Chapter 6, Problem 6B.2BE

(i)

Interpretation Introduction

Interpretation:

Under the given conditions, the standard Gibbs energy of reaction and equilibrium constant at temperature ${25}^{\text{ο}}\text{C}$ has to be calculated.

Concept introduction:

The Gibbs free energy is the energy associated with the chemical reaction that is used to describe whether a chemical reaction is spontaneous or not.  If change in Gibbs free energy is negative then the process is said to be spontaneous otherwise non-spontaneous.  The standard Gibbs energy of reaction is measured at standard conditions.  Equilibrium constant is the ratio of concentrations of products to the concentrations of reactants at equilibrium.

Answer to Problem 6B.2BE

The standard Gibbs free energy and equilibrium constant at temperature $298\text{K}$ are $\underset{\_}{-187944.04Jmo{l}^{-1}}$ and $\underset{\_}{8.47\times 10^{32}}$ respectively.

Explanation of Solution

The standard Gibbs energy of reaction is expressed as,

  ${\Delta }_{\text{r}}{G}^{\text{ο}}={\Delta }_{\text{r}}{H}^{\text{ο}}-T{\Delta }_{\text{r}}{S}^{\text{ο}}$                                                                                   (1)

Where,

• ${\Delta }_{\text{r}}{G}^{\text{ο}}$ is the standard Gibbs energy of reaction.
• ${\Delta }_{\text{r}}{H}^{\text{ο}}$ is the standard enthalpy of reaction.
• ${\Delta }_{\text{r}}{S}^{\text{ο}}$ is the standard entropy of reaction.
• $T$ is the standard temperature.

Rearrange the equation (1) to calculate the standard entropy of formation of ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$.

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

At $\text{298}\text{.15}\text{K}$, it is given that,

The standard Gibbs energy of formation of ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$ is $-73.7\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$ is $-134.1\text{kJ}{\text{mol}}^{-1}$.

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

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

Therefore, the conversion of $-73.7\text{kJ}{\text{mol}}^{-1}$ to $\text{J}{\text{mol}}^{-1}$ is done as,

  $\begin{array}{c}-73.7\text{kJ}{\text{mol}}^{-1}=-73.7\times {10}^3\text{}\text{J}{\text{mol}}^{-1}\\ =-73700\text{J}{\text{mol}}^{-1}\end{array}$

Similarly, the conversion of $-134.1\text{kJ}{\text{mol}}^{-1}$ to $\text{J}{\text{mol}}^{-1}$ is done as,

  $\begin{array}{c}-134.1\text{kJ}{\text{mol}}^{-1}=-134.1\times {10}^3\text{}\text{J}{\text{mol}}^{-1}\\ =-134100\text{J}{\text{mol}}^{-1}\end{array}$

Substitute the values of ${\Delta }_{\text{r}}{G}^{\text{ο}}$, ${\Delta }_{\text{r}}{H}^{\text{ο}}$ and $T$ in equation (2).

  $\begin{array}{c}{\Delta }_{\text{f}}{S}^{\text{ο}}=\frac{-134100\text{J}{\text{mol}}^{-1}-\left(-73700\text{J}{\text{mol}}^{-1}\right)}{298.15\text{K}}\\ =-\frac{60400}{298.15}\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\\ =-202.58\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\end{array}$

Therefore, the standard entropy of formation of ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$ is $-202.58\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The given reaction is,

  ${\text{CH}}_{\text{4}}\left(\text{g}\right)+{\text{3Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{CHCl}}_{\text{3}}\left(\text{l}\right)+\text{3HCl}\left(\text{g}\right)$

The standard enthalpy $\left({\Delta }_{\text{r}}{H}^{\text{ο}}\right)$ for this chemical reaction is expressed as,

  ${\Delta }_{\text{r}}{H}^{\text{ο}}=\left({\Delta }_{\text{f}}{H}_{{\text{CHCl}}_{\text{3}}\left(\text{l}\right)}^{\text{ο}}+3{\Delta }_{\text{f}}{H}_{\text{HCl}\left(\text{g}\right)}^{\text{ο}}-{\Delta }_{\text{f}}{H}_{{\text{CH}}_{\text{4}}\left(\text{g}\right)}^{\text{ο}}-3{\Delta }_{\text{f}}{H}_{{\text{Cl}}_{\text{2}}\left(\text{g}\right)}^{\text{ο}}\right)$                           (3)

It is given that,

The temperature is ${25}^{\text{ο}}\text{C}$.

The standard enthalpy of formation of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is $-74.81\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ is $0\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$ is $-134.1\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of $\text{HCl}\left(\text{g}\right)$ is $-92.31\text{kJ}{\text{mol}}^{-1}$.

Celsius is converted to Kelvin using the equation given below as,

    $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Substitute the temperature values in the above equation as follows.

    $\begin{array}{c}{T}_{}\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =25+273.15\\ =298.15\text{K}\end{array}$

Substitute the values of standard enthalpy of formation of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$, ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$, ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$ and $\text{HCl}\left(\text{g}\right)$ in equation (3).

  $\begin{array}{c}{\Delta }_{\text{r}}{H}^{\text{ο}}=\left(\begin{array}{l}-134.1\text{kJ}{\text{mol}}^{-1}+3\left(-92.31\text{kJ}{\text{mol}}^{-1}\right)-\\ \left(-74.81\text{kJ}{\text{mol}}^{-1}\right)-3\left(0\text{kJ}{\text{mol}}^{-1}\right)\end{array}\right)\\ =\left(-411.03\text{kJ}{\text{mol}}^{-1}+74.81\text{}\text{kJ}{\text{mol}}^{-1}\right)\\ =-336.22\text{kJ}{\text{mol}}^{-1}\end{array}$

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

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

Therefore, the conversion of $-336.22\text{kJ}{\text{mol}}^{-1}$$84.28\text{kJ}{\text{mol}}^{-1}$ to $\text{J}{\text{mol}}^{-1}$ is done as,

  $\begin{array}{c}-336.22\text{kJ}{\text{mol}}^{-1}=-336.22\times {10}^3\text{}\text{J}{\text{mol}}^{-1}\\ =-336220\text{J}{\text{mol}}^{-1}\end{array}$

The standard entropy of the given chemical reaction is calculated as,

  ${\Delta }_{\text{r}}{S}^{\text{ο}}=\left(\Delta S_{{\text{CHCl}}_{\text{3}}\left(\text{l}\right)}^{\text{ο}}+3\Delta S_{\text{HCl}\left(\text{g}\right)}^{\text{ο}}-\Delta S_{{\text{CH}}_{\text{4}}\left(\text{g}\right)}^{\text{ο}}-3\Delta S_{{\text{Cl}}_{\text{2}}\left(\text{g}\right)}^{\text{ο}}\right)$                                       (4)

It is given that,

The standard entropy for ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$ is $-202.58\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The standard entropy for ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is $186.26\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-1}$.

The standard entropy for ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ is $223.07\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-1}$.

The standard entropy for $\text{HCl}\left(\text{g}\right)$ is $186.91\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

Substitute the values of standard entropy of ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$, ${\text{CH}}_{\text{4}}\left(\text{g}\right)$, ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ and $\text{HCl}\left(\text{g}\right)$ in equation (4).

  $\begin{array}{c}{\Delta }_{\text{r}}{S}^{\text{ο}}=\left(\begin{array}{l}-202.58\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}+\left(3\times 186.91\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)-\\ 186.26\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-1}-\left(3\times 223.07\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-1}\right)\end{array}\right)\\ =-497.32\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\end{array}$

Substitute the values of ${\Delta }_{\text{r}}{H}^{\text{ο}}$, ${\Delta }_{\text{r}}{S}^{\text{ο}}$ and $T$ in equation (1).

  $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\text{ο}}=\left(-336220\text{J}{\text{mol}}^{-1}-298.15\text{K}\left(-497.32\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\right)\\ =-336220\text{J}{\text{mol}}^{-1}+148275.958\text{J}{\text{mol}}^{-1}\\ =\underset{\_}{-187944.04Jmo{l}^{-1}}\end{array}$

Therefore, the standard Gibbs free energy of reaction at temperature $298\text{K}$ is $\underset{\_}{-187944.04Jmo{l}^{-1}}$.

The relation between standard Gibbs free energy of reaction and equilibrium constant is expressed as,

  ${\Delta }_{\text{r}}{G}^{\text{ο}}=-RT\ln K_{\text{eq}}$                                                                                        (5)

Where,

• ${\Delta }_{\text{r}}{G}^{\text{ο}}$ is the standard Gibbs free energy of reaction.
• $K_{\text{eq}}$ is the equilibrium constant.
• $R$ is gas constant $\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$.
• $T$ is the temperature.

Rearrange the equation (5) to calculate $K_{\text{eq}}$.

  $\begin{array}{c}\ln K_{\text{eq}}=-\frac{{\Delta }_{\text{r}}{G}^{\text{ο}}}{RT}\\ K_{\text{eq}}=e^{\left(-\frac{{\Delta }_{\text{r}}{G}^{\text{ο}}}{RT}\right)}\end{array}$                                                                                             (6)

Substitute the values of ${\Delta }_{\text{r}}{G}^{\text{ο}}$, $R$ and $T$ in equation (6).

  $\begin{array}{c}{K}_{\text{eq}}=e^{\left(-\frac{\left(-\text{187944}\text{.04}\text{J}{\text{mol}}^{-\text{1}}\right)}{8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 298.15\text{K}}\right)}\\ =e^{\left(75.82\right)}\\ =\underset{\_}{8.47\times 10^{32}}\end{array}$

Therefore, the equilibrium constant at temperature $298\text{K}$ is $\underset{\_}{8.47\times 10^{32}}$.

(ii)

Interpretation Introduction

Interpretation:

Under the given conditions, the standard Gibbs energy of reaction and equilibrium constant at temperature ${50}^{\text{ο}}\text{C}$ has to be calculated.

Concept introduction:

The Gibbs free energy is the energy associated with the chemical reaction that is used to describe whether a chemical reaction is spontaneous or not.  If change in Gibbs free energy is negative then the process is said to be spontaneous otherwise non-spontaneous.  The standard Gibbs energy of reaction is measured at standard conditions.  Equilibrium constant is the ratio of concentrations of products to the concentrations of reactants at equilibrium.

Answer to Problem 6B.2BE

The standard Gibbs free energy and equilibrium constant at temperature ${50}^{\text{ο}}\text{C}$ are $\underset{\_}{-175511.04Jmo{l}^{-1}}$ and $\underset{\_}{2.33\times 10^{28}}$ respectively.

Explanation of Solution

The standard Gibbs energy of reaction is expressed as,

  ${\Delta }_{\text{r}}{G}^{\text{ο}}={\Delta }_{\text{r}}{H}^{\text{ο}}-T{\Delta }_{\text{r}}{S}^{\text{ο}}$                                                                                   (1)

Where,

• ${\Delta }_{\text{r}}{G}^{\text{ο}}$ is the standard Gibbs energy of reaction.
• ${\Delta }_{\text{r}}{H}^{\text{ο}}$ is the standard enthalpy of reaction.
• ${\Delta }_{\text{r}}{S}^{\text{ο}}$ is the standard entropy of reaction.
• $T$ is the standard temperature.

The given reaction is,

  ${\text{CH}}_{\text{4}}\left(\text{g}\right)+{\text{3Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{CHCl}}_{\text{3}}\left(\text{l}\right)+\text{3HCl}\left(\text{g}\right)$

The standard enthalpy $\left({\Delta }_{\text{r}}{H}^{\text{ο}}\right)$ for this chemical reaction is expressed as,

  ${\Delta }_{\text{r}}{H}^{\text{ο}}=\left({\Delta }_{\text{f}}{H}_{{\text{CHCl}}_{\text{3}}\left(\text{l}\right)}^{\text{ο}}+3{\Delta }_{\text{f}}{H}_{\text{HCl}\left(\text{g}\right)}^{\text{ο}}-{\Delta }_{\text{f}}{H}_{{\text{CH}}_{\text{4}}\left(\text{g}\right)}^{\text{ο}}-3{\Delta }_{\text{f}}{H}_{{\text{Cl}}_{\text{2}}\left(\text{g}\right)}^{\text{ο}}\right)$                           (2)

It is given that,

The temperature is ${50}^{\text{ο}}\text{C}$.

The standard enthalpy of formation of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is $-74.81\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ is $0\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$ is $-134.1\text{kJ}{\text{mol}}^{-1}$.

The standard enthalpy of formation of $\text{HCl}\left(\text{g}\right)$ is $-92.31\text{kJ}{\text{mol}}^{-1}$.

Celsius is converted to Kelvin using the equation given below as,

    $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Substitute the temperature values in the above equation as follows.

    $\begin{array}{c}{T}_{}\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =50+273.15\\ =323.15\text{K}\end{array}$

Substitute the values of standard enthalpy of formation of ${\text{CH}}_{\text{4}}\left(\text{g}\right)$, ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$, ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$ and $\text{HCl}\left(\text{g}\right)$ in equation (2).

  $\begin{array}{c}{\Delta }_{\text{r}}{H}^{\text{ο}}=\left(\begin{array}{l}-134.1\text{kJ}{\text{mol}}^{-1}+3\left(-92.31\text{kJ}{\text{mol}}^{-1}\right)-\\ \left(-74.81\text{kJ}{\text{mol}}^{-1}\right)-3\left(0\text{kJ}{\text{mol}}^{-1}\right)\end{array}\right)\\ =\left(-411.03\text{kJ}{\text{mol}}^{-1}+74.81\text{}\text{kJ}{\text{mol}}^{-1}\right)\\ =-336.22\text{kJ}{\text{mol}}^{-1}\end{array}$

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

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

Therefore, the conversion of $-336.22\text{kJ}{\text{mol}}^{-1}$$84.28\text{kJ}{\text{mol}}^{-1}$ to $\text{J}{\text{mol}}^{-1}$ is done as,

  $\begin{array}{c}-336.22\text{kJ}{\text{mol}}^{-1}=-336.22\times {10}^3\text{}\text{J}{\text{mol}}^{-1}\\ =-336220\text{J}{\text{mol}}^{-1}\end{array}$

The standard entropy of the given chemical reaction is calculated as,

  ${\Delta }_{\text{r}}{S}^{\text{ο}}=\left(\Delta S_{{\text{CHCl}}_{\text{3}}\left(\text{l}\right)}^{\text{ο}}+3\Delta S_{\text{HCl}\left(\text{g}\right)}^{\text{ο}}-\Delta S_{{\text{CH}}_{\text{4}}\left(\text{g}\right)}^{\text{ο}}-3\Delta S_{{\text{Cl}}_{\text{2}}\left(\text{g}\right)}^{\text{ο}}\right)$                                       (3)

It is given that,

The standard entropy for ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$ is $-202.58\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

The standard entropy for ${\text{CH}}_{\text{4}}\left(\text{g}\right)$ is $186.26\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-1}$.

The standard entropy for ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ is $223.07\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-1}$.

The standard entropy for $\text{HCl}\left(\text{g}\right)$ is $186.91\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}$.

Substitute the values of standard entropy of ${\text{CHCl}}_{\text{3}}\left(\text{l}\right)$, ${\text{CH}}_{\text{4}}\left(\text{g}\right)$, ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$ and $\text{HCl}\left(\text{g}\right)$ in equation (3).

  $\begin{array}{c}{\Delta }_{\text{r}}{S}^{\text{ο}}=\left(\begin{array}{l}-202.58\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}+\left(3\times 186.91\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)-\\ 186.26\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-1}-3\left(223.07\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-1}\right)\end{array}\right)\\ =-497.32\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\end{array}$

Substitute the values of ${\Delta }_{\text{r}}{H}^{\text{ο}}$, ${\Delta }_{\text{r}}{S}^{\text{ο}}$ and $T$ in equation (1).

  $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\text{ο}}=\left(-336220\text{J}{\text{mol}}^{-1}-323.15\text{K}\left(-497.32\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\right)\\ =-336220\text{J}{\text{mol}}^{-1}+160708.958\text{J}{\text{mol}}^{-1}\\ =\underset{\_}{-175511.04Jmo{l}^{-1}}\end{array}$

Therefore, the standard Gibbs free energy of reaction at temperature ${50}^{\text{ο}}\text{C}$ is $\underset{\_}{-175511.04Jmo{l}^{-1}}$.

The relation between standard Gibbs free energy of reaction and equilibrium constant is expressed as,

  ${\Delta }_{\text{r}}{G}^{\text{ο}}=-RT\ln K_{\text{eq}}$                                                                                        (4)

Where,

• ${\Delta }_{\text{r}}{G}^{\text{ο}}$ is the standard Gibbs free energy of reaction.
• $K_{\text{eq}}$ is the equilibrium constant.
• $R$ is gas constant $\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$.
• $T$ is the temperature.

Rearrange the equation (4) to calculate $K_{\text{eq}}$.

  $\begin{array}{c}\ln K_{\text{eq}}=-\frac{{\Delta }_{\text{r}}{G}^{\text{ο}}}{RT}\\ K_{\text{eq}}=e^{\left(-\frac{{\Delta }_{\text{r}}{G}^{\text{ο}}}{RT}\right)}\end{array}$                                                                                             (5)

Substitute the values of ${\Delta }_{\text{r}}{G}^{\text{ο}}$, $R$ and $T$ in equation (5).

  $\begin{array}{c}{K}_{\text{eq}}=e^{\left(-\frac{\left(-\text{175511}\text{.04}\text{J}{\text{mol}}^{-\text{1}}\right)}{8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 323.15\text{K}}\right)}\\ =e^{\left(65.32\right)}\\ =\underset{\_}{2.33\times 10^{28}}\end{array}$

Therefore, the equilibrium constant at temperature ${50}^{\text{ο}}\text{C}$ is $\underset{\_}{2.33\times 10^{28}}$.