Chapter 12, Problem 12.78PAE

(a)

Interpretation Introduction

Interpretation: Using the values from Appendix E, the equilibrium constant, ${\text{K}}_{\text{P}}$ for the given reaction should be calculated.${\text{H}}_{\text{2}}\left(\text{g}\right){\text{+Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons \text{2HCl(g)}$

Concept introduction:

For a reaction, change in standard Gibbs free energy is calculated as follows:$\Delta {\text{G}}_{\text{reaction}}^{\text{o}}={\displaystyle \sum \Delta {\text{G}}_{\text{product}}^{\text{o}}-{\displaystyle \sum \Delta {\text{G}}_{\text{reactant}}^{\text{o}}}}$

Here, $\Delta {\text{G}}_{\text{product}}^{\text{o}}$ and $\Delta {\text{G}}_{\text{reactant}}^{\text{o}}$ is change in standard Gibbs free energy of product and reactant respectively.

The change in standard Gibbs free energy is related to equilibrium constant as follows:$\Delta {\text{G}}_{\text{reaction}}^{\text{o}}=-\mathrm{RT}\ln \text{K}$

Here, R is Universal gas constant, T is absolute temperature and K is equilibrium constant of the reaction.

Answer to Problem 12.78PAE

Solution:

${\text{K}}_{\text{p}}=\text{2}.\text{6}\times \text{1}{0}^{\text{33}}$

Explanation of Solution

The given reaction is as follows:${\text{H}}_{\text{2}}\left(\text{g}\right){\text{+Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons \text{2HCl(g)}$

First step is to calculate the value of $\Delta {\text{G}}_{\text{reaction}}^{\text{o}}$ from the data given in Appendix E.

From the formula:$\begin{array}{c}{\text{ΔG}}_{\text{reaction}}^{\text{o}}{\text{=2ΔG}}_{\text{HCl}}^{\text{o}}-\left[{\text{ΔG}}_{{\text{H}}_{\text{2}}}^{\text{o}}{\text{+ΔG}}_{{\text{Cl}}_{\text{2}}}^{\text{o}}\right]\\ =2\text{ mol}\times \left(-95.3\text{ kJ/mol}\right)-0\\ =-190.6\text{ kJ}\end{array}$

From the calculated value of $\Delta {\text{G}}_{\text{reaction}}^{\text{o}}$ calculate the value of equilibrium constant ${\text{K}}_{\text{p}}$

${\text{ΔG}}_{\text{reaction}}^{\text{o}}\text{=}-{\text{RTlnK}}_{\text{p}}$

The absolute temperature is 298 .15 K, putting all the values,$\begin{array}{c}-190.6\times {10}^3\text{J}=-\left(8.314\text{ J/mol K}\right)\left(298.15\text{ K}\right)\ln {\text{K}}_{\text{p}}\\ 76.93={\text{lnK}}_{\text{p}}\\ {\text{K}}_{\text{p}}=2.6\times {10}^{33}\end{array}$

(b)

Interpretation Introduction

Interpretation: Using the values from Appendix E, the equilibrium constant, ${\text{K}}_{\text{P}}$ for the given reaction should be calculated.${\text{CH}}_{\text{4}}\left(\text{g}\right){\text{ + H}}_{\text{2}}\text{O }\left(\text{g}\right)\rightleftharpoons \text{ CO }\left(\text{g}\right){\text{ + 3H}}_{\text{2}}\left(\text{g}\right)$

Concept introduction:

For a reaction, change in standard Gibbs free energy is calculated as follows:$\Delta {\text{G}}_{\text{reaction}}^{\text{o}}={\displaystyle \sum \Delta {\text{G}}_{\text{product}}^{\text{o}}-{\displaystyle \sum \Delta {\text{G}}_{\text{reactant}}^{\text{o}}}}$

Here, $\Delta {\text{G}}_{\text{product}}^{\text{o}}$ and $\Delta {\text{G}}_{\text{reactant}}^{\text{o}}$ is change in standard Gibbs free energy of product and reactant respectively.

The change in standard Gibbs free energy is related to equilibrium constant as follows:$\Delta {\text{G}}_{\text{reaction}}^{\text{o}}=-\mathrm{RT}\ln \text{K}$

Here, R is Universal gas constant, T is absolute temperature and K is equilibrium constant of the reaction.

Answer to Problem 12.78PAE

Solution:

${\text{K}}_{\text{p }}=\text{8}.\text{6}\times \text{1}{0}^{-\text{26}}$

Explanation of Solution

The given reaction is as follows:${\text{CH}}_{\text{4}}\left(\text{g}\right){\text{ + H}}_{\text{2}}\text{O }\left(\text{g}\right)\rightleftharpoons \text{ CO }\left(\text{g}\right){\text{ + 3H}}_{\text{2}}\left(\text{g}\right)$ First step is to calculate the value of $\Delta {\text{G}}_{\text{reaction}}^{\text{o}}$ from the data given in Appendix E.

From the formula:$\begin{array}{c}{\text{ΔG}}_{\text{reaction}}^{\text{o}}{\text{=ΔG}}_{\text{CO}}^{\text{o}}+3{\text{ΔG}}_{{\text{H}}_{\text{2}}}^{\text{o}}-\left[{\text{ΔG}}_{{\text{CH}}_4}^{\text{o}}{\text{+ΔG}}_{{\text{H}}_{\text{2}}\text{O}}^{\text{o}}\right]\\ =1\text{ mol}\left(-137\text{ kJ/mol}\right)+3\text{ mol}\left(0\text{ kJ/mol}\right)-\left(1\text{ mol}\left(-51\text{ kJ/mol}\right)+1\text{ mol}\left(-229\text{ kJ/mol}\right)\right)\\ =143\text{ kJ}\end{array}$

From the calculated value of $\Delta {\text{G}}_{\text{reaction}}^{\text{o}}$ calculate the value of equilibrium constant ${\text{K}}_{\text{p}}$

${\text{ΔG}}_{\text{reaction}}^{\text{o}}\text{=}-{\text{RTlnK}}_{\text{p}}$

The absolute temperature is 298 .15 K, putting all the values,

$\begin{array}{c}143\times {10}^3\text{J}=-\left(8.314\text{ J/mol K}\right)\left(298.15\text{ K}\right)\ln {\text{K}}_{\text{p}}\\ 57.98={\text{lnK}}_{\text{p}}\\ {\text{K}}_{\text{p}}=8.6\times {10}^{-26}\end{array}$

(c)

Interpretation Introduction

Interpretation: Using the values from Appendix E, the equilibrium constant, ${\text{K}}_{\text{P}}$ for the given reaction should be calculated.${\text{SO}}_{\text{2}}\left(\text{g}\right){\text{ + Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{ SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{g}\right)$

Concept introduction:

For a reaction, change in standard Gibbs free energy is calculated as follows:$\Delta {\text{G}}_{\text{reaction}}^{\text{o}}={\displaystyle \sum \Delta {\text{G}}_{\text{product}}^{\text{o}}-{\displaystyle \sum \Delta {\text{G}}_{\text{reactant}}^{\text{o}}}}$

Here, $\Delta {\text{G}}_{\text{product}}^{\text{o}}$ and $\Delta {\text{G}}_{\text{reactant}}^{\text{o}}$ is change in standard Gibbs free energy of product and reactant respectively.

The change in standard Gibbs free energy is related to equilibrium constant as follows:$\Delta {\text{G}}_{\text{reaction}}^{\text{o}}=-\mathrm{RT}\ln \text{K}$

Here, R is Universal gas constant, T is absolute temperature and K is equilibrium constant of the reaction.

Answer to Problem 12.78PAE

Solution:

${\text{K}}_{\text{p }}=62.5$

Explanation of Solution

The given reaction is as follows:${\text{SO}}_{\text{2}}\left(\text{g}\right){\text{ + Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{ SO}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(\text{g}\right)$ First step is to calculate the value of $\Delta {\text{G}}_{\text{reaction}}^{\text{o}}$ from the data given in Appendix E.

From the formula:$\begin{array}{c}{\text{ΔG}}_{\text{reaction}}^{\text{o}}{\text{=ΔG}}_{{\text{SO}}_{\text{2}}{\text{Cl}}_{\text{2}}}^{\text{o}}-\left[{\text{ΔG}}_{{\text{SO}}_{\text{2}}}^{\text{o}}{\text{+ΔG}}_{{\text{Cl}}_{\text{2}}}^{\text{o}}\right]\\ =1\text{ mol}\left(-310.45\text{ kJ/mol}\right)-\left(1\text{ mol}\left(-300.2\text{ kJ/mol}\right)+1\text{ mol}\left(0\text{ kJ/mol}\right)\right)\\ =-10.25\text{ kJ}\end{array}$

From the calculated value of $\Delta {\text{G}}_{\text{reaction}}^{\text{o}}$ calculate the value of equilibrium constant ${\text{K}}_{\text{p}}$

${\text{ΔG}}_{\text{reaction}}^{\text{o}}\text{=}-{\text{RTlnK}}_{\text{p}}$

The absolute temperature is 298 .15 K, putting all the values,$\begin{array}{c}-10.25\times {10}^3\text{J}=-\left(8.314\text{ J/mol K}\right)\left(298.15\text{ K}\right)\ln {\text{K}}_{\text{p}}\\ 4.13={\text{lnK}}_{\text{p}}\\ {\text{K}}_{\text{p}}=62.5\end{array}$

(d)

Interpretation Introduction

Interpretation: Using the values from Appendix E, the equilibrium constant, ${\text{K}}_{\text{P}}$ for the given reaction should be calculated.$\text{2HCl }\left(\text{g}\right){\text{ +F}}_2\left(\text{g}\right)\rightleftharpoons \text{2HF}\left(\text{g}\right){\text{+Cl}}_{\text{2}}\left(\text{g}\right)$

Concept introduction:

For a reaction, change in standard Gibbs free energy is calculated as follows:$\Delta {\text{G}}_{\text{reaction}}^{\text{o}}={\displaystyle \sum \Delta {\text{G}}_{\text{product}}^{\text{o}}-{\displaystyle \sum \Delta {\text{G}}_{\text{reactant}}^{\text{o}}}}$

Here, $\Delta {\text{G}}_{\text{product}}^{\text{o}}$ and $\Delta {\text{G}}_{\text{reactant}}^{\text{o}}$ is change in standard Gibbs free energy of product and reactant respectively.

The change in standard Gibbs free energy is related to equilibrium constant as follows:$\Delta {\text{G}}_{\text{reaction}}^{\text{o}}=-\mathrm{RT}\ln \text{K}$

Here, R is Universal gas constant, T is absolute temperature and K is equilibrium constant of the reaction.

Answer to Problem 12.78PAE

Solution:

${\text{K}}_{\text{p }}=2.17\times {10}^{62}$

Explanation of Solution

The given reaction is as follows:$\text{2HCl }\left(\text{g}\right){\text{ +F}}_2\left(\text{g}\right)\rightleftharpoons \text{2HF}\left(\text{g}\right){\text{+Cl}}_{\text{2}}\left(\text{g}\right)$ First step is to calculate the value of $\Delta {\text{G}}_{\text{reaction}}^{\text{o}}$ from the data given in Appendix E.

From the formula:$\begin{array}{c}{\text{ΔG}}_{\text{reaction}}^{\text{o}}{\text{=2ΔG}}_{\text{HF}}^{\text{o}}{\text{+ΔG}}_{{\text{Cl}}_{\text{2}}}^{\text{o}}-\left[{\text{2ΔG}}_{\text{HCl}}^{\text{o}}{\text{+ΔG}}_{{\text{F}}_{\text{2}}}^{\text{o}}\right]\\ =2\text{ mol}\left(-273.2\text{ kJ/mol}\right)+0-\left(\text{2 mol}\left(-95.3\text{ kJ/mol}\right)+0\right)\\ =-355.8\text{ kJ}\end{array}$

From the calculated value of $\Delta {\text{G}}_{\text{reaction}}^{\text{o}}$ calculate the value of equilibrium constant ${\text{K}}_{\text{p}}$

${\text{ΔG}}_{\text{reaction}}^{\text{o}}\text{=}-{\text{RTlnK}}_{\text{p}}$

The absolute temperature is 298 .15 K, putting all the values,$\begin{array}{c}-355.8\times {10}^3\text{J}=-\left(8.314\text{ J/mol K}\right)\left(298.15\text{ K}\right)\ln {\text{K}}_{\text{p}}\\ 143.53={\text{lnK}}_{\text{p}}\\ {\text{K}}_{\text{p}}=2.17\times {10}^{62}\end{array}$