Chapter 15, Problem 15.104AP

Interpretation Introduction

Interpretation: The value of equilibrium constant and ratio of $P_{{\text{SO}}_2}$ to $P_{{\text{SO}}_3}$ for the given reaction is to be calculated.

Concept introduction: The equilibrium constant $\left(K_{\text{c}}\right)$ is expressed as,

$K_{\text{c}}=\frac{{\left[\text{Product}\right]}^y}{{\left[\text{Reactant}\right]}^x}$

To determine: The value of equilibrium constant and ratio of $P_{{\text{SO}}_2}$ to $P_{{\text{SO}}_3}$.

Answer to Problem 15.104AP

Solution

The value of equilibrium constant is $\underset{\_}{8.05\times 10^{-2}}$ and ratio of $P_{{\text{SO}}_2}$ to $P_{{\text{SO}}_3}$ is $\underset{\_}{7.68}$.

Explanation of Solution

Explanation

Given

The balanced chemical reaction is,

$2{\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons 2{\text{SO}}_3\left(g\right)$

The partial pressure of ${\text{O}}_2$ is $0.21\text{atm}$.

Temperature of the reaction is $700°\text{C}$.

The equilibrium of the reaction is calculated by formula,

$K_{\text{c}}=e^{-\frac{\Delta G^{\text{ο}}}{\text{R}\cdot T}}$ (1)

Where,

• $\Delta G^{\text{ο}}$ is the Gibbs free energy of the reaction.
• $\text{R}$ is the universal gas constant.
• $T$ is the temperature of the reaction.

Gibbs free energy of the reaction is calculated by the formula,

$\Delta G^{\text{ο}}=\Delta H-T\Delta S$ (2)

Where,

• $\Delta H$ is the enthalpy change of the reaction.
• $\Delta S$ is the entropy change of the reaction.

The enthalpy change of the given reaction is calculated by the formula,

$\Delta H=2\Delta H_{\text{f}}\left({\text{SO}}_3\right)-\Delta H_{\text{f}}\left(2\left({\text{SO}}_2\right)+{\text{O}}_2\right)$

The enthalpy of formation of ${\text{SO}}_3$ is $-395.7\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{SO}}_{\text{2}}$ is $-296.8\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{O}}_{\text{2}}$ is $0\text{kJ}/\text{mol}$.

Substitute the enthalpy of formation of reactant and product in above formula,

$\begin{array}{c}\Delta H=2\left(-395.7\right)-\left(2\left(-296.8\right)+0\right)\\ =-791.4-\left(-593.6\right)\\ =-197.8\text{kJ}/\text{mol}\end{array}$

Convert $-197.8\text{kJ}/\text{mol}$ to $\text{J}/\text{mol}$.

$\begin{array}{c}-197.8\text{kJ}/\text{mol}=-197.8\text{kJ}/\text{mol}\times 1000\text{J}\\ =-197800\text{J}/\text{mol}\end{array}$

The entropy change of the given reaction is calculated by the formula,

$\Delta S=2\Delta S^{\text{ο}}\left({\text{SO}}_3\right)-\Delta S^{\text{ο}}\left(2\left({\text{SO}}_2\right)+{\text{O}}_2\right)$

The standard entropy of ${\text{SO}}_3$ is $256.8{\text{JK}}^{-1}{\text{mol}}^{-1}$.

The standard entropy of ${\text{SO}}_{\text{2}}$ is $248.2{\text{JK}}^{-1}{\text{mol}}^{-1}$.

The standard entropy of ${\text{O}}_{\text{2}}$ is $205{\text{JK}}^{-1}{\text{mol}}^{-1}$.

Substitute the standard entropies of reactant and product in above formula,

$\begin{array}{c}\Delta S=2\left(256.8\right)-\left(2\left(248.2\right)+205\right)\\ =513.6-\left(701.4\right)\\ =-187.8{\text{JK}}^{-1}{\text{mol}}^{-1}\end{array}$

Convert $700°\text{C}$ to $\text{K}$.

$\begin{array}{c}700°\text{C}=700+273\text{K}\\ =973\text{K}\end{array}$

Substitute the values of enthalpy change, entropy change and temperature in equation (2).

$\begin{array}{c}\Delta G^{\text{ο}}=\left(-197800\right)-973\times \left(-187.8\right)\\ =-197800+182729.4\\ =-15070.6{\text{Jmol}}^{-1}\end{array}$

Substitute the value of Gibbs free energy in equation (1).

$\begin{array}{c}{K}_{\text{c}}=e^{-\frac{\left(-15070.6\right)}{8.314\times 973}}\\ =e^{-\frac{\left(-15070.6\right)}{8089.52}}\\ =e^{1.862}\\ =6.43\end{array}$

The relationship between $K_{\text{p}}$ and $K_{\text{c}}$ is shown by,

$K_{\text{p}}=K_{\text{c}}{\left(\text{R}T\right)}^{\Delta n}$

Where,

• $\Delta n$ is the change in number of moles from reactant to product.

The change in number of moles is calculated by formula,

$\Delta n=n_2-n_1$

Substitute the number of moles of reactant and product in above formula,

$\begin{array}{c}\Delta n=2-3\\ =-1\end{array}$

Substitute the values of $K_{\text{c}}$, universal gas constant, change in number of moles and temperature in above formula.

$\begin{array}{c}{K}_{\text{p}}=6.43{\left(0.08205\times 973\text{K}\right)}^{-1}\\ =6.43\times {\left(79.83\right)}^{-1}\\ =\underset{\_}{8.05\times 10^{-2}}\end{array}$

The expression of equilibrium constant for the given reaction is,

$K_{\text{p}}=\frac{{\left(P_{{\text{SO}}_3\left(g\right)}\right)}^2}{{\left(P_{{\text{SO}}_2\left(g\right)}\right)}^2\left(P_{{\text{O}}_2\left(g\right)}\right)}$

Substitute the values of equilibrium constant and partial pressure of air in above formula,

$\begin{array}{c}8.05\times {10}^{-2}=\frac{{\left(P_{{\text{SO}}_3\left(g\right)}\right)}^2}{{\left(P_{{\text{SO}}_2\left(g\right)}\right)}^2\left(0.21\right)}\\ {\left(\frac{\left(P_{{\text{SO}}_2\left(g\right)}\right)}{\left(P_{{\text{SO}}_3\left(g\right)}\right)}\right)}^2=\frac{1}{8.05\times {10}^{-2}\times \left(0.21\right)}\\ \frac{\left(P_{{\text{SO}}_2\left(g\right)}\right)}{\left(P_{{\text{SO}}_3\left(g\right)}\right)}=\sqrt{0.591\times {10}^2}\\ =\underset{\_}{7.68}\end{array}$

Conclusion

The value of equilibrium constant is $\underset{\_}{8.05\times 10^{-2}}$ and ratio of $P_{{\text{SO}}_2}$ to $P_{{\text{SO}}_3}$ is $\underset{\_}{7.68}$.