Chapter 19, Problem 111IE

(a)

Interpretation Introduction

To determine: The value of equilibrium constant for the given reaction at $298\text{K}$ .

Solution: The value of equilibrium constant for the given reaction is $7.856\times {10}^{15}$ .

Explanation:

Given

The balanced chemical equation for the given reaction is,

${\text{SO}}_2\left(g\right)+2{\text{H}}_2\text{S}\left(g\right)\rightleftharpoons 3\text{S}\left(s\right)+2{\text{H}}_2\text{O}\left(g\right)$

The temperature of the reaction is $298\text{K}$ .

The standard free energy change of the given reaction is calculated by the formula,

$\begin{array}{c}\Delta G^0={\displaystyle \sum n\Delta G^0\left(\text{products}\right)}-{\displaystyle \sum m\Delta G^0\left(\text{reactants}\right)}\\ =\left(2\Delta G_{{\text{H}}_2\text{O}\left(g\right)}^0+3\Delta G_{\text{S}\left(s\right)}^0\right)-\left(\Delta G_{{\text{SO}}_2\left(g\right)}^0+2\Delta G_{{\text{H}}_2\text{S}\left(g\right)}^0\right)\end{array}$

Substitute the values of $\Delta G_{{\text{H}}_2\text{O}\left(g\right)}^0$ , $\Delta G_{\text{S}\left(s\right)}^0$ , $\Delta G_{{\text{SO}}_2\left(g\right)}^0$ and $\Delta G_{{\text{H}}_2\text{S}\left(g\right)}^0$ from appendix C in equation.

$\begin{array}{c}\Delta H^0=\left(2\times -228.57\text{kJ/mol}+3\times 0\text{kJ/mol}\right)-\left(1\times -300.4\text{kJ/mol}+2\times -33.01\text{kJ/mol}\right)\\ =-90.72\text{kJ/mol}\end{array}$

The conversion of free energy from $\text{kJ/mol}$ into $\text{J/mol}$ is done as,

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

Hence, the conversion of free energy from $-90.72\text{kJ/mol}$ into $\text{J/mol}$ is,

$-90.72\text{kJ/mol}=-90.72\times {10}^3\text{J/mol}$

The equilibrium constant for the given reaction is calculated by the formula,

$\begin{array}{l}\Delta G^0=-RT\ln K\\ \ln K=\frac{-\Delta G^0}{RT}\end{array}$

Where,

• $\Delta G^0$ is the standard free energy of the reaction.
• $R$ is the gas constant $\left(8.314\text{J/mol}\cdot \text{K}\right)$ .
• $T$ is the temperature of the reaction.
• $K$ is the equilibrium constant of the reaction.

Substitute the values of $\Delta G^0$ , $T$ and $R$ in above equation.

$\begin{array}{c}\ln K=\frac{-\left(-90.72\times {10}^3\text{J/mol}\right)}{8.314\text{J/mol}\cdot \text{K}\times 298\text{K}}\\ =36.6\end{array}$

The equilibrium constant of the reaction is calculated by taking exponential of the above equation.

$\begin{array}{c}K=e^{36.6}\\ =7.856\times {10}^{15}\end{array}$

Conclusion:

The value of equilibrium constant for the given reaction is $7.856\times {10}^{15}$ .

(b)

Interpretation Introduction

To determine: If the given reaction is feasible to remove ${\text{SO}}_2$ or not.

(c)

Interpretation Introduction

To determine: The equilibrium pressure of ${\text{SO}}_2$ gas at $298\text{K}$ .

(d)

Interpretation Introduction

To determine: If the given reaction is more or less effective at higher temperature.