Chapter 24, Problem 41QP

Interpretation Introduction

Interpretation:

The standard free energy of formation for NO, ${\text{K}}_{\text{P}}{\text{, and K}}_{\text{c}}$ for the reaction are to be determined with given $\Delta \text{G}$ value.

Concept introduction:

The $\Delta {\text{G}}^{\circ }$ for 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 ${\text{K}}_{\text{C}}$ are 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 41QP

Solution: The value of ${\text{K}}_{\text{C}}$, ${\text{K}}_{\text{P}}$, and $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)$ for the given reaction are $4\times {10}^{-31},4\times {10}^{-31}\text{, and 86}\text{.7 kJ/mol,}$ respectively.

Explanation of Solution

Given information: The given reaction is ${\text{N}}_2\left(g\right)+{\text{O}}_2\left(g\right)\rightleftharpoons \text{2NO}\left(g\right)$.

The value of $\Delta {\text{G}}^{\circ }$ from appendix 2 is $\text{173}\text{.4 kJ/mol}$.

The value of standard free energy of formation for NO is calculated by using the relation given below:

$\Delta {\text{G}}^{\circ }=2\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)-\left[\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{N}}_2\right)+\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)\right]$.

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

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

Substitute $0$ for $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{O}}_2\right)$, $0$ for $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left({\text{N}}_2\right)$, and $\text{173}\text{.4 kJ/mol}$ for $\Delta {\text{G}}^{\circ }$ in theabove equation as follows:

$\begin{array}{l}\text{173}\text{.4 kJ/mol}=2\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)-\left[0+0\right]\\ \Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)=86.7\text{ kJ/mol}\text{.}\end{array}$

Therefore, the value of $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)$ is $\text{86}\text{.7 kJ/mol}$.

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

$\begin{array}{l}1\text{ kJ}=1000\text{ J}\text{.}\\ \left(\text{173}\text{.4 kJ/mol }\times \frac{1000\text{ J}}{\text{1 kJ}}\right)=\text{173}\text{.4}\times {10}^3\text{ J/mol}\text{.}\end{array}$

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 $\text{173}\text{.4 }\times {\text{10}}^3\text{J/mol}$ for $\Delta {\text{G}}^{\circ }$, $298\text{ K}$ for $\text{T}$, and $\text{8}\text{.314 J/K}\text{.mol}$ for $\text{R}$ in the above equation as follows:

$\begin{array}{c}{\text{lnK}}_{\text{P}}=-\frac{\left(\text{173}\text{.4 }\times {\text{10}}^3\cancel{\text{J}}\text{/}\cancel{\text{mol}}\right)}{\left(\text{8}\text{.314 }\cancel{\text{J}}\text{/}\cancel{\text{K}}\text{.}\cancel{\text{mol}}\right)\left(298\cancel{\text{K}}\right)}\\ =\frac{-\text{173}\text{.4 }\times {\text{10}}^3}{2477.752}\\ {\text{K}}_{\text{P}}=4\times {10}^{-31}.\end{array}$

Therefore, the value of ${\text{K}}_{\text{P}}$ for the given reaction is $4\times {10}^{-31}$.

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 ${\text{K}}_{\text{C}}$ is same.

Therefore, the value of ${\text{K}}_{\text{C}}$ for the given reaction is $4\times {10}^{-31}$.

Conclusion

The value of $\Delta {\text{G}}_{\text{f}}{}^{\circ }\left(\text{NO}\right)$, ${\text{K}}_{\text{C}},$ and ${\text{K}}_{\text{P}}$ for the given reaction is $4\times {10}^{-31}$.