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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Chem 30324, Spring 2020, Homework 10
Due April 24, 2020
Thermodynamics from scratch.
Let’s calculate the thermodynamic properties of an ideal gas of CO molecules at 1 bar pressure. CO has a rotational constant B = 1.931 cm$^{-1}$ and vibrational frequency ν = 2156.6 cm$^{-1}$ . Suppose you have a 20 dm$^3$ cubic bottle containing 1 mole of CO gas that you can consider to behave ideally.
1. The characteristic temperature $\Theta$ of a particular degree of freedom is the characteristic quantum of energy for the degree of freedom divided by $k_B$ . Calculate the characteristic translational, rotational, and vibrational temperatures of CO.
$\theta_{trans} = \frac{\pi^2\hbar^2}{2mL^2k_B}$
$\theta_{rot} = hcB/k_B$
$\theta_{vib} = hc\tilde{\nu}/k_B$
Step1:  2. Calculate the translational partition function of a CO molecule in the bottle at 298 K. What is the unit of the partition function?
For $T \gg \theta_{trans}$, $\Lambda \ll L$, $q_{trans} = V/\Lambda^3$. $\Lambda = h \left(\frac{\beta}{2\pi m}\right)^{1/2}$.
Step2:  3. Plot the rotational and vibrational partition functions of a CO molecule in the bottle from $T$ =  200  to 2000 K (assume the CO remains a gas over the whole range). Hint
Step3:  4. Plot the total translational, rotational, and vibrational energies of CO in the bottle from $T =$ 200 to 2000 K (assume the CO remains a gas over the whole range). Which (if any) of the three types of motions dominate the total energy?
$U_{trans} = \frac{3}{2}RT$
$U_{rot} = RT$
$U_{vib} = R\frac{\theta_{vib}}{e^{\theta_{vib}/T}-1}$.
Step4:  5. Plot the total translational, rotational, and vibrational constant volume molar heat capacities of CO in the bottle from $T =$ 200 to 2000 K. Which (if any) of the three types of motions dominate the heat capacity?
$C_{V,trans} = \frac{3}{2}R$
$C_{V,rot} = R$
$C_{V,vib} = R\left(\frac{\theta_{vib}}{T}\frac{e^{\theta_{vib}/2T}}{e^{\theta_{vib}/T}-1}\right)^2$
Step5:  6. Plot the total translational, rotational, and vibrational Helmholtz energies of CO in the bottle from $T =$ 200 to 2000 K. Which (if any) of the three types of motions dominate the Helmholtz energy?
$A = U - TS$
$S_{trans} = Rln\left(\frac{e^{5/2}V}{N\Lambda^3}\right)$
$S_{rot} = R(1-ln(\theta_{rot}/T))$
$S_{vib} = R\left(\frac{\theta_{vib}/T}{e^{\theta_{vib}/T}-1}-ln(1-e^{-\theta_{vib}/T})\right)$.
Step6:  7. Use your formulas to calculate $\Delta P$, $\Delta U$, $\Delta A$, and $\Delta S$ associated with isothermally expanding the gas from 20 dm$^3$ to 40 dm$^3$.
T = 298 K. $\Delta U=0$. 
$\Delta P = \frac{RT}{V_2} - \frac{RT}{V_1}$. 
$\Delta S = S_{trans,2} - S_{trans,1}$. 
$A = U-TS$, so, $\Delta A = -T\Delta S$.
Step7:  Reactions from scratch
In 1996, Schneider and co-workers used quantum chemistry to compute the reaction pathway for unimolecular decomposition of trifluoromethanol, a reaction of relevance to the atmospheric degradation of hydrofluorocarbon refrigerants (J. Phys. Chem. 1996, 100, 6097- 6103, doi
Step8:  9. Using the data provided, determine  $\Delta A^{\circ}$(298 K) in kJ mol$^{-1}$, assuming ideal behavior and 1 M standard state.  Recall that $A^\circ=E^\text{elec} + \text{ZPE}-RT\ln(q^\circ)-RT$ and that $q^\circ =(q^\text{trans}/V)q^\text{rot}q^\text{vib}/c^\circ$ in units corresponding with the standard state.
$\Delta{A^{\circ}}= \big{[-k_BT\ln(q_{t}q_{r}q_{v})-k_BT]{COF_2}+[-k_BT\ln(q{t}q_{r}q_{v})-k_BT]{HF}-[-k_BT\ln(q{t}q_{r}q_{v})-k_BT]_{CF_3OH}\big} *N_A + \Delta(E_0)$
$\quad \quad = \Delta(E_0)-RT\ln(Q)-RT$
Step9:  10. Determine $\Delta G^\circ$(298 K).  Recall that $G = A + PV = A + RT$ for an ideal ga.
$\Delta{G^{\circ}} = \Delta{A^{\circ}} + \Delta(PV)$
$\quad = \quad \Delta{A^{\circ}} + RT$
Step10:  11. Determine $\Delta S^{\circ}$(298 K), in J mol$^{-1}$ K$^{-1}$ , assuming a 1 M standard state. Recall that $S = (U - A)/T$.
$\Delta A^{\circ} = \Delta U^{\circ} - T\Delta S^{\circ}$
$\Delta S^{\circ} = \frac{\Delta U^{\circ} - \Delta A^{\circ}}{T}$
Step11:  12. Using the data provided, determine $K_c$ (298 K), assuming a 1 M standard state.  You may either determine from partition functions of from the relationship between $K_c$ and $\Delta G^\circ$.
$\mathrm{A} \rightarrow \mathrm{B + C}$$$K_c(T) = \frac{\frac{q_{\mathrm{B}}}{V}\frac{q_{\mathrm{C}}}{V}}{\frac{q_{\mathrm{A}}}{V}} \frac{1}{c^{\circ}} e^{-\Delta{E(0)}/kT} = e^{-\Delta G^\circ(T)/RT}, \quad \text{where} \quad \frac{q_i}{V} = \frac{q_{trans}}{V}q_{rot}q_{vib}$$
Note
Step12:  13. 1 mole of CF$_3$OH is generated in a 20 L vessel at 298 K and left long enough to come to equilibrium with respect to its decomposition reaction. What is the composition of the gas (concentrations of all the components) at equilibrium (in mol/L)?
1 mol/ 20 L = 0.05 mol/L
$\mathrm{A} \rightarrow \mathrm{B + C}$
$K_c = \frac{x^2}{0.05-x}$, solve for x. Note that x = $[\frac{mol}{L}]$.
Step13:  14. How, directionally, would your answer to Question 13 change if the vessel was at a higher temperature? Use the van'T Hoff relationship to determine the equilibrium constant and equilibrium concentrations at 273 and 323 K. How good was your guess?
From question #8, we know that at 298K,  $\Delta U$ = 18.64 kJ/mol. 
$\Delta H = \Delta U + \Delta (nRT) = \Delta U + RT\Delta n$
Step14:  Since $\Delta H$ is positive, we expect $K \uparrow$ as $T \uparrow$
The Van't Hoff equation states
Step15:  Therefore, the at higher temperatures, the reaction shifts towards the products.
15. How, directionally, would your answer to Question 13 change if the vessel had a volume of 5 L? Redo the calculation at this volume to verify your guess.
If V = 5L, the initial concentration = $\frac{1 mol}{5 L}=.2 M$
$K_c = \frac{x^2}{0.2-x}$
At 298 K, Kc = 2.926 (from problem 12)
Step16:  16. Consult a thermodynamics source (e.g. https
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Python Code:
import numpy as np
import matplotlib.pyplot as plt
hbar = 1.05457e-34 # J*s
h = 6.62607e-34 # J*s
kB = 1.38065e-23 # J/K
m = 28.01*1.6605e-27 # kg/molecule
V = 0.02 # m^3
c = 2.99792e10 # cm/s
B = 1.931 # cm^-1
v = 2156.6 # cm^-1
T_trans = np.pi**2*hbar**2/2/m/V**(2/3)/kB
T_rot = h*c*B/kB
T_vib = h*c*v/kB
print('Characteristic temperatures of CO:')
print('T_trans = {0:.3e} K, T_rot = {1:.3f} K, T_vib = {2:0.2f} K.'.format(T_trans,T_rot,T_vib))
Explanation: Chem 30324, Spring 2020, Homework 10
Due April 24, 2020
Thermodynamics from scratch.
Let’s calculate the thermodynamic properties of an ideal gas of CO molecules at 1 bar pressure. CO has a rotational constant B = 1.931 cm$^{-1}$ and vibrational frequency ν = 2156.6 cm$^{-1}$ . Suppose you have a 20 dm$^3$ cubic bottle containing 1 mole of CO gas that you can consider to behave ideally.
1. The characteristic temperature $\Theta$ of a particular degree of freedom is the characteristic quantum of energy for the degree of freedom divided by $k_B$ . Calculate the characteristic translational, rotational, and vibrational temperatures of CO.
$\theta_{trans} = \frac{\pi^2\hbar^2}{2mL^2k_B}$
$\theta_{rot} = hcB/k_B$
$\theta_{vib} = hc\tilde{\nu}/k_B$
End of explanation
Lamda = h*(1/(kB*298*2*np.pi*m))**0.5
print(Lamda)
q_trans = V/Lamda**3
print('The translational partition function of a CO molecule in the bottle at 298 K is {:.4E}.'.format(q_trans))
print('It is dimensionless.')
Explanation: 2. Calculate the translational partition function of a CO molecule in the bottle at 298 K. What is the unit of the partition function?
For $T \gg \theta_{trans}$, $\Lambda \ll L$, $q_{trans} = V/\Lambda^3$. $\Lambda = h \left(\frac{\beta}{2\pi m}\right)^{1/2}$.
End of explanation
T = np.linspace(200,2000,1000) # r = R/a_0
q_rot = T/T_rot
q_vib = 1/(1-np.exp(-T_vib/T))
plt.plot(T,q_rot)
plt.xlabel('T (K)')
plt.ylabel('$q_{rot}$')
plt.title('The rotational partition function of a CO molecule')
plt.show()
plt.plot(T,q_vib)
plt.xlabel('T (K)')
plt.ylabel('$q_{vib}$')
plt.title('The vibrational partition function of a CO molecule')
plt.show()
Explanation: 3. Plot the rotational and vibrational partition functions of a CO molecule in the bottle from $T$ =  200  to 2000 K (assume the CO remains a gas over the whole range). Hint: Use your answer to Problem 1 to simplify calculating the rotational partition function.
$q_{rot} = \frac{1}{\sigma}\frac{T}{\theta_{rot}} = \frac{T}{\theta_{rot}}$
$q_{vib} = \frac{1}{1-e^{-\theta_{vib}/T}}$
End of explanation
R = 8.31447 # J/(mol*K)
U_trans = 1.5*R*T
U_rot = R*T
U_vib = R*T_vib/(np.exp(T_vib/T)-1)
plt.plot(T,U_trans,label='U_trans')
plt.plot(T,U_rot,label='U_rot')
plt.plot(T,U_vib,label='U_vib')
plt.legend()
plt.xlabel('T (K)')
plt.ylabel('Internal Energy (J/mol)')
plt.title('Internal energies of CO in the bottle')
plt.show()
Explanation: 4. Plot the total translational, rotational, and vibrational energies of CO in the bottle from $T =$ 200 to 2000 K (assume the CO remains a gas over the whole range). Which (if any) of the three types of motions dominate the total energy?
$U_{trans} = \frac{3}{2}RT$
$U_{rot} = RT$
$U_{vib} = R\frac{\theta_{vib}}{e^{\theta_{vib}/T}-1}$.
End of explanation
Cv_trans = np.linspace(1.5*R,1.5*R,1000)
Cv_rot = np.linspace(R,R,1000)
Cv_vib = R*(T_vib/T*np.exp(T_vib/2./T)/(np.exp(T_vib/T)-1))**2
plt.plot(T,Cv_trans,label='Cv_trans')
plt.plot(T,Cv_rot,label='Cv_rot')
plt.plot(T,Cv_vib,label='Cv_vib')
plt.legend()
plt.xlabel('T (K)')
plt.ylabel('Heat Capacity (J/mol/K)')
plt.title('Constant volume molar heat capacities of CO in the bottle ')
plt.show()
print('While translational motion contributes the most to the molar heat capacity of CO,')
print('it does not dominate over rotational and vibrational motion.')
Explanation: 5. Plot the total translational, rotational, and vibrational constant volume molar heat capacities of CO in the bottle from $T =$ 200 to 2000 K. Which (if any) of the three types of motions dominate the heat capacity?
$C_{V,trans} = \frac{3}{2}R$
$C_{V,rot} = R$
$C_{V,vib} = R\left(\frac{\theta_{vib}}{T}\frac{e^{\theta_{vib}/2T}}{e^{\theta_{vib}/T}-1}\right)^2$
End of explanation
NA = 6.022e23
S_trans = R*np.log(np.exp(2.5)*V/NA/Lamda**3)
S_rot = R*(1-np.log(T_rot/T))
S_vib = R*(T_vib/T/(np.exp(T_vib/T)-1)-np.log(1-np.exp(-T_vib/T)))
A_trans = U_trans-T*S_trans
A_rot = U_rot-T*S_rot
A_vib = U_vib-T*S_vib
plt.plot(T,A_trans,label='A_trans')
plt.plot(T,A_rot,label='A_rot')
plt.plot(T,A_vib,label='A_vib')
plt.legend()
plt.xlabel('T (K)')
plt.ylabel('Helmholtz Energy (J/mol)')
plt.title('Helmholtz energies of CO in the bottle')
plt.show()
Explanation: 6. Plot the total translational, rotational, and vibrational Helmholtz energies of CO in the bottle from $T =$ 200 to 2000 K. Which (if any) of the three types of motions dominate the Helmholtz energy?
$A = U - TS$
$S_{trans} = Rln\left(\frac{e^{5/2}V}{N\Lambda^3}\right)$
$S_{rot} = R(1-ln(\theta_{rot}/T))$
$S_{vib} = R\left(\frac{\theta_{vib}/T}{e^{\theta_{vib}/T}-1}-ln(1-e^{-\theta_{vib}/T})\right)$.
End of explanation
V2 = 0.04 # m^3
deltaP = R*298*(1/V2-1/V)
deltaS = R*np.log(np.exp(2.5)*V2/NA/Lamda**3) - R*np.log(np.exp(2.5)*V/NA/Lamda**3)
deltaA = -deltaS*298
print('Delta P = {0:.3f} Pa, Delta U = 0, Delta A = {1:.3f} J/mol, and Delta S = {2:.3f} J/mol/K.'.format(deltaP,deltaA,deltaS))
Explanation: 7. Use your formulas to calculate $\Delta P$, $\Delta U$, $\Delta A$, and $\Delta S$ associated with isothermally expanding the gas from 20 dm$^3$ to 40 dm$^3$.
T = 298 K. $\Delta U=0$. 
$\Delta P = \frac{RT}{V_2} - \frac{RT}{V_1}$. 
$\Delta S = S_{trans,2} - S_{trans,1}$. 
$A = U-TS$, so, $\Delta A = -T\Delta S$.
End of explanation
import numpy as np
T = 298 # K
k = 1.38065e-23 # J/K
R = 8.31447 # J/(mol*K)
Na = 6.0221e23  # 1/mol
c = 6.0221e26  # 1/m^3, conversion factor of 1mol/L = 6.02e26 particles/m^3
autokJ = 2625.50
Eelec = [-412.90047 ,-312.57028 ,-100.31885 ] # kJ/mol
ZPE = [0.02889 ,0.01422 ,0.00925 ] # kJ/mol
dE0 = ((Eelec[1] + ZPE[1] + Eelec[2] + ZPE[2]) - (Eelec[0] + ZPE[0]))* autokJ # kJ/mol
u_trans = [3.7,3.7,3.7] #kJ/mol
u_rot = [3.7,3.7,2.5] #kJ/mol
u_vib = [4.3,1.2,0] #kJ/mol
dU = dE0 + (u_trans[1]+u_rot[1]+u_vib[1])+(u_trans[2]+u_rot[2]+u_vib[2])-(u_trans[0]+u_rot[0]+u_vib[0])#kJ/mol
print("delta_U = %.2f kJ/mol"%dU)
Explanation: Reactions from scratch
In 1996, Schneider and co-workers used quantum chemistry to compute the reaction pathway for unimolecular decomposition of trifluoromethanol, a reaction of relevance to the atmospheric degradation of hydrofluorocarbon refrigerants (J. Phys. Chem. 1996, 100, 6097- 6103, doi:10.1021/jp952703m): $$\mathrm{CF_3OH\rightarrow COF_2 + HF}$$
Following are some of the reported results, computed at 298 K:
|              | CF$_3$OH| C(O)F$_2$ | HF |    |
|:--------------|---------:|-----------:|----:|----:| 
| $E^\text{elec}$ | -412.90047 | -312.57028 | -100.31885 | (Hartree) |
| ZPE             | 0.02889 | 0.01422 | 0.00925 | (Hartree) |
| $U^\text{trans}$ | 3.7 | 3.7| 3.7 | (kJ mol$^{-1}$) |
| $U^\text{rot }$ | 3.7 | 3.7| 2.5 | (kJ mol$^{-1}$) |
| $U^\text{vib}$ | 4.3 | 1.2 | 0 |  (kJ mol$^{-1}$) |
| $q^{\text{trans}}/V$ | $7.72\times 10^{32}$ | $1.59\times 10^{32}$ |$8.65\times 10^{31}$ | (m$^{-3}$) |
| $q^\text{rot}$ | 61830 | 679 | 9.59 | |
| $q^\text{vib}$ | 2.33 | 1.16 | 1 | |
8. Using the data provided, determine $\Delta U^{\circ}$(298 K)), in kJ mol$^{-1}$, assuming ideal behavior and 1 M standard state.  Recall that $U(T)$ is the sum of the contributions of all degrees of freedom.
Remember that $E_0$ is contained in $\Delta{U}^{\circ},\Delta{A}^{\circ},$ and $\Delta{G}^{\circ}.$ For example:
$$\Delta U^{\circ} =\sum_{product}U_{trans} + U_{rot} + U_{vib} + (E_{elec} + ZPE) - \sum_{reactant} U_{trans} + U_{rot} + U_{vib} + (E_{elec} + ZPE)$$$$=\sum_{product}U_{trans} + U_{rot} + U_{vib} + E_0 - \sum_{reactant} U_{trans} + U_{rot} + U_{vib} + E_0$$
$\Delta U^{\circ} =\Delta(U_{trans} + U_{rot} + U_{vib}) + \Delta(E_0)$
End of explanation
q_trans = [7.72e32/c,1.59e32/c,8.65e31/c] # change translational partition functions from 1/m3 to mol/l std state
q_rot = [61830,679,9.59] # unitless
q_vib = [2.33,1.16,1] # unitless    
Q = (q_trans[1]*q_rot[1]*q_vib[1])*(q_trans[2]*q_rot[2]*q_vib[2])/(q_trans[0]*q_rot[0]*q_vib[0]) # total partition 
dA = dE0 + (-R*T*np.log(Q)- R*T)/1000 #kJ/mol
print("Q = %.2f"%Q)
print("delta_E0 = %.2f"%dE0)
print("delta_A = %.2f kJ/mol"%dA)
Explanation: 9. Using the data provided, determine  $\Delta A^{\circ}$(298 K) in kJ mol$^{-1}$, assuming ideal behavior and 1 M standard state.  Recall that $A^\circ=E^\text{elec} + \text{ZPE}-RT\ln(q^\circ)-RT$ and that $q^\circ =(q^\text{trans}/V)q^\text{rot}q^\text{vib}/c^\circ$ in units corresponding with the standard state.
$\Delta{A^{\circ}}= \big{[-k_BT\ln(q_{t}q_{r}q_{v})-k_BT]{COF_2}+[-k_BT\ln(q{t}q_{r}q_{v})-k_BT]{HF}-[-k_BT\ln(q{t}q_{r}q_{v})-k_BT]_{CF_3OH}\big} *N_A + \Delta(E_0)$
$\quad \quad = \Delta(E_0)-RT\ln(Q)-RT$
End of explanation
dG = dA + R*T/1000 #kJ/mol
print("delta_G = %.2f kJ/mol"%dG)
Explanation: 10. Determine $\Delta G^\circ$(298 K).  Recall that $G = A + PV = A + RT$ for an ideal ga.
$\Delta{G^{\circ}} = \Delta{A^{\circ}} + \Delta(PV)$
$\quad = \quad \Delta{A^{\circ}} + RT$
End of explanation
dS = 1000*(dU - dA)/T
print("delta_S = %.2f J/mol K"%dS)
Explanation: 11. Determine $\Delta S^{\circ}$(298 K), in J mol$^{-1}$ K$^{-1}$ , assuming a 1 M standard state. Recall that $S = (U - A)/T$.
$\Delta A^{\circ} = \Delta U^{\circ} - T\Delta S^{\circ}$
$\Delta S^{\circ} = \frac{\Delta U^{\circ} - \Delta A^{\circ}}{T}$
End of explanation
Kc = Q *np.exp(-dE0*1000/(R*T)) # Kc equation from lecture notes
print('Kc = %.3f (unitless). '%(Kc))
Kc = np.exp(-dG*1000/(R*T))
print('Kc = %.3f (unitless). '%(Kc))
Explanation: 12. Using the data provided, determine $K_c$ (298 K), assuming a 1 M standard state.  You may either determine from partition functions of from the relationship between $K_c$ and $\Delta G^\circ$.
$\mathrm{A} \rightarrow \mathrm{B + C}$$$K_c(T) = \frac{\frac{q_{\mathrm{B}}}{V}\frac{q_{\mathrm{C}}}{V}}{\frac{q_{\mathrm{A}}}{V}} \frac{1}{c^{\circ}} e^{-\Delta{E(0)}/kT} = e^{-\Delta G^\circ(T)/RT}, \quad \text{where} \quad \frac{q_i}{V} = \frac{q_{trans}}{V}q_{rot}q_{vib}$$
Note: $K_c(T)$ is formally unitless but "remembers" that it refers to 1 M standard state.
End of explanation
from sympy import * 
x = symbols('x',positive=True)
c = solve(x**2-(0.05-x)*Kc,x)
print('At equilibrium, CF3OH = %.2E mol/L, COF2 = %.5f mol/L, HF = %.5f mol/L.'%(0.05-c[0],c[0],c[0]))
print('At equilibrium, CF3OH = %.2E mol, COF2 = %.5f mol, HF = %.5f mol.'%((0.05-c[0])*20,c[0]*20,c[0]*20))
Explanation: 13. 1 mole of CF$_3$OH is generated in a 20 L vessel at 298 K and left long enough to come to equilibrium with respect to its decomposition reaction. What is the composition of the gas (concentrations of all the components) at equilibrium (in mol/L)?
1 mol/ 20 L = 0.05 mol/L
$\mathrm{A} \rightarrow \mathrm{B + C}$
$K_c = \frac{x^2}{0.05-x}$, solve for x. Note that x = $[\frac{mol}{L}]$.
End of explanation
dn = 2-1
R = 8.314/1000 #kJ/mol K
T = 298 #K
dH = dU+dn*R*T #kJ/mol
print("dH =",round(dH,3),"kJ/mol")
Explanation: 14. How, directionally, would your answer to Question 13 change if the vessel was at a higher temperature? Use the van'T Hoff relationship to determine the equilibrium constant and equilibrium concentrations at 273 and 323 K. How good was your guess?
From question #8, we know that at 298K,  $\Delta U$ = 18.64 kJ/mol. 
$\Delta H = \Delta U + \Delta (nRT) = \Delta U + RT\Delta n$
End of explanation
K1 = 2.926
T1 = 298 #K
T2 = 273 #K
K2 = K1*np.exp(-dH/R*(1/T2-1/T1))
print('K=', round(K2,4), 'at 273 K.')
x = symbols('x',positive=True)
c = solve(x**2-(0.05-x)*K2,x)
print('At equilibrium, CF3OH = %.2E mol/L, COF2 = %.5f mol/L, HF = %.5f mol/L.'%(0.05-c[0],c[0],c[0]))
K1 = 2.926
T1 = 298 #K
T2 = 323 #K
K2 = K1*np.exp(-dH/R*(1/T2-1/T1))
print('K=', round(K2,4), 'at 323 K.')
x = symbols('x',positive=True)
c = solve(x**2-(0.05-x)*K2,x)
print('At equilibrium, CF3OH = %.2E mol/L, COF2 = %.5f mol/L, HF = %.5f mol/L.'%(0.05-c[0],c[0],c[0]))
Explanation: Since $\Delta H$ is positive, we expect $K \uparrow$ as $T \uparrow$
The Van't Hoff equation states:
$ln\frac{K(T_2)}{K(T_1)}=\frac{-\Delta H^o}{R}(\frac{1}{T_2}-\frac{1}{T_1})$
We know from question 12, K = 2.926 at 298 K.
End of explanation
T = 298 #K
R = 8.314
Kc = np.exp(-dG*1000/(R*T))
print('At 298 K, Kc = %.3f (unitless). '%(Kc))
x = symbols('x',positive=True)
c = solve(x**2-(0.2-x)*Kc,x)
print('At equilibrium, CF3OH = %.2E mol/L, COF2 = %.5f mol/L, HF = %.5f mol/L.'%(0.2-c[0],c[0],c[0]))
print('At equilibrium, CF3OH = %.2E mol, COF2 = %.5f mol, HF = %.5f mol.'%((0.2-c[0])*5,c[0]*5,c[0]*5))
print('At a smaller volume, the concentration of products increases, but the number of moles decreases.')
Explanation: Therefore, the at higher temperatures, the reaction shifts towards the products.
15. How, directionally, would your answer to Question 13 change if the vessel had a volume of 5 L? Redo the calculation at this volume to verify your guess.
If V = 5L, the initial concentration = $\frac{1 mol}{5 L}=.2 M$
$K_c = \frac{x^2}{0.2-x}$
At 298 K, Kc = 2.926 (from problem 12)
End of explanation
T = 298 #K
#All values were taken from NIST
#Methanol
Hm = -205 #kJ/mol
Sm = .2399 #kJ/mol K
Gm = Hm - T*Sm #kJ/mol
#Hydrogen
Hh = 0
Sh = .13068 #J/mol K
Gh = Hh - T*Sh #kJ/mol
#Formaldehyde
Hf = -108.6 #kJ/mol
Sf = .21895 #kJ/mol K
Gf = Hf - T*Sf #kJ/mol
delta_H = Hf+Hh-Hm #kJ/mol
delta_S = Sf+Sh-Sm #kJ/mol K
delta_G = Gf+Gh-Gm #kJ/mol
print('Delta H =',delta_H,'kJ/mol.')
print('Delta S =',delta_S,'kJ/mol K.')
print('Delta G =',delta_G,'kJ/mol.')
print('Therefore, by substituting F with H makes the reaction less favorable.')
Explanation: 16. Consult a thermodynamics source (e.g. https://webbook.nist.gov/chemistry/) to determine $\Delta H^\circ$(298 K), $\Delta S^\circ$(298 K), and $\Delta G^\circ$(298 K) for the homologous reaction CH$_3$OH (g)$\rightarrow$ H$_2$O (g) + H$_2$CO (g).  Does the substitution of F by H make the reaction more or less favorable?
End of explanation
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