Chapter 10, Problem 10.84PAE

Methane can be produced from CO and H₂.The process might be done in two steps, as shown below, with each step carried out in a separate reaction vessel within the production plant.

Reaction 1 $CO\left(g\right)+{\text{2H}}_{\text{2}}\left(\text{g}\right)\to {\text{CH}}_{\text{3}}\text{OH}(\mathcal{l})$

$\Delta \text{S}°$ = -532 J/K

Reaction 2 ${\text{CH}}_{\text{3}}\text{OH}(\mathcal{l})\to {\text{CH}}_{\text{4}}\left(\text{g}\right)+\frac{1}{2}{O}_2(g)$

   $\Delta \text{S}°$= +162 J/K

NOTE: You should be able to work this problem without using any additional tabulated data.

(a) Calculate $\Delta \text{H}°$ for reaction 1.

(b) Calculate $\Delta {\text{G}}_f°$ for CO(g).

(c) Calculate S° for O₂(g).

(d) At what temperatures is reaction 1 spontaneous?

(e) Suggest a reason why these two steps would need to be carried out separately.

Substance	$\Delta {\text{H}}_f°$ (kJ mol-1)	$\Delta {\text{G}}_f°$ (kJ mol-1)	$\text{S}°$ (J mol-1 K-1)
CO(s)	-110.5		197.674
CH3OH(    $\mathcal{l}$ )	-238.7	-166.4	126.8
CH4(g)	-74.8		186.2

Interpretation Introduction

Interpretation:

Thermodynamic entropy, enthalpy and Gibb’s free energy as asked must be calculated for the reactions given.

Concept Introduction:

Enthalpy change for a process is determined as:

$\Delta H^{\text{o}}={\displaystyle \sum _i{v}_i}\Delta H_{\text{f}}{}^{\text{o}}{\left(\text{product}\right)}_i-{\displaystyle \sum _j{v}_j}\Delta H_{\text{f}}{}^{\text{o}}{\left(\text{reactant}\right)}_j$

where v is for stoichiometric coefficients

Similarly entropy change for a process is determined as:

$\Delta S^{\text{o}}={\displaystyle \sum _i{v}_i}{S}^{\text{o}}{\left(\text{product}\right)}_i-{\displaystyle \sum _j{v}_j}{S}^{\text{o}}{\left(\text{reactant}\right)}_j$

where v is for stoichiometric coefficients and $S^{\text{o}}$ is for absolute entropies of reactants and products..

Gibb’s free energy is a state function which predicts whether a process is spontaneous or not at conditions of constant pressure and temperature. Gibb’s free energy change for a process at constant temperature is defined as:

$\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\Delta S^{\text{o}}$

where $\Delta S$ is the total change in the entropy of the system and $\Delta H$ is the total change in the enthalpy of the system. For a spontaneous process $\Delta G$ must be negative.

Similar to enthalpy and entropy, standard Gibb’s free energy change can be calculated as:

$\Delta G^{\text{o}}={\displaystyle \sum _i{v}_i}\Delta G_{\text{f}}{}^{\text{o}}{\left(\text{product}\right)}_i-{\displaystyle \sum _j{v}_j}\Delta G_{\text{f}}{}^{\text{o}}{\left(\text{reactant}\right)}_j$

Also $\Delta H_{\text{f}}{}^{\text{o}}$ and $\Delta G_{\text{f}}{}^{\text{o}}$ for elements in their standard states is zero.

Answer to Problem 10.84PAE

Solution:

a)$\Delta H^{\text{o}}$ for reaction 1 is $-128.2{\text{ kJ mol}}^{-1}$.

b)$\Delta G_{\text{f}}{}^{\text{o}}$ for $\text{CO}\left(\text{g}\right)$ is $-137.1{\text{ kJ mol}}^{-1}$.

c)$S^{\text{o}}$ for ${\text{O}}_2\left(\text{g}\right)$ is $205.2{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

d) Hence reaction 1 is spontaneous at lower temperatures.

e) Since reaction 1 is spontaneous at lower temperatures and reaction 2 is spontaneous at higher temperatures, they must be carried out separately.

a)

Explanation of Solution

Given reaction 1 is:

$\text{CO}\left(\text{g}\right)+2{\text{ H}}_2\left(\text{g}\right)\to {\text{CH}}_3\text{OH}\left(l\right)$

Calculate enthalpy change the above reaction that is $\Delta H^{\text{o}}$, enthalpy of formation for hydrogen would be zero.

$\begin{array}{c}\Delta H^{\text{o}}=\Delta H_{\text{f}}{}^{\text{o}}\left[{\text{CH}}_3\text{OH}\left(l\right)\right]-\Delta H_{\text{f}}{}^{\text{o}}\left[\text{CO}\left(\text{g}\right)\right]-2\Delta H_{\text{f}}{}^{\text{o}}\left[{\text{H}}_2\left(\text{g}\right)\right]\\ =-238.7{\text{ kJ mol}}^{-1}-\left(-110.5{\text{ kJ mol}}^{-1}\right)-0\\ =-128.2{\text{ kJ mol}}^{-1}\end{array}$

b)

Using the value of enthalpy change of reaction 1 calculated in part a) and the given entropy change calculate the Gibb’s free energy change for reaction 1. Since we are using standard values the temperature will be 298 K.

$\begin{array}{l}\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\Delta S^{\text{o}}\\ =-128.2{\text{ kJ mol}}^{-1}-\left(298\text{ K}\times -332{\text{ J K}}^{-1}{\text{ mol}}^{-1}\right)\\ =-128.2\times {10}^3{\text{ J mol}}^{-1}+98936{\text{ J mol}}^{-1}\\ =-29.3\times {10}^3{\text{ J mol}}^{-1}\\ =-29.3{\text{ kJ mol}}^{-1}\end{array}$

Now use the formula described above in the concept introduction to calculate $\Delta G_{\text{f}}{}^{\text{o}}$ for $\text{CO}\left(\text{g}\right)$.

$\begin{array}{c}\Delta G^{\text{o}}=\Delta G_{\text{f}}{}^{\text{o}}\left[{\text{CH}}_3\text{OH}\left(l\right)\right]-\Delta G_{\text{f}}{}^{\text{o}}\left[\text{CO}\left(\text{g}\right)\right]-2\Delta G_{\text{f}}{}^{\text{o}}\left[{\text{H}}_2\left(\text{g}\right)\right]\\ -29.3{\text{ kJ mol}}^{-1}=-166.4{\text{ kJ mol}}^{-1}-\Delta G_{\text{f}}{}^{\text{o}}\left[\text{CO}\left(\text{g}\right)\right]-0\\ \Delta G_{\text{f}}{}^{\text{o}}\left[\text{CO}\left(\text{g}\right)\right]=-137.1{\text{ kJ mol}}^{-1}\end{array}$

c)

Given reaction 2 is:

${\text{CH}}_3\text{OH}\left(l\right)\to {\text{CH}}_4\left(\text{g}\right)+1/2{\text{ O}}_2\left(\text{g}\right)$

The entropy change for the above reaction is $\Delta S^{\text{o}}=+162{\text{ J K}}^{-1}{\text{ mol}}^{-1}$, calculate $S^{\text{o}}$ for ${\text{O}}_2\left(\text{g}\right)$.

$\begin{array}{c}\Delta S^{\text{o}}=\frac{1}{2}{S}^{\text{o}}\left[{\text{O}}_2\left(\text{g}\right)\right]+S^{\text{o}}\left[{\text{CH}}_4\left(\text{g}\right)\right]-S^{\text{o}}\left[{\text{CH}}_3\text{OH}\left(l\right)\right]\\ +162{\text{ J K}}^{-1}{\text{ mol}}^{-1}=\frac{1}{2}{S}^{\text{o}}\left[{\text{O}}_2\left(\text{g}\right)\right]+186.2{\text{ J K}}^{-1}{\text{ mol}}^{-1}-126.8{\text{ J K}}^{-1}{\text{ mol}}^{-1}\\ \frac{1}{2}{S}^{\text{o}}\left[{\text{O}}_2\left(\text{g}\right)\right]=102.6{\text{ J K}}^{-1}{\text{ mol}}^{-1}\\ S^{\text{o}}\left[{\text{O}}_2\left(\text{g}\right)\right]=205.2{\text{ J K}}^{-1}{\text{ mol}}^{-1}\end{array}$

d)

Gibb’s free energy is a state function which predicts whether a process is spontaneous or not at conditions of constant pressure and temperature. Gibb’s free energy change for a process at constant temperature is defined as:

$\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\Delta S^{\text{o}}$

Where $\Delta S$ is the total change in the entropy of the system and $\Delta H$ is the total change in the enthalpy of the system. For a spontaneous process $\Delta G$ must be negative.

Given reaction 1 is:

$\text{CO}\left(\text{g}\right)+2{\text{ H}}_2\left(\text{g}\right)\to {\text{CH}}_3\text{OH}\left(l\right)$

The entropy change for the above reaction is $\Delta S^{\text{o}}=-332{\text{ J K}}^{-1}{\text{ mol}}^{-1}$. The value of enthalpy change of reaction 1 calculated in part a) is $-128.2{\text{ kJ mol}}^{-1}$. Since both $\Delta H<0$ and $\Delta S<0$ Gibb’s free energy depends on temperature. At higher temperatures the term $T\Delta S$ dominates, so the process will be spontaneous at lower temperatures where $\Delta H$ dominates.

Hence reaction 1 is spontaneous at lower temperatures.

e)

As explained above in part c) reaction 1 is spontaneous at lower temperatures. Similarly given reaction 2 is:

${\text{CH}}_3\text{OH}\left(l\right)\to {\text{CH}}_4\left(\text{g}\right)+1/2{\text{ O}}_2\left(\text{g}\right)$

The entropy change for the above reaction is $\Delta S^{\text{o}}=+162{\text{ J K}}^{-1}{\text{ mol}}^{-1}$

. Calculate enthalpy change for this reaction:

$\begin{array}{c}\Delta H^{\text{o}}=\frac{1}{2}\Delta H_{\text{f}}{}^{\text{o}}\left[{\text{O}}_2\left(\text{g}\right)\right]+\Delta H_{\text{f}}{}^{\text{o}}\left[{\text{CH}}_4\left(\text{g}\right)\right]-\Delta H_{\text{f}}{}^{\text{o}}\left[{\text{CH}}_3\text{OH}\left(l\right)\right]\\ =0+\left(-74.8{\text{ kJ mol}}^{-1}\right)-\left(-238.7{\text{ kJ mol}}^{-1}\right)\\ =164{\text{ kJ mol}}^{-1}\end{array}$

Gibb’s free energy is a state function which predicts whether a process is spontaneous or not at conditions of constant pressure and temperature. Gibb’s free energy change for a process at constant temperature is defined as:

$\Delta G^{\text{o}}=\Delta H^{\text{o}}-T\Delta S^{\text{o}}$

Where $\Delta S$ is the total change in the entropy of the system and $\Delta H$ is the total change in the enthalpy of the system. For a spontaneous process $\Delta G$ must be negative.

Now since both $\Delta H>0$ and $\Delta S>0$ Gibb’s free energy again depends on temperature. At higher temperatures the term $T\Delta S$ dominates, so the process will be spontaneous at higher temperatures where $\Delta S$ dominates.

Since reaction 1 is spontaneous at lower temperatures and reaction 2 is spontaneous at higher temperatures, they must be carried out separately.