Chapter 6, Problem 6A.8AE

(i)

Interpretation Introduction

Interpretation:

The value of ${\Delta }_{\text{r}}G$ when $Q=0.010$ has to be calculated.

Concept Introduction:

The Gibbs free energy of the reaction is the key factor of the identification of the direction of spontaneity of the reaction. For the reaction to be spontaneous, the Gibbs free energy of the reaction must be negative.

Answer to Problem 6A.8AE

The Gibbs free energy for the given reaction when $Q=0.010$ is $\underset{\_}{-44296.83Jmo{l}^{-1}}$.

Explanation of Solution

The given reaction is shown below.

    ${\text{N}}_{\text{2}}+{\text{3H}}_{\text{2}}\to 2{\text{NH}}_{\text{3}}$

The reaction quotient for the given reaction is $0.010$.

It is given that the standard Gibbs free energy for the above reaction is $-32.9\text{kJ}{\text{mol}}^{\text{-1}}$ at $298\text{K}$.

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

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

Therefore, the conversion of $\text{-32}\text{.9}\text{kJ}{\text{mol}}^{\text{-1}}$ to $\text{J}{\text{mol}}^{\text{-1}}$ is done as,

    $-32.9\text{kJ}{\text{mol}}^{\text{-1}}=-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}$

Therefore, the standard Gibbs free energy for the above reaction is $-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}$ at $298\text{K}$.

The value of ${\Delta }_{\text{r}}G$ when $Q=0.010$ is calculated by the following formula.

    ${\Delta }_{\text{r}}G={\Delta }_{\text{r}}{G}^{\Theta }+RT\ln Q\left(1\right)$

Where,

$Q$ is the reaction quotient.

${\Delta }_{\text{r}}G$ is the Gibbs energy of the given reaction for the given value of $Q$.

${\Delta }_{\text{r}}{G}^{\Theta }$ is the standard Gibbs free energy of the given reaction.

$R$ is the gas constant.

$T$ is the given temperature.

The value of gas constant $\left(R\right)$ is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Substitute the value of ${\Delta }_{\text{r}}{G}^{\Theta },R,T$ and $Q$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}G=-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \ln \left(0.010\right)\\ =-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \left(-4.60\right)\\ =-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}-11396.83\text{J}{\text{mol}}^{\text{-1}}\\ =\underset{\_}{-44296.83Jmo{l}^{-1}}\end{array}$

Hence, the Gibbs free energy for the given reaction when $Q=0.010$ is $\underset{\_}{-44296.83Jmo{l}^{-1}}$.

(ii)

Interpretation Introduction

Interpretation:

The value of ${\Delta }_{\text{r}}G$ when $Q=1.0$ has to be calculated.

Concept Introduction:

Refer to (i)

Answer to Problem 6A.8AE

The Gibbs free energy for the given reaction when $Q=1.0$ is $\underset{\_}{-32900Jmo{l}^{-1}}$.

Explanation of Solution

The reaction quotient for the given reaction is $1.0$.

The standard Gibbs free energy for the above reaction is $-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}$ at $298\text{K}$.

The value of gas constant $\left(R\right)$ is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Substitute the value of ${\Delta }_{\text{r}}{G}^{\Theta },R,T$ and $Q$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}G=-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \ln \left(1.0\right)\\ =-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \left(0\right)\\ =-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+0\text{J}{\text{mol}}^{\text{-1}}\\ =\underset{\_}{-32900Jmo{l}^{-1}}\end{array}$

Hence, the Gibbs free energy for the given reaction when $Q=1.0$ is $\underset{\_}{-32900Jmo{l}^{-1}}$.

(iii)

Interpretation Introduction

Interpretation:

The value of ${\Delta }_{\text{r}}G$ when $Q=10.0$ has to be calculated.

Concept introduction:

Refer to (i)

Answer to Problem 6A.8AE

The Gibbs free energy for the given reaction when $Q=10.0$ is $\underset{\_}{-27201.59Jmo{l}^{-1}}$.

Explanation of Solution

The reaction quotient for the given reaction is $10.0$.

The standard Gibbs free energy for the above reaction is $-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}$ at $298\text{K}$.

The value of gas constant $\left(R\right)$ is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Substitute the value of ${\Delta }_{\text{r}}{G}^{\Theta },R,T$ and $Q$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}G=-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \ln \left(10.0\right)\\ =-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \left(2.30\right)\\ =-32900\text{J}{\text{mol}}^{\text{-1}}+5698.41\text{J}{\text{mol}}^{\text{-1}}\\ =\underset{\_}{-27201.59Jmo{l}^{-1}}\end{array}$

Hence, the Gibbs free energy for the given reaction when $Q=10.0$ is $\underset{\_}{-27201.59Jmo{l}^{-1}}$.

(iv)

Interpretation Introduction

Interpretation:

The value of ${\Delta }_{\text{r}}G$ when $Q=100000$ has to be calculated.

Concept Introduction:

Refer to (i)

Answer to Problem 6A.8AE

The Gibbs free energy for the given reaction when $Q=100000$ is $\underset{\_}{-4375.9Jmo{l}^{-1}}$.

Explanation of Solution

The reaction quotient for the given reaction is $100000$.

The standard Gibbs free energy for the above reaction is $-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}$ at $298\text{K}$.

The value of gas constant $\left(R\right)$ is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Substitute the value of ${\Delta }_{\text{r}}{G}^{\Theta },R,T$ and $Q$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}G=-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \ln \left(100000\right)\\ =-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \left(11.51\right)\\ =-32900\text{J}{\text{mol}}^{\text{-1}}+28524.10\text{J}{\text{mol}}^{\text{-1}}\\ =\underset{\_}{-4375.9Jmo{l}^{-1}}\end{array}$

Hence, the Gibbs free energy for the given reaction when $Q=100000$ is $\underset{\_}{-4375.9Jmo{l}^{-1}}$.

(v)

Interpretation Introduction

Interpretation:

The value of ${\Delta }_{\text{r}}G$ has to be calculated.

Concept Introduction:

Refer to (i)

Answer to Problem 6A.8AE

The Gibbs free energy for the given reaction when $Q=1000000$ is $\underset{\_}{-1315.26Jmo{l}^{-1}}$.

The actual value of the equilibrium constant for the given reaction is $\underset{\_}{5.7\times 10^5}$.

Explanation of Solution

The reaction quotient for the given reaction is $1000000$.

The standard Gibbs free energy for the above reaction is $-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}$ at $298\text{K}$.

The value of gas constant $\left(R\right)$ is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Substitute the value of ${\Delta }_{\text{r}}{G}^{\Theta },R,T$ and $Q$ in equation (1).

    $\begin{array}{c}{\Delta }_{\text{r}}G=-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \ln \left(1000000\right)\\ =-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}+8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \left(13.81\right)\\ =-32900\text{J}{\text{mol}}^{\text{-1}}+34215.26\text{J}{\text{mol}}^{\text{-1}}\\ =\underset{\_}{-1315.26Jmo{l}^{-1}}\end{array}$

Hence, the Gibbs free energy for the given reaction when $Q=1000000$ is $\underset{\_}{-1315.26Jmo{l}^{-1}}$.

The value of equilibrium constant $\left(K\right)$ is calculated by the following formula.

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

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

    $\begin{array}{c}{\Delta }_{\text{r}}{G}^{\Theta }=-8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 298\text{K}\times \ln \left(K\right)\\ -32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}=2477.572\text{J}{\text{mol}}^{\text{-1}}\times \ln \left(K\right)\\ \frac{-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}}{2477.572\text{J}{\text{mol}}^{\text{-1}}}=\ln \left(K\right)\end{array}$

Rearrange the above equation as,

    $\begin{array}{c}\ln \left(K\right)=\frac{-32.9\times {10}^3\text{J}{\text{mol}}^{\text{-1}}}{2477.572\text{J}{\text{mol}}^{\text{-1}}}\\ =13.27\\ K={\text{e}}^{13.27}\\ =\underset{\_}{5.7\times 10^5}\end{array}$

Hence, the actual value of the equilibrium constant for the given reaction is $\underset{\_}{5.7\times 10^5}$.