Chapter 17, Problem 76E

Hydrogen sulfide can be removed from natural gas by the reaction

$2{\text{H}}_{\text{2}}\text{S}\left(g\right)+{\text{SO}}_{\text{2}}\left(g\right)\rightleftharpoons 3\text{S}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Calculate ∆G° and K (at 298 K) for this reaction. Would this reaction be favored at a high or low temperature?

Interpretation Introduction

Interpretation: The reaction between ${\text{H}}_{\text{2}}\text{S}$ and ${\text{SO}}_{\text{2}}$ is given. The value of $\Delta G°$ and $K$ at a given temperature is to be calculated. The temperature conditions at which the stated reaction will be favored is to be stated.

Concept introduction: Equilibrium constant $K$, is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, then the free energy change will be,

$\begin{array}{c}\Delta G=\text{0}\\ Q=K\end{array}$

The expression for free energy change is,

$\Delta G°=-RT\ln \left(K\right)$

The expression for standard Gibbs free energy, $\Delta G°$, is,

$\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}$

Answer to Problem 76E

Answer

The value of $\Delta G°$ for the given reaction is $\underset{\_}{-90kJ}$.

The value of $K$ for the given reaction is $\underset{\_}{5.94\times 10^{15}}$.

The forward reaction will be favored at a low temperature, whereas the reverse reaction will be favored at a high temperature.

Explanation of Solution

Explanation

The reaction that takes place is,

${\text{2H}}_{\text{2}}\text{S}\left(g\right)+{\text{SO}}_{\text{2}}\left(g\right)\stackrel{}{\to }\text{3S}\left(s\right)+{\text{2H}}_{\text{2}}\text{O}\left(g\right)$

Refer Appendix $\text{4}$.

The value of $\Delta G°\left(\text{kJ/mol}\right)$ for the given reactant and product is,

Molecules	$\Delta G°\left(\text{kJ/mol}\right)$
${\text{SO}}_{\text{2}}(g)$	$-\text{300}$
$\text{S}(g)$	$\text{0}$
${\text{H}}_{\text{2}}\text{S}(g)$	$-\text{34}$
${\text{H}}_{\text{2}}\text{O}(g)$	$-\text{229}$

The formula of $\Delta G°$ is,

$\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}$

Where,

• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta G{°}_{\left(\text{product}\right)}$ is the free energy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta G{°}_{\left(\text{reactant}\right)}$ is the free energy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute the required values from the table in the above equation.

$\begin{array}{c}\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}\\ =\left[\text{3}\left(\text{0}\right)+\text{2}\left(-\text{229}\right)-\left\{\text{2}\left(-\text{34}\right)+\left(-\text{300}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-90kJ}\end{array}$

The value of $\Delta G°$ is $-\text{90}\text{kJ}$.

Temperature is $\text{298}\text{K}$.

The conversion of kilo-joule $\left(\text{kJ}\right)$ into joule $\left(\text{J}\right)$ is done as,

$\text{1}\text{kJ}={\text{10}}^{\text{3}}\text{J}$

Hence,

The conversion of $-\text{90}\text{kJ}$ into joule is,

$\begin{array}{c}-\text{198}\text{kJ}=\left(-\text{90}\text{kJ}\times {\text{10}}^{\text{3}}\right)\text{J}\\ =-\text{90}\text{kJ}\times {\text{10}}^{\text{3}}\text{J}\end{array}$

Formula

The expression for free energy change is,

$\Delta G=\Delta G°+RT\ln \left(Q\right)$

Where,

• $\Delta G$ is the free energy change for a reaction at specified pressure.
• $R$ is the gas law constant $(\text{8}\text{.3145}\text{J/K}\cdot \text{mol})$.
• $T$ is the absolute temperature.
• $Q$ is the reaction constant.

Since, the reaction is at equilibrium, the free energy change,

$\begin{array}{c}\Delta G=\text{0}\\ Q=K\end{array}$

Where,

$K$ is the equilibrium constant.

Substitute these values in the free energy change expression.

$\begin{array}{c}\Delta G=\Delta G°+RT\ln \left(Q\right)\\ \Delta G°=-RT\ln \left(K\right)\\ \ln \left(K\right)=-\frac{\Delta G°}{RT}\end{array}$

Substitute the values of $R,\Delta G°$ and $T$ in the above expression.

$\begin{array}{c}\ln \left(K\right)=-\frac{\Delta G°}{RT}\\ =-\frac{-\text{90}\times {\text{10}}^{\text{3}}\text{J}}{(\text{8}\text{.3145}\text{J/K})\left(298\text{K}\right)}\\ K=\underset{\_}{5.94\times 10^{15}}\end{array}$

The value of $\Delta H°\left(\text{kJ/mol}\right)$ for the given reactant and product is,

Molecules	$\Delta H°\left(\text{kJ/mol}\right)$
${\text{SO}}_{\text{2}}(g)$	$-\text{297}$
$\text{S}(g)$	$\text{0}$
${\text{H}}_{\text{2}}\text{S}(g)$	$-\text{21}$
${\text{H}}_{\text{2}}\text{O}(g)$	$-\text{242}$

The formula of $\Delta H°$ is,

$\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}$

Where,

• $\Delta H°$ is the standard enthalpy of reaction.
• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta H{°}_{\left(\text{product}\right)}$ is the standard enthalpy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta H{°}_{\left(\text{reactant}\right)}$ is the standard enthalpy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta H°=\sum n_{\text{p}}\Delta H{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H{°}_{\left(\text{reactant}\right)}\\ =\left[\text{3}\left(\text{0}\right)+\text{2}\left(-\text{242}\right)-\left\{\text{2}\left(-\text{21}\right)+\left(-\text{297}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-145kJ}\end{array}$

A forward reaction will be favored at low temperature if,

$\Delta H°<\text{0}$

The value of $\Delta S°\left(\text{J/K}\cdot \text{mol}\right)$ for the given reactant and product is,

Molecules	$\Delta S°\left(\text{J/K}\cdot \text{mol}\right)$
${\text{SO}}_{\text{2}}(g)$	$\text{248}$
$\text{S}(g)$	$\text{32}$
${\text{H}}_{\text{2}}\text{S}(g)$	$\text{206}$
${\text{H}}_{\text{2}}\text{O}(g)$	$\text{189}$

The formula of $\Delta S°$ is,

$\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}$

Where,

• $\Delta S°$ is the standard enthalpy of reaction.
• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the  number of moles each reactant.
• $\Delta S{°}_{\left(\text{product}\right)}$ is the standard enthalpy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta S{°}_{\left(\text{reactant}\right)}$ is the standard enthalpy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta S°=\sum n_{\text{p}}\Delta S{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S{°}_{\left(\text{reactant}\right)}\\ =\left[\text{3}\left(\text{32}\right)+\text{2}\left(\text{189}\right)-\left\{\text{2}\left(\text{206}\right)+\left(\text{248}\right)\right\}\right]\text{J/K}\cdot \text{mol}\\ =\underset{\_}{-186kJ}\end{array}$

A reverse reaction will be favored at high temperature if,

$\Delta S°<\text{0}$

Conclusion

Conclusion

The calculated value of $\Delta G°$ for the given reaction is $\underset{\_}{-90kJ}$ and the value of $K$ for the given reaction is $\underset{\_}{5.94\times 10^{15}}$. The forward reaction will be favored at a low temperature, whereas the reverse reaction will be favored at a high temperature