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Calculating Reaction Thermodynamics for Combustion of Methane

Overview

Hydrocarbon resources constitute the largest source of energy production, contributing to over 85% of the world's energy!  Consider the combustion of methane:

Rxn (1)

In this activity, you will calculate the ΔHrxn, ΔGrxn, and ΔSrxn from first principles and then compare your calculated results with values determined using Hess's Law and enthalpies of formation data:

Table 1: Thermodynamic values at 298 K and 1 atm

 S (J/molK) ΔHf (kJ/mol) ΔGf (kJ/mol) CH4 186.16 -74.85 -50.79 O2 205.03 0 0 CO2 -393.52 -394.38 213.69 H2O -241.82 -228.59 188.72



Initialize

 > $\mathrm{restart};\mathrm{with}\left(\mathrm{QuantumChemistry}\right):\mathrm{Digits}≔15:\mathrm{with}\left(\mathrm{ScientificConstants}\right):\mathrm{with}\left(\mathrm{LinearAlgebra}\right):$





Calculate Thermodynamics of Reactants and Products

You will calculate ΔS, ΔH, and ΔG for Rxn (1) using the following expression:

where F refers to values of S, H, or G, and n refers to stoichiometric coefficients in Rxn (1).   In this section, you will use the Thermodynamics function to calculate S, H, and G for each reactant and product.  For each, you will need to enter the following information:

name = "methane", "oxygen", "carbondioxide", or "water"  (do not use spaces between carbon and dioxide).

symm_num = 12 for methane,     = 2 for oxygen, carbon dioxide, and water.

energy_method = HartreeFock, DensityFunctional, Coupled Cluster, etc.

energy_basis = "sto-3g", "6-31g", "cc-pvdz", etc.

freq_method = HartreeFock or DensityFunctional  (Note, only Hartree-Fock seems to work in a timely manner)

energy_basis = "sto-3g", "6-31g", "cc-pvdz", etc. (Note, only sto-3g seems to work in a timely manner)

freq_scaling = float  (0.8905 for HF,    0.9613 for DFT)

 > $\mathrm{molec_name}≔"oxygen";$
 ${\mathrm{molec_name}}{≔}{"oxygen"}$ (3.1)
 >
 ${\mathrm{molec_label}}{≔}{2}$ (3.2)
 > $\mathrm{symm_num}≔2;$
 ${\mathrm{symm_num}}{≔}{2}$ (3.3)
 > $\mathrm{energy_method}≔\mathrm{HartreeFock};$
 ${\mathrm{energy_method}}{≔}{\mathrm{HartreeFock}}$ (3.4)
 > $\mathrm{energy_basis}≔"sto-3g";$
 ${\mathrm{energy_basis}}{≔}{"sto-3g"}$ (3.5)
 > $\mathrm{freq_method}≔\mathrm{HartreeFock};$
 ${\mathrm{freq_method}}{≔}{\mathrm{HartreeFock}}$ (3.6)
 > $\mathrm{freq_basis}≔"sto-3g";$
 ${\mathrm{freq_basis}}{≔}{"sto-3g"}$ (3.7)
 > $\mathrm{frequency_scaling}≔0.8905;$
 ${\mathrm{frequency_scaling}}{≔}{0.89050000}$ (3.8)
 >
 ${\mathrm{molec_spin}}{≔}{0}$ (3.9)

 > $\mathrm{molec}≔\mathrm{MolecularData}\left(\mathrm{molec_name},"geometry3d"\right);$
 ${\mathrm{molec}}{≔}\left[\left[{"O"}{,}{-0.61600000}{,}{0}{,}{0}\right]{,}\left[{"O"}{,}{0.61600000}{,}{0}{,}{0}\right]\right]$ (3.10)
 > $\mathrm{data}≔\mathrm{Thermodynamics}\left(\mathrm{molec},\mathrm{energy_method},\mathrm{basis}=\mathrm{energy_basis},\mathrm{symmetry_number}=\mathrm{symm_num},\mathrm{freq_scaling}=\mathrm{frequency_scaling},\mathrm{spin}=2\cdot \mathrm{molec_spin}\right);$
 ${\mathrm{data}}{≔}{table}{}\left(\left[{\mathrm{electronic_energy}}{=}{-}{3.87395968}{}{{10}}^{{8}}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧{,}{\mathrm{energy}}{=}{-}{3.87368491}{}{{10}}^{{8}}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧{,}{\mathrm{enthalpy}}{=}{-}{3.87366013}{}{{10}}^{{8}}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧{,}{{\mathrm{\theta }}}_{{C}}{=}{1.99753627}{}⟦{K}⟧{,}{\mathrm{zpe}}{=}{10634.27017022}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧{,}{\mathrm{free_energy}}{=}{-}{3.87481576}{}{{10}}^{{8}}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧{,}{{\mathrm{\theta }}}_{{B}}{=}{1.99753627}{}⟦{K}⟧{,}{\mathrm{entropy}}{=}{387.60228332}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧{,}{\mathrm{heat_capacity}}{=}{20.90118890}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧\right]\right)$ (3.11)
 >
 ${{\mathrm{enthalpy}}}_{{2}}{≔}{-}{387366.01251576}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$ (3.12)
 > $\mathrm{gibbs}\left[\mathrm{molec_label}\right]≔\frac{\mathrm{data}\left['\mathrm{free_energy}'\right]}{1000.};$
 ${{\mathrm{gibbs}}}_{{2}}{≔}{-}{387481.57613653}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$ (3.13)
 > $\mathrm{entropy}\left[\mathrm{molec_label}\right]≔\mathrm{data}\left['\mathrm{entropy}'\right];$
 ${{\mathrm{entropy}}}_{{2}}{≔}{387.60228332}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧$ (3.14)
 > 

Repeat for each reactant and product before moving onto next section.



Calculate DS, DH, and DG

 > $\mathrm{ΔS}≔\mathrm{entropy}\left[3\right]+2\cdot \mathrm{entropy}\left[4\right]-\mathrm{entropy}\left[1\right]-2\cdot \mathrm{entropy}\left[2\right];$
 ${\mathrm{ΔS}}{≔}{{\mathrm{entropy}}}_{{3}}{+}{2}{}{{\mathrm{entropy}}}_{{4}}{-}{{\mathrm{entropy}}}_{{1}}{-}{775.20456663}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧$ (4.1)
 > $\mathrm{ΔH}≔\mathrm{enthalpy}\left[3\right]+2\cdot \mathrm{enthalpy}\left[4\right]-\mathrm{enthalpy}\left[1\right]-2\cdot \mathrm{enthalpy}\left[2\right];$
 ${\mathrm{ΔH}}{≔}{{\mathrm{enthalpy}}}_{{3}}{+}{2}{}{{\mathrm{enthalpy}}}_{{4}}{-}{{\mathrm{enthalpy}}}_{{1}}{+}{774732.02503151}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$ (4.2)
 > $\mathrm{ΔG}≔\mathrm{gibbs}\left[3\right]+2\cdot \mathrm{gibbs}\left[4\right]-\mathrm{gibbs}\left[1\right]-2\cdot \mathrm{gibbs}\left[2\right];$
 ${\mathrm{ΔG}}{≔}{{\mathrm{gibbs}}}_{{3}}{+}{2}{}{{\mathrm{gibbs}}}_{{4}}{-}{{\mathrm{gibbs}}}_{{1}}{+}{774963.15227306}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧$ (4.3)
 > 

Compare with values determined using Hess's law and S, ΔHf, and ΔGf from Table 1:

 >
 ${\mathrm{expDH}}{≔}{-802.31000000}$ (4.4)
 >
 ${\mathrm{expDG}}{≔}{-800.77000000}$ (4.5)
 > $\mathrm{expDS}≔213.69+2\cdot 188.72-186.16-2\cdot 205.03;$
 ${\mathrm{expDS}}{≔}{-5.09000000}$ (4.6)
 > $\mathrm{entropy}\left[2\right]$
 ${387.60228332}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧$ (4.7)

 S (J/molK) ΔHf (kJ/mol) ΔGf (kJ/mol) CH4 186.16 -74.85 -50.79 O2 205.03 0 0 CO2 213.69 -393.52 -394.38 H2O 188.72 -241.82 -228.59