved in the results and discussion section. These can be used to reward particularly insightful comments and/or explanations, on the basis of good/bad presentation, use of scientific language, or for other reasons.

The final task of the lab (task 10 - MSD diffusion simulations) is NOT part of the lab this year. The section has been left on the wiki for any interested student to give it a go. However, it should NOT be submitted as part of your lab report, and will not be marked.

==Getting Help==

Please feel free to ask any demonstrators for help during the lab sessions - that is what we are here for. Questions can also be asked via email.

The member of academic staff responsible for this exercise is Professor Fernando Bresme (f.bresme@imperial.ac.uk).

==Structure of this Experiment==

This experimental manual has been broken up into a number of subsections. To help you plan your time it is suggested you complete the following at these times:

Monday (morning session): Theory - Introduction to molecular dynamics simulations

Monday (afternoon session): Theory + Equilibration (submit your files for running)

Tuesday (morning session): Equilibration (analyse your files)

Tuesday (afternoon session): Running simulations under specific conditions (submit and read)

Thursday (morning session): Checkpoint for progress. Submit your input files for the radial distribution function and analyse your equation of state from the previous section

Thursday (afternoon session).

Friday (morning session): Analyse MSD diffusion simulations

Friday (afternoon session): Analyse MSD diffusion simulations. Report write-up.

Direct links to each of them may be found below. You should attempt them in order, and you should complete all of them to finish the experiment.
# [[Third year simulation experiment/Files to download|Downloading the Files]]
# [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]]
# [[Third_year_simulation_experiment/Equilibration|Equilibration]]
# [[Third_year_simulation_experiment/Running_simulations_under_specific_conditions|Running simulations under specific conditions]]
# [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]]
# [[Third year simulation experiment/Dynamical properties and the diffusion coefficient|Dynamical properties and the diffusion coefficient]]

== Introduction Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# Give a brief account on what molecular dynamics is, and (broadly speaking) what it aims to achieve. [5]
# Give two examples of instances where molecular dynamics simulations have/can be used (in a helpful way) instead of or in synergy with experimental techniques. In each case explain why. [10] (''Please consider only chemically relevant examples. E.g. To study geochemical processes which occur under extreme temperature/pressure conditions, and therefore cannot be studied experimentally.'')
# Explain what is meant by a thermodynamic ensemble, and the term "conserved quantity". [2]
# What is the Ergodic hypothesis? Explain its relevance to molecular dynamics simulations. [3]

== Conclusion Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# In this lab, you have used the Lennard-Jones potential exclusively. Give one example of a system that can be described well using only LJ potentials, and one that cannot. In each case explain why. [4]
# What cut-off have you been using in your simulations? What are the advantages and disadvantages for using a shorter/longer cut-off? [3]
# What are finite size effects? Do you think they are significant in the simulations you have performed? Why? [3]
# Algorithms such as SHAKE and RATTLE (holonomic constraints) allow MD simulations to be performed while fixing bond lengths. Why is this desirable? [3] ''(Hint: think about the timestep.)''
# In the simulations you have performed ergodicity has not been an issue. Describe a system in which "brute force" MD struggles to achieve ergodic sampling. Describe one "enhanced sampling" technique that can be used to overcome this. [7]

Third year simulation experiment

2020-10-03T07:52:43Z

Fdp18: /* Introduction */

'''<big>This is the optional experiment which may be chosen by any third year student. If you are looking for the compulsory simulation experiment for students studying chemistry with molecular physics, you will find it [[Third_year_CMP_compulsory_experiment|here]].</big>'''

==Introduction==

Computer simulation is widely used to study a huge variety of chemical phenomena, from the behaviour of exotic materials under extreme conditions to protein folding and the properties of biological systems such as lipid membranes. In this experiment, we hope to give you a gentle introduction to one of the most powerful methods for the simulation of chemical systems, '''molecular dynamics simulation'''.

This course closely follows some of the ideas introduced in Professor Bresme's statistical thermodynamics lecture course. Do not worry if you are doing this experiment before the lectures begin, everything that you need to know to complete the experiment will be explained in these instructions. We will begin with a brief overview of the fundamental theory behind the method, before you start running your own simulations of a simple liquid using the college's [dx.doi.org/10.14469/hpc/2232/ high performance computing] facilities.

At the end of this experiment, you will have performed your own simulations using state of the art software packages used by researchers all around the world, and used those simulations to calculate both structural and dynamic properties of a simple liquid. You will have seen how the concepts of statistical physics introduced by Professor Bresme are needed to calculate thermodynamic quantities such as temperature and pressure in computer simulations, and you will see how computers can be used to validate those concepts.

All of the information that you need to complete the experiment is provided in these wiki pages. We have also tried to provide links to external resources and relevant textbooks where possible — unless explicitly stated, reading these resources '''is not required'''; they are provided only as further information for those interested in the subject.

==Assessment==

At the end of this experiment you must submit a report (pdf format via turnitin) and a zip file with inputs and outputs. The report should be structured:

* Introduction Questions (20% of the total mark)
* Results and Discussion (60% of total mark)
* Conclusion Questions (20% of total mark)
* References

Relevant supplementary material can be added at the end of the report so long as it supports your discussion. Please limit the introduction and conclusion sections (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font). It is certainly possible to answer the introduction and conclusion questions with under this length limit, and as with most scientific writing, shorter concise but well thought out answers are preferable. We have not imposed a hard word limit because, equally, you should spend the majority of this lab learning, not editing and rewriting answers to just satisfy the word limit.

Please note that five "floating" marks have been reserved in the results and discussion section. These can be used to reward particularly insightful comments and/or explanations, on the basis of good/bad presentation, use of scientific language, or for other reasons.

The final task of the lab (task 10 - MSD diffusion simulations) is NOT part of the lab this year. The section has been left on the wiki for any interested student to give it a go. However, it should NOT be submitted as part of your lab report, and will not be marked.

==Getting Help==

Please feel free to ask any demonstrators for help during the lab sessions - that is what we are here for. Questions can also be asked via email.

The member of academic staff responsible for this exercise is Professor Fernando Bresme (f.bresme@imperial.ac.uk).

==Structure of this Experiment==

This experimental manual has been broken up into a number of subsections. To help you plan your time it is suggested you complete the following at these times:

Monday (morning session): Theory - Introduction to molecular dynamics simulations

Monday (afternoon session): Theory + Equilibration (submit your files for running)

Tuesday (morning session): Equilibration (analyse your files)

Tuesday (afternoon session): Running simulations under specific conditions (submit and read)

Thursday (morning session): Checkpoint for progress. Submit your input files for the radial distribution function and analyse your equation of state from the previous section

Thursday (afternoon session).

Friday (morning session): Analyse MSD diffusion simulations

Friday (afternoon session): Analyse MSD diffusion simulations. Report write-up.

Direct links to each of them may be found below. You should attempt them in order, and you should complete all of them to finish the experiment.
# [[Third year simulation experiment/Files to download|Downloading the Files]]
# [[Third year simulation experiment/Introduction to molecular dynamics simulation|Introduction to molecular dynamics simulation]]
# [[Third_year_simulation_experiment/Equilibration|Equilibration]]
# [[Third_year_simulation_experiment/Running_simulations_under_specific_conditions|Running simulations under specific conditions]]
# [[Third year simulation experiment/Structural properties and the radial distribution function|Structural properties and the radial distribution function]]
# [[Third year simulation experiment/Dynamical properties and the diffusion coefficient|Dynamical properties and the diffusion coefficient]]

== Introduction Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# Give a brief account on what molecular dynamics is, and (broadly speaking) what it aims to achieve. [5]
# Give two examples of instances where molecular dynamics simulations have/can be used (in a helpful way) instead of or in synergy with experimental techniques. In each case explain why. [10] (''Please consider only chemically relevant examples. E.g. To study geochemical processes which occur under extreme temperature/pressure conditions, and therefore cannot be studied experimentally.'')
# Explain what is meant by a thermodynamic ensemble, and the term "conserved quantity". [2]
# What is the Ergodic hypothesis? Explain its relevance to molecular dynamics simulations. [3]

== Conclusion Questions ==
Please limit your answers to this section (which are now just answers to questions) to a maximum of ~1.5 pages at size 12 font Times New Roman, or equivalent (e.g. if you are using latex or simply prefer another font).
# In this lab, you have used the Lennard-Jones potential exclusively. Give one example of a system that can be described well using only LJ potentials, and one that cannot. In each case explain why. [4]
# What cut-off have you been using in your simulations? What are the advantages and disadvantages for using a shorter/longer cut-off? [3]
# What are finite size effects? Do you think they are significant in the simulations you have performed? Why? [3]
# Algorithms such as SHAKE and RATTLE (holonomic constraints) allow MD simulations to be performed while fixing bond lengths. Why is this desirable? [3] ''(Hint: think about the timestep.)''
# In the simulations you have performed ergodicity has not been an issue. Describe a system in which "brute force" MD struggles to achieve ergodic sampling. Describe one "enhanced sampling" technique that can be used to overcome this. [7]

MRD:mb7418

2020-05-30T07:53:03Z

Fdp18: /* EXERCISE 1: H + H2 system */

== EXERCISE 1: H + H2 system ==

{{fontcolor1|purple|To me it looks like you understood what you are doing, although you have problems expressing yourself in a clear way sometimes. for example, "The transition state is the minimum energy required for the reactants to convert into the products" is wrong - that would be the activation energy, not the transition state. (What is the difference? look at the absolute energies...) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:50, 30 May 2020 (BST)}}

===Dynamics from the transition state region===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? {{fontcolor1|red| This text is repeated below which makes this section difficult to read. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} Potential energy surface diagrams show the potential energy the system has according to the coordinates of the atoms in the system.<ref>Chemistry LibreTexts. 2020. ''2.6: Potential Energy Surfaces''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/2%3A_Chemical_Kinetics/2.06%3A_Potential_Energy_Surfaces</nowiki>> [Accessed 19 May 2020].</ref> Figure 1 shows a potential energy surface diagram representing the reaction, equation 1: A + BC → AB + C.
{| class="wikitable"
|+Table 1. Molecules distances (r) and momenta (p)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|230
|<nowiki>-5.2</nowiki>
|-
|r1=r(BC)
|74
|0
|}
In this case r(1)=r(BC) and r(2)=r(AB). BC is a hydrogen molecule consisting of two hydrogen atoms, B and C. Atom A is a hydrogen atom which has energy equal to or more than the activation energy, thus reaching the transition state. The transition state structure consists of atom A forming a bond with molecule BC while the BC bond is slightly dissociating. After the transition state, the bond between AB would form making a new molecule and the BC bond would be fully dissociated.
* Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
[[File:PE with TS.png|left|thumb|[[Figure]] 1. Potential energy surface diagram with transition state labelled ]]
In a potential energy surface diagram, the transition state is mathematically defined as the first-order saddle point<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref>:

∂V(r1)/∂(r1)=0 and ∂V(r2)/∂r2=0

∂2V(r)/∂(r1)2 > 0 and ∂2V(r)/∂(r2)2 < 0

[∂2V(r)/∂(r1)2]*[ ∂2V(r)/∂(r2)2] - [ ∂2V/∂(r1)(r2)]2 < 0

Hessian matrix is shown in Table 2, resulting in DET = -0.999698 < 0
{| class="wikitable"
|+Table 1. Hessian matrix for the transition state
!
!f(x)
!f(y)
|-
|f(x)
|<nowiki>+0.707</nowiki>
|<nowiki>-0.707</nowiki>
|-
|f(y)
|<nowiki>-0.707</nowiki>
|<nowiki>-0.707</nowiki>
|}
In Figure 1, the black wavy lines shows that the molecule is vibrating. This is the minimum energy path required for the reactants to react and turn into the products via the transition state.<ref>Atkins, de Paula, Keeler: Physical Chemistry, 11th Edition. Section 18D.3 Potential Energy Surfaces (PESs)</ref> This suggests that the transition state is the peak in the minimum energy path. The transition state is the minimum energy required for the reactants to convert into the products. Hence once the transition state is reached, if the reaction was successful the products would be formed. Figure 1 shows the location of the transition state in the potential energy diagram.

The transition state can be distinguished from a local minimum of the potential energy surface because it is on the diagonal axis. {{fontcolor1|red| Because it is also a maximum (minimum AND maximum in orthogonal directions.. i.e a saddle point) therefore distinguishable from a local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} It is the saddle point.This is when everything in the system is stationary when there is no momentum.

===Locating the transition state r1=r2===
* Q2. Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

<nowiki> </nowiki>When the transition state is formed, due to all the atoms being hydrogen, this suggest that the transition state is symmetrical. This means that the bond distance between A and B is the same as the bond distance of B and C. This can be used to locate the transition state when the momentum is 0. Table 3 shows an estimate of the transition state position (rts).
{| class="wikitable"
|+Table 3. Distances and momenta to locate the transition state
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|90.75
|0
|}
[[File:Mb7418 dis vs time ts.png|thumb|[[Figure 2]]. Distance vs time plot of the transition state ]]
Figure 2 shows an Intermolecular distance vs time graph resulted from the initial conditions from Table 2. The line is flat, this suggest that everything in the system is stationary. Whereas, if it wasn't the transition state, in the intermolecular distance vs time graph you'd see oscillations representing the vibrational movements. The more oscillations would indicate the more vibrational movements. Hence the transition state is a flat line as there isn't any. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:33, 29 May 2020 (BST)}}
[[File:Mb7418 ts contour.png|thumb|[[Figure 3]]. Potential energy diagram as a contour plot of the transition state]]
The contour plot shown in Figure 3 shows the transition state as a red cross. This is the saddle point of the potential surface. Whenever r1 and r2 are the same, the point would be anywhere across the diagonal axis. This contour plot helped estimate the transition state position as the trajectory shows that the AB and BC bond lengths are the same. This is the point at which reactants have to overcome to react into the products.

The minimum energy path can differ depending on the relative momentum of the atoms/molecules and their vibrational excitation energy. The transition state of the reaction won't change its position. The reactants require sufficient translational kinetic energy and vibrational energy to overcome the transition state (saddle point) in order to react and turn into the products.

===Trajectories from r1 = rts+δ, r2 = rts ===
* Q3. Comment on how the ''mep'' and the trajectory you just calculated differ. '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''' and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1
The trajectory of the minimum energy path can be visualised slightly near the transition state, this is when atom A is approaching molecule BC to form a bond with it, A + BC.
{| class="wikitable"
|+Table 4. Initial conditions slightly displaced from transition state
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|91.75
|0
|}
The initial conditions shown in Table 4 gives the potential energy diagram in Figure 4 and 5 when the calculation type is MEP. The minimum energy path shown by the black line isn't wavy. [[File:Mb7418 mep.png|left|thumb|[[Figure 4]]. MEP calculation diagram]]
[[File:Mb7418 mep 1.png|thumb|234x234px|[[Figure 5]]. MEP calculation diagram bottom view]]However, when the same initial conditions are applied when the calculation type is Dynamic, the minimum energy path is wavy shown in Figure 6. This shows that the MEP calculation gives a different result of the reaction trajectory. The molecule is constantly moving at a velocity since there is no external force acting upon the reactants. This is called inertia motion.<ref>Evans, M.G. and Polanyi, M., 1938. Inertia and driving force of chemical reactions. ''Transactions of the Faraday Society'', ''34'', pp.11-24.</ref> Whereas the potential energy diagram produced from the Dynamics calculation takes into account the reactants inertia by representing the pathway as non-linear. Additionally, the MEP calculation diagram has a smaller trajectory than the Dynamics calculation.The reason for this is beacause the MEP calculation sets the momentum to zero at every step whereas the dynamic calculation the momenta is developed classically<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., 8.3.1, Prentice-Hall, 1998.</ref>. The MEP calculation only takes the gradient into account. The amount of steps there are corresponds to the number of times it is calculating what the gradient is at that point. The only thing relative to the MEP calculation is the gradient. Whereas with dynamics, it takes into account how long it takes for the gradient to change. The momentum (velocity) is changing each time instead of setting it to zero every step. Hence it's longer as after the products are formed, they are still have motion such as vibrational and kinetic. {{fontcolor1|red| Yes. This does show an understanding of MEP vs Dynamics. Just to clarify - the amount of steps is the number of times the system moves during the simulation - this was a bit confused in your response. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:36, 29 May 2020 (BST)}} [[File:Mb7418 dyn 1.png|thumb|[[Figure 6]]. Dynamics calculation bottom view]]

== Reactive and unreactive trajectories ==

===Interpreting trajectories===

* Q4. For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.

{| class="wikitable"
|+Table 5. Reactive and unreactive trajectories
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Reactive
|The starting point (represented by the red cross) starts at 200 pm between atoms A and B. Then the distance AB decreases. Atoms A and B are approaching each other and once it reaches the transition state, the AB distance and BC distance are the same and all the atoms are stationary. Then the AB bond is formed and atom C repels from atom B. The molecule AB is constantly vibrating. This could suggest that some translational kinetic energy got converted into vibrational energy after the transition state. Hence after the trajectory has oscillations after it passes the transition state.
|[[File:Mb7418 6.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The trajectory has some oscillations, showing that the reactants are constantly vibrating. Hence it has vibrational energy. However, it doesn't go over the transition state so it doesn't react into the products. This suggests that the reactants approach each other but then they repel. While they do that they oscillate as they get closer. It ends up with a high AB distance (essentially the starting point of the reaction), this shows that the reactants haven't reacted due to the AB molecule not forming. This could be due to not having enough translational kinetic energy.
|[[File:Mb7418 2.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The starting point starts at 200 pm between atoms A and B. Then the the atoms got closer and reaches the transition state. This is when the distance AB and BC are equal. Then the AB distance remains continuous while the BC distance begins to increase. The BC bond breaks and the AB bond forms. It only crosses the transition state once. It ends up in the products as the distance AB is short (molecule containing atoms A and B bonding) and BC is distance is long (reactant molecule BC bond has been broken).
|[[File:Mb7418 3.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|The whole trajectory shows that it's unreactive because the starting point and ending point are similar. It started at a high AB distance value and also ended at a high AB distance value. This suggests this is end unreactive.
At the starting point, AB are far apart. Then they begin approaching each other and reach the transition state. However, they repel from each other and then start approaching each other again. This happens again and then the distance AB begins to increase again. The BC bond is formed again, resulting in the reactant molecule BC at the end.
|[[File:Mb7418 4.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|This crosses the transition point more than once. It's reactive as it's end point is at a high BC distance and low AB distance, suggesting that the BC bond broke and the AB bond has formed resulting in AB + B.

From the starting point, the AB distance starts to decrease. This suggests atom A and B approach each other, but then the AB distance begins to increase again so the A and B atoms repel. There is a moment when the distance between atoms A and B is the same as the distance between B and C (transition state) Then the B and C atoms approach each other and then repel B continuously. While atoms A and B get closer forming a bond. The molecule AB vibrates while atom C continuously repels away from it.
|[[File:Mb7418 5.png|centre|thumb|200x200px]]
|}

{{fontcolor1|red| OK. What do you conclude from the table? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:40, 29 May 2020 (BST)}}

=== Transition state theory ===
* Q5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
The three assumptions of the transition state theory<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Chapter 10 pg. 20, Prentice-Hall, 1998</ref> are:

1) The transition state will only be crossed over once. Once the transition state it reached, the products will be formed.

2) During the transition state, any motion along the reaction coordinate can be separate from the other motions and treated classically as a translation. The concentration of the transition state is in equilibrium with the reactants. (concentration constant relating them both, there's a Maxwell-Boltzmann distribution between the reactants and the transition state at any given time) This is defined as the Quasi equilibrium. It's assumed that the kinetic energy is predicted by the Boltzmann distribution. The transition states that are turning into products are distributed among their states according to the Maxwell-Boltzmann Laws<ref>Rowlinson*, J.S., 2005. The Maxwell–Boltzmann distribution. ''Molecular Physics'', ''103''(21-23), pp.2821-2828.</ref> even in absence of an equilibrium between the reactant and product molecules.

'''3)''' There's minimum effects of quantum tunneling.

The quantum tunneling effect<ref>Abyss.uoregon.edu. 2020. ''Quantum Tunneling''. [online] Available at: <<nowiki>http://abyss.uoregon.edu/~js/glossary/quantum_tunneling.html</nowiki>> [Accessed 22 May 2020].</ref> is when the reactants can overcome the barrier by moving across it to turn into the products. The transition state theory doesn't include quantum tunneling. If quantum tunneling was included, it would increase the rate of reaction as more reactants will turn into products. There are only classical motions in the transition state theory. Hence it is suggested that quantum tunneling isn't included in the model used in Table 5 because it's assumed there's a faster alternative route used.

Additionally, in the Table 5 model the transition state theory wasn't true because the transition state was crossed over more than once in the last two reactions in Table 5, this could decrease the rate of reaction due to the products being able to cross the transition state again and go back into the reactants. Therefore, the transition state theory is an over-estimate in comparison to the model used. {{fontcolor1|red| Yes. Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:41, 29 May 2020 (BST)}}

== EXERCISE 2: F - H - H system ==

=== PES inspection ===
* Q6. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?
{| class="wikitable"
|+Table 6. Initial conditions for Figure 7 (F + H2 → HF + H)
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|70
|0
|}

[[File:Mb7418 F + H2 exo edit CORRECT.png|left|thumb|[[Figure 7]]. F + H2 -> HF + H exothermic potential surface diagram]]
In the following equations, A = atom F, B = atom H, C = atom H.

The initial conditions in Table 6 gives the potential energy diagram for Figure 7. The reason for this is because the F and H bond is very far apart, suggesting that it's not bonded whereas the hydrogen atoms are closer together, suggesting a F atom and H2 molecule are in the initial conditions.

Equation 2: F + H2 → HF + H is an exothermic reaction<ref name=":0">Duff, J.W. and Truhlar, D.G., 1975. Effect of curvature of the reaction path on dynamic effects in endothermic chemical reactions and product energies in exothermic reactions. ''The Journal of Chemical Physics'', ''62''(6), pp.2477-2491.</ref> shown in Figure 7. The energy of the reactants is higher than the energy of the products. This shows that this is an exothermic reaction due to the process releasing energy. This suggests that the bond strength of H2 is weaker than the bond strength of HF as no additional energy was required to break its bond. This reaction has an early transition state<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref> due to the the fluorine atom being highly electro-negative so it attacks the hydrogen molecule very quickly.
{| class="wikitable"
|+Table 7. Initial conditions for Figure 8 (HF + H → F + H2)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|70
|0
|-
|r(BC)
|500
|0
|}
[[File:Mb7418 NEW CORRECT ENDO.png|left|thumb|[[Figure 8]]. HF + H -> F + H2 endothermic potential surface diagram]]
The initial conditions in Table 7 gives the potential energy diagram for Figure 8. The reason for this is because the F and H atoms are very close together compared to both the hydrogens. This suggests that a HF molecule and one H atom are in the initial conditions.

Equation 3: HF + H → F + H2 is an endothermic reaction<ref name=":0" /> shown in Figure 8. The energy of the reactants is lower than the energy of the products. This suggests that this reaction is endothermic as the process used additional energy to break the H-F bond to form the H-H bond. Therefore, in the potential energy surface diagram there will be a lot of vibrational energy before the transition state compared to after. This also suggest that the HF bond is stronger than the HH bond as it requires additional energy to break its bond. This will give a late transition state.The translational energy could get converted into vibrational energy in order to break the bond between the strong H-F bond<ref name=":1">Polanyi, J.C. and Tardy, D., 1969. Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF+ H. ''The Journal of Chemical Physics'', ''51''(12), pp.5717-5719.</ref>. This favours the endothermic reaction because the reactants position will be aligned with the direction of the minimum energy path.''''' '''''

* Q7. Locate the approximate position of the transition state.
{| class="wikitable"
|+Table 8. Transition state position
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|180
|0
|-
|r(BC)
|74.5
|0
|}
The transition state position shown in Table 8 was estimated by looking at the initial conditions in Table 6 for the exothermic F + H2 → HF + H reaction. Due to exothermic reactions having early transition states, this suggested that the bond distance BC (the hydrogen atoms) would be small as it's the reactant. The initial bond distance of AB was reduced until the hessian matrix, shown in Table 9, gave a value of Det= -1.0 <0 which proved it was a saddle point. Additionally, there was no force. This also suggested it was the transition state.
{| class="wikitable"
|+Table 9. Hessian matrix for Transition state
|AB direction
|<nowiki>+1.000</nowiki>
|<nowiki>-0.024</nowiki>
|-
|BC direction
|<nowiki>-0.024</nowiki>
|<nowiki>-1.000</nowiki>
|}
[[File:Mb7418 dis vs time ts ex2.png|thumb|[[Figure 9]]. Intermolecular distance vs time graph for transition state for equations two and three]]
Additionally, Figure 9 shows the intermolecular distance vs time graph. This is a flat line, this could be due to everything in the system being stationary. There are no oscillations showing that there are any vibrational motions. This suggest that it's at the initial conditions in Table 8 represent the transition state position.

* Q8. Report the activation energy for both reactions.
Equation 2: F + H2 → HF + H the activation energy value is 1.0 kJ/mol.

Equation 3: HF + H → F + H2 the activation energy value is 126.7 kJ/mol.

The activation energy for equation 2 was calculated by getting the potential energy for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 10. Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(435.056) = 1.075 kJ/mol
{| class="wikitable"
|+Table 10. Initial conditions for equation 2
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|74.5
|0
|}
The activation energy for equation 3 was calculated by getting the potential energy for for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 11. r(AB) which is the HF bond distance = 92 ppm<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7664393&charge=0</nowiki>> [Accessed 22 May 2020].</ref>'''. '''The r(BC) bond which is the HH bond distance was set very high because the hydrogen atoms aren't bonded in the reactants.

Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(560.698) = 126.7 kJ/mol
{| class="wikitable"
|+Table 11. Initial conditions for equation 3
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|92
|0
|-
|r(BC)
|500
|0
|}

{{fontcolor1|red| Good work. well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:28, 29 May 2020 (BST)}}

=== Reaction dynamics ===
* Q9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.
{| class="wikitable"
|+Table 12. Initial conditions for reactive trajectory
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|175
|<nowiki>-1.6</nowiki>
|-
|r(BC)
|74
|0.3
|}
[[File:Mb7418 THIS reactive.png|left|thumb|[[Figure 10]]. Reactive trajectory]]
The step number was increased to 4000 and the initial conditions in Table 12 gives the potential energy surface diagram in Figure 10. This shows a reactive trajectory for the exothermic reaction F + H2 → HF + H as it passes the transition state as well as having many oscillations. The end point is also far away from the starting point.

The starting point (reactants) has a high energy level compared to the products due to being exothermic<ref name=":0" />. Hence the potential energy converts into translational kinetic energy. Therefore, there are many oscillations in Figure 10 due to energy being conserved.

This can be explained by the difference in distribution of the energy levels between an excited and relaxed state. The excited state will have an occupied ground state (v=0) as well as an occupied first state (v=1) where v=vibrational number. Whereas in the relaxed state only the ground state will be occupied. This approach uses the harmonic oscillator<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of N2 Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7727379&charge=0</nowiki>> [Accessed 18 May 2020].</ref>. When energy is released, in the excited state there will be two transitions, v(0)→v(1) and v(1)→v(2), the fundamental peak and overtone. In the ground state it will only have transitions from v(0)→v(1).

Therefore, in an IR spectrum<ref>Chemistry LibreTexts. 2020. ''5.5: The Harmonic Oscillator And Infrared Spectra''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.05%3A_The_Harmonic_Oscillator_and_Infrared_Spectra</nowiki>> [Accessed 22 May 2020].</ref>, for the excited state there will be two peaks corresponding to the overtone and fundamental peak. The overtone peak intensity will be smaller than the fundamental's peak intensity. Hence there are many oscillations due to the overtone decreasing and then increasing in intensity while the fundamental increases and then decreases in intensity.

Therefore, this could be confirmed experimentally by Infrared spectroscopy (IR).
* Q10. 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.
[[File:Mb7418 last Q.png|left|thumb|[[Figure 11]]. Reactive potential energy surface contour plot]]
The position of the transition state affects whether the reaction is either exothermic or endothermic. An endothermic reaction will have a late transition state, suggesting that the vibrational energy efficiency will be higher<ref>Guo, H. and Liu, K., 2020. ''Control Of Chemical Reactivity By Transition-State And Beyond''.</ref>. The reason for this is because if it had a higher translational kinetic energy, the reactants wouldn't follow the minimum energy path and will just collide with each other but then immediately repel resulting in being back at the starting point. Hence, the translational kinetic energy converts into vibrational energy<ref name=":1" /> instead so the reactants collide and are aligned with the minimum energy pathway. The direction of the path taken is shown in Figure 11. The initial conditions in Table 12 was used to give the reactive trajectory. This suggests that endothermic reactions have higher vibrational energy to make their reactions more efficient so it's able to overcome the transition state. Hence there are a lot of oscillations before the transition state is reached because it requires more efficient vibrational energy in order to reach the transition state.

Whereas if the reaction has an early transition state, it will be an exothermic reaction. This suggest that there will be a higher translational kinetic energy compared to vibrational energy. This means that there will be less oscillations before it reaches the transition state as shown in Figure 11.

Overall, an efficient endothermic reaction will have a higher vibrational energy whereas an efficient exothermic reaction will have a higher translational kinetic energy. {{fontcolor1|red| Great. Please see a couple of comments of mine earlier on in the report. Otherwise I think this is a good job. Thank you especially for the extensive bibliography and for referencing throughout the text, good practice. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 21:40, 29 May 2020 (BST)}}

MRD:mb7418

2020-05-30T07:50:12Z

Fdp18:

== EXERCISE 1: H + H2 system ==

{{fontcolor1|red|To me it looks like you understood what you are doing, although you have problems expressing yourself in a clear way sometimes. for example, "The transition state is the minimum energy required for the reactants to convert into the products" is wrong - that would be the activation energy, not the transition state. (What is the difference? look at the absolute energies...) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:50, 30 May 2020 (BST)}}

===Dynamics from the transition state region===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? {{fontcolor1|red| This text is repeated below which makes this section difficult to read. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} Potential energy surface diagrams show the potential energy the system has according to the coordinates of the atoms in the system.<ref>Chemistry LibreTexts. 2020. ''2.6: Potential Energy Surfaces''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/2%3A_Chemical_Kinetics/2.06%3A_Potential_Energy_Surfaces</nowiki>> [Accessed 19 May 2020].</ref> Figure 1 shows a potential energy surface diagram representing the reaction, equation 1: A + BC → AB + C.
{| class="wikitable"
|+Table 1. Molecules distances (r) and momenta (p)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|230
|<nowiki>-5.2</nowiki>
|-
|r1=r(BC)
|74
|0
|}
In this case r(1)=r(BC) and r(2)=r(AB). BC is a hydrogen molecule consisting of two hydrogen atoms, B and C. Atom A is a hydrogen atom which has energy equal to or more than the activation energy, thus reaching the transition state. The transition state structure consists of atom A forming a bond with molecule BC while the BC bond is slightly dissociating. After the transition state, the bond between AB would form making a new molecule and the BC bond would be fully dissociated.
* Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
[[File:PE with TS.png|left|thumb|[[Figure]] 1. Potential energy surface diagram with transition state labelled ]]
In a potential energy surface diagram, the transition state is mathematically defined as the first-order saddle point<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref>:

∂V(r1)/∂(r1)=0 and ∂V(r2)/∂r2=0

∂2V(r)/∂(r1)2 > 0 and ∂2V(r)/∂(r2)2 < 0

[∂2V(r)/∂(r1)2]*[ ∂2V(r)/∂(r2)2] - [ ∂2V/∂(r1)(r2)]2 < 0

Hessian matrix is shown in Table 2, resulting in DET = -0.999698 < 0
{| class="wikitable"
|+Table 1. Hessian matrix for the transition state
!
!f(x)
!f(y)
|-
|f(x)
|<nowiki>+0.707</nowiki>
|<nowiki>-0.707</nowiki>
|-
|f(y)
|<nowiki>-0.707</nowiki>
|<nowiki>-0.707</nowiki>
|}
In Figure 1, the black wavy lines shows that the molecule is vibrating. This is the minimum energy path required for the reactants to react and turn into the products via the transition state.<ref>Atkins, de Paula, Keeler: Physical Chemistry, 11th Edition. Section 18D.3 Potential Energy Surfaces (PESs)</ref> This suggests that the transition state is the peak in the minimum energy path. The transition state is the minimum energy required for the reactants to convert into the products. Hence once the transition state is reached, if the reaction was successful the products would be formed. Figure 1 shows the location of the transition state in the potential energy diagram.

The transition state can be distinguished from a local minimum of the potential energy surface because it is on the diagonal axis. {{fontcolor1|red| Because it is also a maximum (minimum AND maximum in orthogonal directions.. i.e a saddle point) therefore distinguishable from a local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} It is the saddle point.This is when everything in the system is stationary when there is no momentum.

===Locating the transition state r1=r2===
* Q2. Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.

<nowiki> </nowiki>When the transition state is formed, due to all the atoms being hydrogen, this suggest that the transition state is symmetrical. This means that the bond distance between A and B is the same as the bond distance of B and C. This can be used to locate the transition state when the momentum is 0. Table 3 shows an estimate of the transition state position (rts).
{| class="wikitable"
|+Table 3. Distances and momenta to locate the transition state
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|90.75
|0
|}
[[File:Mb7418 dis vs time ts.png|thumb|[[Figure 2]]. Distance vs time plot of the transition state ]]
Figure 2 shows an Intermolecular distance vs time graph resulted from the initial conditions from Table 2. The line is flat, this suggest that everything in the system is stationary. Whereas, if it wasn't the transition state, in the intermolecular distance vs time graph you'd see oscillations representing the vibrational movements. The more oscillations would indicate the more vibrational movements. Hence the transition state is a flat line as there isn't any. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:33, 29 May 2020 (BST)}}
[[File:Mb7418 ts contour.png|thumb|[[Figure 3]]. Potential energy diagram as a contour plot of the transition state]]
The contour plot shown in Figure 3 shows the transition state as a red cross. This is the saddle point of the potential surface. Whenever r1 and r2 are the same, the point would be anywhere across the diagonal axis. This contour plot helped estimate the transition state position as the trajectory shows that the AB and BC bond lengths are the same. This is the point at which reactants have to overcome to react into the products.

The minimum energy path can differ depending on the relative momentum of the atoms/molecules and their vibrational excitation energy. The transition state of the reaction won't change its position. The reactants require sufficient translational kinetic energy and vibrational energy to overcome the transition state (saddle point) in order to react and turn into the products.

===Trajectories from r1 = rts+δ, r2 = rts ===
* Q3. Comment on how the ''mep'' and the trajectory you just calculated differ. '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''' and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1
The trajectory of the minimum energy path can be visualised slightly near the transition state, this is when atom A is approaching molecule BC to form a bond with it, A + BC.
{| class="wikitable"
|+Table 4. Initial conditions slightly displaced from transition state
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|91.75
|0
|}
The initial conditions shown in Table 4 gives the potential energy diagram in Figure 4 and 5 when the calculation type is MEP. The minimum energy path shown by the black line isn't wavy. [[File:Mb7418 mep.png|left|thumb|[[Figure 4]]. MEP calculation diagram]]
[[File:Mb7418 mep 1.png|thumb|234x234px|[[Figure 5]]. MEP calculation diagram bottom view]]However, when the same initial conditions are applied when the calculation type is Dynamic, the minimum energy path is wavy shown in Figure 6. This shows that the MEP calculation gives a different result of the reaction trajectory. The molecule is constantly moving at a velocity since there is no external force acting upon the reactants. This is called inertia motion.<ref>Evans, M.G. and Polanyi, M., 1938. Inertia and driving force of chemical reactions. ''Transactions of the Faraday Society'', ''34'', pp.11-24.</ref> Whereas the potential energy diagram produced from the Dynamics calculation takes into account the reactants inertia by representing the pathway as non-linear. Additionally, the MEP calculation diagram has a smaller trajectory than the Dynamics calculation.The reason for this is beacause the MEP calculation sets the momentum to zero at every step whereas the dynamic calculation the momenta is developed classically<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., 8.3.1, Prentice-Hall, 1998.</ref>. The MEP calculation only takes the gradient into account. The amount of steps there are corresponds to the number of times it is calculating what the gradient is at that point. The only thing relative to the MEP calculation is the gradient. Whereas with dynamics, it takes into account how long it takes for the gradient to change. The momentum (velocity) is changing each time instead of setting it to zero every step. Hence it's longer as after the products are formed, they are still have motion such as vibrational and kinetic. {{fontcolor1|red| Yes. This does show an understanding of MEP vs Dynamics. Just to clarify - the amount of steps is the number of times the system moves during the simulation - this was a bit confused in your response. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:36, 29 May 2020 (BST)}} [[File:Mb7418 dyn 1.png|thumb|[[Figure 6]]. Dynamics calculation bottom view]]

== Reactive and unreactive trajectories ==

===Interpreting trajectories===

* Q4. For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.

{| class="wikitable"
|+Table 5. Reactive and unreactive trajectories
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Reactive
|The starting point (represented by the red cross) starts at 200 pm between atoms A and B. Then the distance AB decreases. Atoms A and B are approaching each other and once it reaches the transition state, the AB distance and BC distance are the same and all the atoms are stationary. Then the AB bond is formed and atom C repels from atom B. The molecule AB is constantly vibrating. This could suggest that some translational kinetic energy got converted into vibrational energy after the transition state. Hence after the trajectory has oscillations after it passes the transition state.
|[[File:Mb7418 6.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The trajectory has some oscillations, showing that the reactants are constantly vibrating. Hence it has vibrational energy. However, it doesn't go over the transition state so it doesn't react into the products. This suggests that the reactants approach each other but then they repel. While they do that they oscillate as they get closer. It ends up with a high AB distance (essentially the starting point of the reaction), this shows that the reactants haven't reacted due to the AB molecule not forming. This could be due to not having enough translational kinetic energy.
|[[File:Mb7418 2.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The starting point starts at 200 pm between atoms A and B. Then the the atoms got closer and reaches the transition state. This is when the distance AB and BC are equal. Then the AB distance remains continuous while the BC distance begins to increase. The BC bond breaks and the AB bond forms. It only crosses the transition state once. It ends up in the products as the distance AB is short (molecule containing atoms A and B bonding) and BC is distance is long (reactant molecule BC bond has been broken).
|[[File:Mb7418 3.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|The whole trajectory shows that it's unreactive because the starting point and ending point are similar. It started at a high AB distance value and also ended at a high AB distance value. This suggests this is end unreactive.
At the starting point, AB are far apart. Then they begin approaching each other and reach the transition state. However, they repel from each other and then start approaching each other again. This happens again and then the distance AB begins to increase again. The BC bond is formed again, resulting in the reactant molecule BC at the end.
|[[File:Mb7418 4.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|This crosses the transition point more than once. It's reactive as it's end point is at a high BC distance and low AB distance, suggesting that the BC bond broke and the AB bond has formed resulting in AB + B.

From the starting point, the AB distance starts to decrease. This suggests atom A and B approach each other, but then the AB distance begins to increase again so the A and B atoms repel. There is a moment when the distance between atoms A and B is the same as the distance between B and C (transition state) Then the B and C atoms approach each other and then repel B continuously. While atoms A and B get closer forming a bond. The molecule AB vibrates while atom C continuously repels away from it.
|[[File:Mb7418 5.png|centre|thumb|200x200px]]
|}

{{fontcolor1|red| OK. What do you conclude from the table? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:40, 29 May 2020 (BST)}}

=== Transition state theory ===
* Q5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
The three assumptions of the transition state theory<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Chapter 10 pg. 20, Prentice-Hall, 1998</ref> are:

1) The transition state will only be crossed over once. Once the transition state it reached, the products will be formed.

2) During the transition state, any motion along the reaction coordinate can be separate from the other motions and treated classically as a translation. The concentration of the transition state is in equilibrium with the reactants. (concentration constant relating them both, there's a Maxwell-Boltzmann distribution between the reactants and the transition state at any given time) This is defined as the Quasi equilibrium. It's assumed that the kinetic energy is predicted by the Boltzmann distribution. The transition states that are turning into products are distributed among their states according to the Maxwell-Boltzmann Laws<ref>Rowlinson*, J.S., 2005. The Maxwell–Boltzmann distribution. ''Molecular Physics'', ''103''(21-23), pp.2821-2828.</ref> even in absence of an equilibrium between the reactant and product molecules.

'''3)''' There's minimum effects of quantum tunneling.

The quantum tunneling effect<ref>Abyss.uoregon.edu. 2020. ''Quantum Tunneling''. [online] Available at: <<nowiki>http://abyss.uoregon.edu/~js/glossary/quantum_tunneling.html</nowiki>> [Accessed 22 May 2020].</ref> is when the reactants can overcome the barrier by moving across it to turn into the products. The transition state theory doesn't include quantum tunneling. If quantum tunneling was included, it would increase the rate of reaction as more reactants will turn into products. There are only classical motions in the transition state theory. Hence it is suggested that quantum tunneling isn't included in the model used in Table 5 because it's assumed there's a faster alternative route used.

Additionally, in the Table 5 model the transition state theory wasn't true because the transition state was crossed over more than once in the last two reactions in Table 5, this could decrease the rate of reaction due to the products being able to cross the transition state again and go back into the reactants. Therefore, the transition state theory is an over-estimate in comparison to the model used. {{fontcolor1|red| Yes. Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:41, 29 May 2020 (BST)}}

== EXERCISE 2: F - H - H system ==

=== PES inspection ===
* Q6. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?
{| class="wikitable"
|+Table 6. Initial conditions for Figure 7 (F + H2 → HF + H)
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|70
|0
|}

[[File:Mb7418 F + H2 exo edit CORRECT.png|left|thumb|[[Figure 7]]. F + H2 -> HF + H exothermic potential surface diagram]]
In the following equations, A = atom F, B = atom H, C = atom H.

The initial conditions in Table 6 gives the potential energy diagram for Figure 7. The reason for this is because the F and H bond is very far apart, suggesting that it's not bonded whereas the hydrogen atoms are closer together, suggesting a F atom and H2 molecule are in the initial conditions.

Equation 2: F + H2 → HF + H is an exothermic reaction<ref name=":0">Duff, J.W. and Truhlar, D.G., 1975. Effect of curvature of the reaction path on dynamic effects in endothermic chemical reactions and product energies in exothermic reactions. ''The Journal of Chemical Physics'', ''62''(6), pp.2477-2491.</ref> shown in Figure 7. The energy of the reactants is higher than the energy of the products. This shows that this is an exothermic reaction due to the process releasing energy. This suggests that the bond strength of H2 is weaker than the bond strength of HF as no additional energy was required to break its bond. This reaction has an early transition state<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref> due to the the fluorine atom being highly electro-negative so it attacks the hydrogen molecule very quickly.
{| class="wikitable"
|+Table 7. Initial conditions for Figure 8 (HF + H → F + H2)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|70
|0
|-
|r(BC)
|500
|0
|}
[[File:Mb7418 NEW CORRECT ENDO.png|left|thumb|[[Figure 8]]. HF + H -> F + H2 endothermic potential surface diagram]]
The initial conditions in Table 7 gives the potential energy diagram for Figure 8. The reason for this is because the F and H atoms are very close together compared to both the hydrogens. This suggests that a HF molecule and one H atom are in the initial conditions.

Equation 3: HF + H → F + H2 is an endothermic reaction<ref name=":0" /> shown in Figure 8. The energy of the reactants is lower than the energy of the products. This suggests that this reaction is endothermic as the process used additional energy to break the H-F bond to form the H-H bond. Therefore, in the potential energy surface diagram there will be a lot of vibrational energy before the transition state compared to after. This also suggest that the HF bond is stronger than the HH bond as it requires additional energy to break its bond. This will give a late transition state.The translational energy could get converted into vibrational energy in order to break the bond between the strong H-F bond<ref name=":1">Polanyi, J.C. and Tardy, D., 1969. Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF+ H. ''The Journal of Chemical Physics'', ''51''(12), pp.5717-5719.</ref>. This favours the endothermic reaction because the reactants position will be aligned with the direction of the minimum energy path.''''' '''''

* Q7. Locate the approximate position of the transition state.
{| class="wikitable"
|+Table 8. Transition state position
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|180
|0
|-
|r(BC)
|74.5
|0
|}
The transition state position shown in Table 8 was estimated by looking at the initial conditions in Table 6 for the exothermic F + H2 → HF + H reaction. Due to exothermic reactions having early transition states, this suggested that the bond distance BC (the hydrogen atoms) would be small as it's the reactant. The initial bond distance of AB was reduced until the hessian matrix, shown in Table 9, gave a value of Det= -1.0 <0 which proved it was a saddle point. Additionally, there was no force. This also suggested it was the transition state.
{| class="wikitable"
|+Table 9. Hessian matrix for Transition state
|AB direction
|<nowiki>+1.000</nowiki>
|<nowiki>-0.024</nowiki>
|-
|BC direction
|<nowiki>-0.024</nowiki>
|<nowiki>-1.000</nowiki>
|}
[[File:Mb7418 dis vs time ts ex2.png|thumb|[[Figure 9]]. Intermolecular distance vs time graph for transition state for equations two and three]]
Additionally, Figure 9 shows the intermolecular distance vs time graph. This is a flat line, this could be due to everything in the system being stationary. There are no oscillations showing that there are any vibrational motions. This suggest that it's at the initial conditions in Table 8 represent the transition state position.

* Q8. Report the activation energy for both reactions.
Equation 2: F + H2 → HF + H the activation energy value is 1.0 kJ/mol.

Equation 3: HF + H → F + H2 the activation energy value is 126.7 kJ/mol.

The activation energy for equation 2 was calculated by getting the potential energy for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 10. Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(435.056) = 1.075 kJ/mol
{| class="wikitable"
|+Table 10. Initial conditions for equation 2
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|74.5
|0
|}
The activation energy for equation 3 was calculated by getting the potential energy for for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 11. r(AB) which is the HF bond distance = 92 ppm<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7664393&charge=0</nowiki>> [Accessed 22 May 2020].</ref>'''. '''The r(BC) bond which is the HH bond distance was set very high because the hydrogen atoms aren't bonded in the reactants.

Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(560.698) = 126.7 kJ/mol
{| class="wikitable"
|+Table 11. Initial conditions for equation 3
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|92
|0
|-
|r(BC)
|500
|0
|}

{{fontcolor1|red| Good work. well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:28, 29 May 2020 (BST)}}

=== Reaction dynamics ===
* Q9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.
{| class="wikitable"
|+Table 12. Initial conditions for reactive trajectory
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|175
|<nowiki>-1.6</nowiki>
|-
|r(BC)
|74
|0.3
|}
[[File:Mb7418 THIS reactive.png|left|thumb|[[Figure 10]]. Reactive trajectory]]
The step number was increased to 4000 and the initial conditions in Table 12 gives the potential energy surface diagram in Figure 10. This shows a reactive trajectory for the exothermic reaction F + H2 → HF + H as it passes the transition state as well as having many oscillations. The end point is also far away from the starting point.

The starting point (reactants) has a high energy level compared to the products due to being exothermic<ref name=":0" />. Hence the potential energy converts into translational kinetic energy. Therefore, there are many oscillations in Figure 10 due to energy being conserved.

This can be explained by the difference in distribution of the energy levels between an excited and relaxed state. The excited state will have an occupied ground state (v=0) as well as an occupied first state (v=1) where v=vibrational number. Whereas in the relaxed state only the ground state will be occupied. This approach uses the harmonic oscillator<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of N2 Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7727379&charge=0</nowiki>> [Accessed 18 May 2020].</ref>. When energy is released, in the excited state there will be two transitions, v(0)→v(1) and v(1)→v(2), the fundamental peak and overtone. In the ground state it will only have transitions from v(0)→v(1).

Therefore, in an IR spectrum<ref>Chemistry LibreTexts. 2020. ''5.5: The Harmonic Oscillator And Infrared Spectra''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.05%3A_The_Harmonic_Oscillator_and_Infrared_Spectra</nowiki>> [Accessed 22 May 2020].</ref>, for the excited state there will be two peaks corresponding to the overtone and fundamental peak. The overtone peak intensity will be smaller than the fundamental's peak intensity. Hence there are many oscillations due to the overtone decreasing and then increasing in intensity while the fundamental increases and then decreases in intensity.

Therefore, this could be confirmed experimentally by Infrared spectroscopy (IR).
* Q10. 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.
[[File:Mb7418 last Q.png|left|thumb|[[Figure 11]]. Reactive potential energy surface contour plot]]
The position of the transition state affects whether the reaction is either exothermic or endothermic. An endothermic reaction will have a late transition state, suggesting that the vibrational energy efficiency will be higher<ref>Guo, H. and Liu, K., 2020. ''Control Of Chemical Reactivity By Transition-State And Beyond''.</ref>. The reason for this is because if it had a higher translational kinetic energy, the reactants wouldn't follow the minimum energy path and will just collide with each other but then immediately repel resulting in being back at the starting point. Hence, the translational kinetic energy converts into vibrational energy<ref name=":1" /> instead so the reactants collide and are aligned with the minimum energy pathway. The direction of the path taken is shown in Figure 11. The initial conditions in Table 12 was used to give the reactive trajectory. This suggests that endothermic reactions have higher vibrational energy to make their reactions more efficient so it's able to overcome the transition state. Hence there are a lot of oscillations before the transition state is reached because it requires more efficient vibrational energy in order to reach the transition state.

Whereas if the reaction has an early transition state, it will be an exothermic reaction. This suggest that there will be a higher translational kinetic energy compared to vibrational energy. This means that there will be less oscillations before it reaches the transition state as shown in Figure 11.

Overall, an efficient endothermic reaction will have a higher vibrational energy whereas an efficient exothermic reaction will have a higher translational kinetic energy. {{fontcolor1|red| Great. Please see a couple of comments of mine earlier on in the report. Otherwise I think this is a good job. Thank you especially for the extensive bibliography and for referencing throughout the text, good practice. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 21:40, 29 May 2020 (BST)}}

MRD:01504509

2020-05-27T07:58:56Z

Fdp18: /* Molecular Reaction Dynamics: Applications to Triatomic systems */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

{{fontcolor1|red|There are some important issues with understanding, see the comments below. In some cases I think you understood the concepts, but were not able to articulate your thoughts in a clear and concise way. I recommend that, when you're finished with a report, you give it to a peer/family member/... to read over, it can be as simple as deleting a repeated work that makes one stumble over a sentence. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:58, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again. {{fontcolor1|red|If it returns to the reactants - is it reactive then? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:32, 27 May 2020 (BST)}}
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

{{fontcolor1|red|OK, I see the problem. This is really important - the 5th row is the one with the ''highest'' energy. The values are negative - i.e. the one with the least negative value is the highest. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:35, 27 May 2020 (BST)}}

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions. {{fontcolor1|red|I don't understand this last sentence. Quantum tunneling does not change the barrier, nor does the rate of reactions. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.

{{fontcolor1|red|be careful not to confuse "transition state" and "reaction rate". The transition state stays the same in all cases, and it doesn't even have to be encountered in a "real" trajectory to be reactive. what is overestimated (by TST) is the reaction rate. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:

{{fontcolor1|red|ok - however, some sentences are hard to read and understand, especially the first one. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:41, 27 May 2020 (BST)}}

{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

{{fontcolor1|red|How did you arrive at this value? I'm looking for a description for what you did, similar to the H-H-H case. With the information given above, I could not reproduce your experiment... [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:43, 27 May 2020 (BST)}}

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

{{fontcolor1|red|Good value and convincing confirmation that the points are stationary, although I still don't see how you arrived there. I know you must have done the right thing because of the values you present, but you need to show what you did. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:44, 27 May 2020 (BST)}}

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

{{fontcolor1|red|ok [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:51, 27 May 2020 (BST)}}

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

{{fontcolor1|red|This is not a good example, as the trajectory has not yet reached the interesting region. It might or might not react - you simply can't tell that yet. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:54, 27 May 2020 (BST)}}

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

{{fontcolor1|red|Unfortunately, this is also not a good example - the trajectory ends in the reactant's region. I would expect the reactants to be formed fully again, although you have not simulated for long enough to make that statement definitively. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:56, 27 May 2020 (BST)}}

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:56:14Z

Fdp18: /* Question 10 */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again. {{fontcolor1|red|If it returns to the reactants - is it reactive then? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:32, 27 May 2020 (BST)}}
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

{{fontcolor1|red|OK, I see the problem. This is really important - the 5th row is the one with the ''highest'' energy. The values are negative - i.e. the one with the least negative value is the highest. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:35, 27 May 2020 (BST)}}

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions. {{fontcolor1|red|I don't understand this last sentence. Quantum tunneling does not change the barrier, nor does the rate of reactions. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.

{{fontcolor1|red|be careful not to confuse "transition state" and "reaction rate". The transition state stays the same in all cases, and it doesn't even have to be encountered in a "real" trajectory to be reactive. what is overestimated (by TST) is the reaction rate. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:

{{fontcolor1|red|ok - however, some sentences are hard to read and understand, especially the first one. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:41, 27 May 2020 (BST)}}

{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

{{fontcolor1|red|How did you arrive at this value? I'm looking for a description for what you did, similar to the H-H-H case. With the information given above, I could not reproduce your experiment... [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:43, 27 May 2020 (BST)}}

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

{{fontcolor1|red|Good value and convincing confirmation that the points are stationary, although I still don't see how you arrived there. I know you must have done the right thing because of the values you present, but you need to show what you did. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:44, 27 May 2020 (BST)}}

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

{{fontcolor1|red|ok [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:51, 27 May 2020 (BST)}}

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

{{fontcolor1|red|This is not a good example, as the trajectory has not yet reached the interesting region. It might or might not react - you simply can't tell that yet. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:54, 27 May 2020 (BST)}}

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

{{fontcolor1|red|Unfortunately, this is also not a good example - the trajectory ends in the reactant's region. I would expect the reactants to be formed fully again, although you have not simulated for long enough to make that statement definitively. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:56, 27 May 2020 (BST)}}

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:54:19Z

Fdp18: /* answer */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again. {{fontcolor1|red|If it returns to the reactants - is it reactive then? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:32, 27 May 2020 (BST)}}
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

{{fontcolor1|red|OK, I see the problem. This is really important - the 5th row is the one with the ''highest'' energy. The values are negative - i.e. the one with the least negative value is the highest. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:35, 27 May 2020 (BST)}}

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions. {{fontcolor1|red|I don't understand this last sentence. Quantum tunneling does not change the barrier, nor does the rate of reactions. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.

{{fontcolor1|red|be careful not to confuse "transition state" and "reaction rate". The transition state stays the same in all cases, and it doesn't even have to be encountered in a "real" trajectory to be reactive. what is overestimated (by TST) is the reaction rate. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:

{{fontcolor1|red|ok - however, some sentences are hard to read and understand, especially the first one. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:41, 27 May 2020 (BST)}}

{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

{{fontcolor1|red|How did you arrive at this value? I'm looking for a description for what you did, similar to the H-H-H case. With the information given above, I could not reproduce your experiment... [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:43, 27 May 2020 (BST)}}

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

{{fontcolor1|red|Good value and convincing confirmation that the points are stationary, although I still don't see how you arrived there. I know you must have done the right thing because of the values you present, but you need to show what you did. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:44, 27 May 2020 (BST)}}

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

{{fontcolor1|red|ok [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:51, 27 May 2020 (BST)}}

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

{{fontcolor1|red|This is not a good example, as the trajectory has not yet reached the interesting region. It might or might not react - you simply can't tell that yet. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:54, 27 May 2020 (BST)}}

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:51:33Z

Fdp18: /* answer */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again. {{fontcolor1|red|If it returns to the reactants - is it reactive then? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:32, 27 May 2020 (BST)}}
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

{{fontcolor1|red|OK, I see the problem. This is really important - the 5th row is the one with the ''highest'' energy. The values are negative - i.e. the one with the least negative value is the highest. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:35, 27 May 2020 (BST)}}

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions. {{fontcolor1|red|I don't understand this last sentence. Quantum tunneling does not change the barrier, nor does the rate of reactions. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.

{{fontcolor1|red|be careful not to confuse "transition state" and "reaction rate". The transition state stays the same in all cases, and it doesn't even have to be encountered in a "real" trajectory to be reactive. what is overestimated (by TST) is the reaction rate. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:

{{fontcolor1|red|ok - however, some sentences are hard to read and understand, especially the first one. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:41, 27 May 2020 (BST)}}

{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

{{fontcolor1|red|How did you arrive at this value? I'm looking for a description for what you did, similar to the H-H-H case. With the information given above, I could not reproduce your experiment... [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:43, 27 May 2020 (BST)}}

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

{{fontcolor1|red|Good value and convincing confirmation that the points are stationary, although I still don't see how you arrived there. I know you must have done the right thing because of the values you present, but you need to show what you did. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:44, 27 May 2020 (BST)}}

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

{{fontcolor1|red|ok [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:51, 27 May 2020 (BST)}}

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:44:39Z

Fdp18: /* answer */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again. {{fontcolor1|red|If it returns to the reactants - is it reactive then? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:32, 27 May 2020 (BST)}}
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

{{fontcolor1|red|OK, I see the problem. This is really important - the 5th row is the one with the ''highest'' energy. The values are negative - i.e. the one with the least negative value is the highest. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:35, 27 May 2020 (BST)}}

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions. {{fontcolor1|red|I don't understand this last sentence. Quantum tunneling does not change the barrier, nor does the rate of reactions. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.

{{fontcolor1|red|be careful not to confuse "transition state" and "reaction rate". The transition state stays the same in all cases, and it doesn't even have to be encountered in a "real" trajectory to be reactive. what is overestimated (by TST) is the reaction rate. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:

{{fontcolor1|red|ok - however, some sentences are hard to read and understand, especially the first one. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:41, 27 May 2020 (BST)}}

{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

{{fontcolor1|red|How did you arrive at this value? I'm looking for a description for what you did, similar to the H-H-H case. With the information given above, I could not reproduce your experiment... [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:43, 27 May 2020 (BST)}}

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

{{fontcolor1|red|Good value and convincing confirmation that the points are stationary, although I still don't see how you arrived there. I know you must have done the right thing because of the values you present, but you need to show what you did. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:44, 27 May 2020 (BST)}}

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:43:04Z

Fdp18: /* answer */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again. {{fontcolor1|red|If it returns to the reactants - is it reactive then? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:32, 27 May 2020 (BST)}}
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

{{fontcolor1|red|OK, I see the problem. This is really important - the 5th row is the one with the ''highest'' energy. The values are negative - i.e. the one with the least negative value is the highest. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:35, 27 May 2020 (BST)}}

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions. {{fontcolor1|red|I don't understand this last sentence. Quantum tunneling does not change the barrier, nor does the rate of reactions. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.

{{fontcolor1|red|be careful not to confuse "transition state" and "reaction rate". The transition state stays the same in all cases, and it doesn't even have to be encountered in a "real" trajectory to be reactive. what is overestimated (by TST) is the reaction rate. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:

{{fontcolor1|red|ok - however, some sentences are hard to read and understand, especially the first one. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:41, 27 May 2020 (BST)}}

{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

{{fontcolor1|red|How did you arrive at this value? I'm looking for a description for what you did, similar to the H-H-H case. With the information given above, I could not reproduce your experiment... [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:43, 27 May 2020 (BST)}}

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:41:27Z

Fdp18: /* answer: */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again. {{fontcolor1|red|If it returns to the reactants - is it reactive then? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:32, 27 May 2020 (BST)}}
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

{{fontcolor1|red|OK, I see the problem. This is really important - the 5th row is the one with the ''highest'' energy. The values are negative - i.e. the one with the least negative value is the highest. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:35, 27 May 2020 (BST)}}

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions. {{fontcolor1|red|I don't understand this last sentence. Quantum tunneling does not change the barrier, nor does the rate of reactions. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.

{{fontcolor1|red|be careful not to confuse "transition state" and "reaction rate". The transition state stays the same in all cases, and it doesn't even have to be encountered in a "real" trajectory to be reactive. what is overestimated (by TST) is the reaction rate. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:

{{fontcolor1|red|ok - however, some sentences are hard to read and understand, especially the first one. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:41, 27 May 2020 (BST)}}

{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:39:00Z

Fdp18: /* Question 5 */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again. {{fontcolor1|red|If it returns to the reactants - is it reactive then? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:32, 27 May 2020 (BST)}}
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

{{fontcolor1|red|OK, I see the problem. This is really important - the 5th row is the one with the ''highest'' energy. The values are negative - i.e. the one with the least negative value is the highest. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:35, 27 May 2020 (BST)}}

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions. {{fontcolor1|red|I don't understand this last sentence. Quantum tunneling does not change the barrier, nor does the rate of reactions. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.

{{fontcolor1|red|be careful not to confuse "transition state" and "reaction rate". The transition state stays the same in all cases, and it doesn't even have to be encountered in a "real" trajectory to be reactive. what is overestimated (by TST) is the reaction rate. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:39, 27 May 2020 (BST)}}

===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:
{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:35:04Z

Fdp18: /* Question 5 */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again. {{fontcolor1|red|If it returns to the reactants - is it reactive then? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:32, 27 May 2020 (BST)}}
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

{{fontcolor1|red|OK, I see the problem. This is really important - the 5th row is the one with the ''highest'' energy. The values are negative - i.e. the one with the least negative value is the highest. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:35, 27 May 2020 (BST)}}

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions.

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.
===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:
{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:32:29Z

Fdp18: /* answer */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again. {{fontcolor1|red|If it returns to the reactants - is it reactive then? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:32, 27 May 2020 (BST)}}
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions.

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.
===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:
{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:31:10Z

Fdp18: /* answer */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{{fontcolor1|red|In the whole table below, "AC" is never a product (and also not a reactant) - take a look at the right column. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:31, 27 May 2020 (BST)}}

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again.
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions.

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.
===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:
{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:29:37Z

Fdp18: /* answer */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. {{fontcolor1|red|Carefully look at your trajectories again. The reactants are AB in this case. (applies to other assignments as well) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:29, 27 May 2020 (BST)}} [[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again.
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions.

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.
===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:
{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:26:59Z

Fdp18: /* answer */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). {{fontcolor1|red|Think about this carefully. We are simulating in classical mechanics - is what you just described possible? To make such a controversial statement, you need to provide the barrier height and then show that the atoms did not have enough energy to overcome it, but still did. (Which really should not happen) [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:26, 27 May 2020 (BST)}} If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side.[[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again.
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions.

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.
===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:
{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

===== answer =====
[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

==== Question 8 ====
Report the activation energy for both reactions.

===== answer =====
[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.

MRD:01504509

2020-05-27T07:23:12Z

Fdp18: /* answer */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

{{fontcolor1|red|General remarks: Please do get in the habit of defining all your quantities (What is V? what is A,B,C?), giving proper references, and also labelling and referencing your Figures (i.e. adding a caption with e.g. 'Figure 1', but also referring to every figure in the text). Sometimes you have mentioned, e.g. 'figure 1' in the text, but the figures are not labelled, so I cannot assign them. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:20, 27 May 2020 (BST)}}

=== EXERCISE 1: H + H2 system ===

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

===== answer =====
[[File:Qq11.png|centre|thumb|783x783px]]
[[File:Qq12.png|centre|thumb|500x500px]]

On the PES (Potential Energy Surface) diagram, the transition state has the highest potential energy {{fontcolor1|red|That is not correct - just take a look at the PES: there are many points with higher energy, and in the 3D surface you even clipped some. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:15, 27 May 2020 (BST)}} and is represented by the saddle point thus the gradient is zero (∂V('''ri''')/∂'''ri'''=0) as the maximum point in one direction, or we say it is a minimum along with the symmetric stretch coordinates, but a maximum along the asymmetric stretching coordinates. Although the saddle points mean that the gradient/slope is zero which can represent both local maximum and minimum, this can be distinguished by its second derivative since the second derivative at the point of the transition state(maximum) will be a negative value (negative curvature) while for the local minima, it is a positive value (positive curvature). {{fontcolor1|red|That is correct - but how do you use the second derivatives to define the saddle point? You only defined local maxima and minima so far. [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:17, 27 May 2020 (BST)}}

==== Question 2 ====

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

===== answer =====
[[File:615b.png|left|thumb|500x500px]]
[[File:615a.png|thumb|centre|480x480px]]

[[File:615c.png|thumb|499x499px|left]]
[[File:615d.png|centre|thumb|561x561px]]

The estimate of the transition state position should be about 90.8pm. H-H-H should be the transition state as a triatomic molecule since it shows the highest energy. For the initial conditions, the distance r1=r2=100pm(figure 2), The trajectory shows periodic symmetric vibrations(oscillations), but it might fall off {{fontcolor1|red|What do you mean by "fall off"? [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:18, 27 May 2020 (BST)}} if the motion is slight. As can be seen from the three plots, when r=91pm, the three atoms are stationary. At r=100.8pm or r=80.9pm, the three atoms move to the greatest amplitude spending similar times. When the distance decreases to 80pm(figure 2), we can see that the atoms will first get further; When the distance increases(to 120pm), the atoms will first get closer(figure 3) all of which shows that the transition state (triatomic molecule) will be formed at a position of r=91pm and the trajectory formed will not fall off since the slope is zero at this stationary point. {{fontcolor1|red|I assume you mean "fall off the transition state". While you can imagine the PES like a landscape, you should aware that this is only to help you understand it - the atoms do not "move in this valley". [[User:Fdp18|Fdp18]] ([[User talk:Fdp18|talk]]) 08:23, 27 May 2020 (BST)}}

==== Question 3 ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

===== answer =====
[[File:Qq31.png|centre|thumb|500x500px]]
[[File:Qq32.png|centre|thumb|500x500px]]

The two graphs show a minimum energy pathway (MEP) and a dynamic trajectory with the same distance condition: r1 = 92pm, r2 = 91pm and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. The '''MEP''' describes infinitely slow motion which has the property that any point on the '''path''' is at an '''energy minimum''' in all directions perpendicular to the '''path'''. This '''path''' passes through at least one first-order saddle point. The '''MEP''' can also be described as the union of steepest descent '''paths''' from the saddle points to the minima. For Figure 1, MEP shows that distance B-C decreases, while distance A-B increases at the pattern as that of distance A-C with a decreasing gradient and eventually reaches velocity zero. For Figure 2, Dynamic shows that the distance B-C keeps oscillating showing the oscillations of the atoms. Distance A-B is almost constant at first but increases sharply later on. Since the velocity is zero as we concluded in MEP, the momentum and kinetic energy will also be zero( p=mv and KE=1/2mv^2). Along with the increase of the distance B-C compared with the transition state, the reaction pathway will favor the product BC. Thus, if we change the condition to r1 = 91pm, r2 = 92pm, the only change is that the pathway will favor the reactant AB. The directions that atoms move can decide which product will be formed because an opposite sign in the momenta will result in the opposite direction in velocity.

==== Question 4 ====
Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

===== answer =====
Conclusion:

1. Particles oscillate if they are close together: When they are too close, repulsion occurs; When they move a bit further away, the attraction will then occur, and the higher the KE will result in stronger oscillation.

2. Atoms with energy lower than the activation energy are still able to make the reaction happen(passes the transition state). If the kinetic energy is out of a specific range, the products and reactants will be unstable, which maybe due to the barrier recrossing.

3. The atom at a long distance from the other two atoms experiences no force, thus constant momentum.

{| class="wikitable"
!
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1 
 

|}
!p2/ g.mol-1.pm.fs-1
!Etot
!
{| class="wikitable"
!Reactive? 
 
|}
!Description of the dynamics
!Illustration of the trajectory(1)
!Illustration of the trajectory(2)
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side.[[File:Qq41.png|thumb|323x323px]]
|[[File:Qq4111.png|thumb]]
|[[File:Qq42.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The reactive trajectory moves along reactants BC side then stops at the TS without reaching the product AB, and comes back to its starting point[[File:Qq43.png|thumb|301x301px]]
|[[File:Qq422.png|thumb]]
|[[File:Qq44.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side[[File:Qq45.png|thumb|280x280px]]
|[[File:Qq433.png|thumb]]
|[[File:Qq46.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Reactive
|
{| class="wikitable"
|The reactive trajectory moves along the reactants BC side, passes the transition state then reaches the product AC side. Then the trajectory returns back to the reactants passing the transition state again.
|}
[[File:Qq47.png|thumb|300x300px]]
|[[File:Qq444.png|thumb]]
|[[File:Qq48.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|
{| class="wikitable"
|<nowiki>-10.6</nowiki> 
 
|}
|<nowiki>-349.477</nowiki>
|Reactive
|The reactive trajectory moves along the reactants BC side, passes the transition state(for a few tims) then reaches the product AC side[[File:Qq49.png|thumb|319x319px]]
{| class="wikitable"
|
|}
|[[File:Qq455.png|thumb]]
|[[File:Qq410.png|thumb]]
|}

==== Question 5 ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

For systems with different momenta thus kinetic energy, it was thought that only systems with enough energy to overcome the activation energy can proceed to the product. However, the table below with the data for different momenta of the hydrogen molecules and hydrogen atoms has shown that this hypothesis is not appropriate, because we cannot make a conclusion between the reaction pathways and the kinetic energy'''.''' For example, in the second row the system Etot is greater than in the first case, however, the reaction trajectory shows that no product is formed and in the 5th row, the system Etot is the lowest of all but still shows a reaction trajectory which passes the transition state and forms product.

'''Transition state theory''' (TST), instead, can be used to explain the observations in the table below which explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes.[2]

There are several assumptions for TST:

1. The reactants are in quasi-equilibrium with the transition state

2. Once a reaction has passed the transition state, it cannot reform the reactants

3. The motion of the reaction can be treated classically, ignoring quantum effects such as quantum tunneling

4. Energy is distributed in accordance with the Maxwell-Boltzmann distribution

Another theory is '''Quantum Tunnelling''' effect, stating that a subatomic particle's probability disappears from one side of a potential barrier and appears on the other side without any probability current (flow) appearing inside the well. Quantum tunneling is not predicted by the laws of classical mechanics where surmounting a potential barrier requires enough potential energy. In this case, the quantum tunnelling effect contributes more to the energy barrier instead of the rate of reactions.

In the 4th and 5th row, the trajectory moves back to the initial state when it passes the transition state, which contradicts TST, thus there is an overestimation of the transition state in these two situations.
===== answer =====

=== EXERCISE 2: F - H - H system ===

==== Question 6 ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

===== answer: =====
[[File:Qq6.png|centre|thumb|543x543px]]

The initial conditions of the PES were set to r1=200 pm and r2=100pm as MEP, showing that when the distance H-F decreases the distance between increases along with the more negative energy. For the formation of HF using F+H2, it should be classified as an exothermic reaction since the energy is released. For H+HF, the distance HF decreases, and the distance between the two H atoms increases, thus relatively higher energy will be needed to break the bond between H and F, making HF + H reaction endothermic. For an exothermic reaction, the energy released during bond formation will be larger than the energy required to break bonds, so the energy released for HF formation is higher than the energy needed to break the H-H bond. For the endothermic, the energy released in the process of the H2 bond formation is less than the energy needed to break the H-F bond. Therefore, the bond strength of HF is stronger than that of H2, this agrees with the experimental values:
{| class="wikitable"
!Bond
!Bond length
!Bond energy/kJ/mol
|-
|H-H
|74
|432
|-
|H-F
|92
|565
|}
[1]

==== Question 7 ====
Locate the approximate position of the transition state.

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[[File:Qq7.png|thumb|500x500px|centre]]

It is clear that the molecule is not symmetric. When we change rAB, no resultant forces will act on the system and the distances between the atoms are constant over time, confirming that transition state has been found. The figure above shows that at the transition state, rAB(HH)=181pm and rBC(HF)=74.5pm, which can be used to determine whichever the reactant or the product the transition state favors. For example, in an ''exothermic'' reaction the transition state is closer in energy to the reactants than to the products. Therefore, the transition state will be more geometrically similar to the reactants than to the products. In contrast, however, in an ''endothermic'' reaction the transition state is closer in energy to the ''products'' than to the reactants. So, according to Hammond’s postulate the structure of the transition state would resemble the products more than the reactants[3], thus the transition state favors H2 and F.

==== Question 8 ====
Report the activation energy for both reactions.

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[[File:Qq81.png|centre|thumb|518x518px]]
[[File:Qq82.png|centre|thumb|526x526px]]
[[File:Qq83.png|centre|thumb|572x572px]]

The three figures above show the plot of the energies of reactant(H2+F), product(HF+H), and transition state thus can be used to determine the activation energy. The activation energy is the minimum energy required to overcome for the reaction to happen, thus it is the difference between the potential energy of the transition state and the initial state. Therefore, the activation of the F+H2 reaction is 1.08 KJ mol-1 and the activation energy of HF+H reaction is 126.7KJ mol-1.

==== Question 9 ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

===== answer =====
[[File:Qq91.png|centre|thumb|500x500px]]
[[File:Qq92.png|centre|thumb|507x507px]]

To make a reactive trajectory, the initial conditions were set as rHH=74pm and rHF=170pm for the F+H2 reaction. As an exothermic reaction, the energy was released for forming H-F. For figure 1, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=3g.mol-1.pm.fs-1 showing that the HF is not formed so not so much potential energy is transferred to kinetic energy to overcome the activation energy. For figure 2, the reactant has enough energy to overcome the activation energy thus HF is formed, the momenta was set to: pHF=-0.1g.mol-1.pm.fs-1 and pHH=20g.mol-1.pm.fs-1. Greater amount of kinetic energy has been transferred from potential energy, around +398 KJ mol-1.

Therefore, we can conclude that energy is released with the formation of HF bond where potential energy has been transferred to kinetic energy. The mean kinetic energy is related to its temperature, thus the increase in kinetic energy can be visualized by the heat released(temperature rise). We can use IR-Spectrometry to investigate the reaction and the heat release since the energy can be in the form of vibrational energy, and molecules will be in an excited state when H-F bond is formed.

==== Question 10 ====
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.

===== answer =====
Based on Polanyi's rules, we know that the energy requirement for a reaction to occur depends on the activation barrier location (transition state location). The translational energy is more effective than vibrational energy when the transition state resembles reactant more, while the vibrational energy is more effective if a transition state is closer to the product. The products of exothermic reactions normally have more vibrational energy than translational energy and characterized by early TS, the products should have more translational energy as it is more effective than having a lot of vibrational energy to overcome the energy barrier. Vice versa for endothermic reactions.

Two examples below support Polanyi's rules:
[[File:Qq101.png|thumb|centre]]

F + H2

rAB=225pm, rBC=74pm, pHF=-0.1g.mol-1.pm.fs-1 and pHH=6.1g.mol-1.pm.fs-1

This is an unreactive trajectory as the trajectory cannot pass the transition state and returns to its initial state. Reactants have relatively high vibrational energy with high amplitude but low translational energy.

[[File:Qq102.png|thumb|centre]]

H + HF

rAB=225pm, rBC=90pm, pHF=20g.mol-1.pm.fs-1 and pHH=-0.5g.mol-1.pm.fs-1

This is a reactive trajectory since it passes the transition state without returning to the starting material. The large amplitude indicates that the reactants' vibrational energy is large. Trajectory passes through the transition state. While the amplitude of vibration decreases as the trajectory reaches the product region.

=== Reference ===
# Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958
# IUPAC, ''Compendium of Chemical Terminology'', 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "transition state theory". doi:10.1351/goldbook.T06470
# Carey FA, Sundberg R (2007). ''Advanced Organic Chemistry Part A:Structure and Mechanisms''. Norwell: Springer.