Chapter 24, Problem 63QP

Interpretation Introduction

Interpretation:

The value of ${\text{K}}_{\text{C}}{\text{, K}}_{\text{p}}\text{ and }\Delta {\text{G}}^{\circ }$ for the given reaction is to be determined from the given data.

Concept introduction:

The $\Delta {\text{G}}^{\circ }$ of the reaction is calculated by the expression as:

$\Delta {\text{G}}^{\circ }=\sum \Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{products}\right)-\sum \Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{reactant}\right)$.

Here, $\Delta {\text{G}}^{\circ }$ is the change in standard Gibbs free energy, $\sum \Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{products}\right)$ is the sum of standard Gibbs free energy of products, and $\sum \Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{reactant}\right)$ is the sum of standard Gibbs free energy of reactants.

The value of ${\text{K}}_{\text{P}}$ is calculated by using the relation given below:

${\text{lnK}}_{\text{P}}=-\frac{\Delta {\text{G}}^{\circ }}{\text{RT}}$.

Here, $\Delta {\text{G}}^{\circ }$ is the change in standard Gibbs free energy, $\text{R}$ is the gas constant, and $\text{T}$ is the temperature.

The value of ${\text{K}}_{\text{P}}$ and ${\text{K}}_{\text{C}}$ is same if there is no change in the number of moles in the reaction.

The relationship between kilojoules and joules can be expressed as: $1\text{ kJ}=1000\text{ J}$.

To convert kilojoules to joules, conversion factor is $\frac{1000\text{ J}}{\text{1 kJ}}$

Answer to Problem 63QP

Solution: The values of ${\text{K}}_{\text{C}}{\text{, K}}_{\text{p}}\text{, and }\Delta {\text{G}}^{\circ }$ for the given reaction are $6\times {10}^{34},6\times {10}^{34}\text{ and }-198.3\text{ kJ/mol,}$ respectively.

Explanation of Solution

Given information: The given reaction is $\text{NO}\left(g\right){\text{+O}}_3\left(g\right)\to {\text{NO}}_2\left(g\right)+{\text{O}}_2\left(g\right)$.

The value of $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{NO}}_2\right)$ from appendix two is $\text{51}\text{.8 kJ/mol}$.

The value of $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)$ from appendix two is $\text{86}\text{.7 kJ/mol}$.

The value of $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_3\right)$ from appendix two is $\text{163}\text{.4 kJ/mol}$.

The value of $\Delta {\text{G}}^{\circ }$ for the given reaction is calculated by using the relation given below:

$\Delta {\text{G}}^{\circ }=\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{NO}}_2\right)+\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)-\left[\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)+\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_3\right)\right]$

Here, $\Delta {\text{G}}^{\circ }$ is the change in standard Gibbs free energy, $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{NO}}_2\right)$ is the change in standard Gibbs free energy of ${\text{NO}}_2$, $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$ is the change in standard Gibbs free energy of ${\text{O}}_2$, $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)$ is the change in standard Gibbs free energy of $\text{NO}$ and $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_3\right)$ is the change in standard Gibbs free energy of ${\text{O}}_3$.

The value of change in standard Gibbs free energy for atoms in their standard state is zero. In the reaction, ${\text{O}}_2$ is present in its standard state. Thus, the value of $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$ is zero.

Substitute $0$ for $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$, $\text{163}\text{.4 kJ/mol}$ for $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_3\right)$, $\text{51}\text{.8 kJ/mol}$ for $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{NO}}_2\right)$, and $\text{86}\text{.7 kJ/mol}$ for $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)$ in the equation as:

$\begin{array}{c}\Delta {\text{G}}^{\circ }=\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{NO}}_2\right)+\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)-\left[\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)+\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_3\right)\right]\\ =\left(1\right)\left(\text{51}\text{.8 kJ/mol}\right)+\left(0\right)-\left[\left(1\right)\left(\text{86}\text{.7 kJ/mol}\right)+\left(1\right)\left(\text{163}\text{.4 kJ/mol}\right)\right]\\ =-198.3\text{ kJ/mol}\text{.}\end{array}$

Therefore, the value of $\Delta {\text{G}}^{\circ }$ for the given reaction is $-198.3\text{ kJ/mol}$.

The value of $\Delta {\text{G}}^{\circ }$ is converted to $\text{J/mol}$ by using the relation:

$\begin{array}{l}1\text{ kJ}=1000\text{ J;}\\ \left(-198.3\text{ kJ/mol }\times \frac{1000\text{ J}}{\text{1 kJ}}\right)=-198.3\times {10}^3\text{ J/mol}\text{.}\end{array}$

The value of ${\text{K}}_{\text{P}}$ is calculated by using the relation given below:

${\text{lnK}}_{\text{P}}=-\frac{\Delta {\text{G}}^{\circ }}{\text{RT}}$.

Here, $\Delta {\text{G}}^{\circ }$ is the change in standard Gibbs free energy, $\text{R}$ is the gas constant, and $\text{T}$ is the temperature.

Substitute $-198.3\times {10}^3\text{ J/mol}$ for $\Delta {\text{G}}^{\circ }$, $298\text{ K}$ for $\text{T}$, $\text{8}\text{.314 J/K}\text{.mol}$ for $\text{R}$ in the above equation as follows:

$\begin{array}{c}{\text{lnK}}_{\text{P}}=-\frac{\left(-198.3\times {10}^3\text{ J/mol}\right)}{\left(\text{8}\text{.314 J/K}\text{.mol}\right)\left(298\text{ K}\right)}\\ =\frac{198.3\times {10}^3\text{ J/mol}}{\left(\text{8}\text{.314 J/K}\text{.mol}\right)\left(298\text{ K}\right)}\\ {\text{K}}_{\text{P}}=6\times {10}^{34}.\end{array}$

Therefore, the value of ${\text{K}}_{\text{P}}$ for the given reaction is $6\times {10}^{34}$.

In the given reaction, the number of moles of gaseous atoms is two on both sides of the reaction, and thus there is no change in the number of moles.

Hence, the value of ${\text{K}}_{\text{P}}$ and ${\text{K}}_{\text{C}}$ are same.

Conclusion

The value of ${\text{K}}_{\text{C}}$, ${\text{K}}_{\text{P}},$ and $\Delta {\text{G}}^{\circ }$ for the given reactionis $6\times {10}^{34}$