Chapter 6, Problem 6A.1ST

Interpretation Introduction

Interpretation:

The equilibrium constant for the reaction ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)\to {\text{2NO}}_{\text{2}}\left(\text{g}\right)$ at $\text{298}\text{K}$ has to be evaluated.

Concept Introduction:

Equilibrium constant of a reaction is the function of temperature only.  It remains the same for a given reaction at a given temperature.  It is derived from the ratio of either partial pressures of the components or activities and is dimensionless.

Answer to Problem 6A.1ST

The equilibrium constant for the reaction, ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)\to {\text{2NO}}_{\text{2}}\left(\text{g}\right)$ at $\text{298}\text{K}$ is calculated as $\underset{\_}{0.148}$.

Explanation of Solution

The standard reaction Gibbs energy of a reaction $\left({\Delta }_{\text{r}}{G}^o\right)$ is given as,

    ${\Delta }_{\text{r}}{G}^o=n_{\text{P}}{\Delta }_{\text{f}}{G}_{\text{P}}^{\text{ο}}+n_{\text{R}}{\Delta }_{\text{f}}{G}_{\text{R}}^{\text{ο}}\left(1\right)$

Where,

$n_{\text{P}}$ is the number of moles of the product.

$n_{\text{R}}$ is the number of moles of the reactant.

${\Delta }_{\text{f}}{G}_{\text{P}}^{\text{ο}}$ is the Gibbs energy of formation of the product.

${\Delta }_{\text{f}}{G}_{\text{R}}^{\text{ο}}$ is the Gibbs energy of formation of the reactant.

The given reaction is,

    ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)\to {\text{2NO}}_{\text{2}}\left(\text{g}\right)$

The number of moles of the product, ${\text{NO}}_{\text{2}}\left(\text{g}\right)$ $\left(n_{\text{P}}\right)$ is $2$.

The number of moles of the reactant, ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)$$\left(n_{\text{R}}\right)$ is $1$.

The Gibbs energy of formation of product, ${\text{NO}}_{\text{2}}\left(\text{g}\right)$, ${\Delta }_{\text{f}}{G}_{\text{P}}^{\text{ο}}$ is $\text{51}\text{.31}\text{kJ}{\text{mol}}^{\text{-1}}$.

The Gibbs energy of formation of reactant, ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\left(\text{g}\right)$, ${\Delta }_{\text{f}}{G}_{\text{R}}^{\text{ο}}$ is $97.89\text{kJ}{\text{mol}}^{\text{-1}}$.

Substitute the value of $n_{\text{P}}$, $n_{\text{R}}$, ${\Delta }_{\text{f}}{G}_{\text{P}}^{\text{ο}}$ and ${\Delta }_{\text{f}}{G}_{\text{R}}^{\text{ο}}$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\text{ο}}=\text{2}\text{mol×51}\text{.31}\text{kJ}{\text{mol}}^{\text{-1}}-\text{1}\text{mol×97}\text{.89}\text{kJ}{\text{mol}}^{\text{-1}}\\ =\text{4}\text{.73}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

The equilibrium constant of a reaction $\left(K\right)$ is related to reaction Gibbs energy $\left({\Delta }_{\text{r}}{G}^o\right)$ by the expression,

    ${\Delta }_{\text{r}}{G}^o=-RT\ln K\left(2\right)$

Where,

$R$ is the gas constant $\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)$.

$T$ is the temperature.

The temperature is $\text{298}\text{K}$.

The reaction Gibbs energy is $\text{4}\text{.73}\text{kJ}{\text{mol}}^{\text{-1}}$.

The conversion of $\text{kJ}{\text{mol}}^{\text{-1}}$ to $\text{J}{\text{mol}}^{\text{-1}}$ is done as,

    $\text{1}\text{kJ}{\text{mol}}^{\text{-1}}\text{=1000}\text{J}{\text{mol}}^{\text{-1}}$

Hence, $\text{4}\text{.73}\text{kJ}{\text{mol}}^{\text{-1}}$ is converted to $\text{J}{\text{mol}}^{\text{-1}}$ as,

    $\begin{array}{c}4.73\text{kJ}{\text{mol}}^{\text{-1}}=\text{4}\text{.73}\times \text{1000}\text{J}{\text{mol}}^{\text{-1}}\\ =4730\text{J}{\text{mol}}^{\text{-1}}\end{array}$

Substitute the value of ${\Delta }_{\text{r}}{G}^o$, $R$ and $T$ in equation (2).

    $\begin{array}{c}\text{4730}\text{J}{\text{mol}}^{\text{-1}}\text{=}-\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{×298}\text{K}\times \ln K\\ \ln K=-\frac{\text{4730}\text{J}{\text{mol}}^{\text{-1}}}{\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\text{×298}\text{K}}\\ K=e^{\left(-1.909\right)}\\ =\underset{\_}{0.148}\end{array}$

Hence, the value of equilibrium constant is $\underset{\_}{0.148}$.