Chapter 3, Problem 3.46AP

Interpretation Introduction

(a)

Interpretation:

The percentage of $2,2-$ dimethylpropane and pentane in an equilibrium mixture at $25°\text{C}$ is to be stated.

Concept introduction:

The product of natural log of equilibrium constant, gas constant and specific temperature gives the value of free energy for the reaction. Free energy of a reaction is expressed in $\text{J}/\text{mol}$ or $\text{kJ}/\text{mol}$.

Answer to Problem 3.46AP

The percentage of $2,2-$ dimethylpropane is $94.1\%$ and the pentane is $5.9\%$ in the mixture.

Explanation of Solution

It is given that standard free energy $\left(\Delta G^{\text{ο}}\right)$ for $2,2-$ dimethylpropane and pentane is $6.86\text{kJ}{\text{mol}}^{-1}$.

Assume that $2,2-$ dimethylpropane is A and pentane is B.

Then the equilibrium between A and B is shown below.

$\text{A}\rightleftarrows \text{B}$

The formula to calculate the $K_{eq}$ is shown below.

$\Delta G^{\text{ο}}=-2.3RT\log K_{eq}$

Where

• $R$ is gas constant.
• $T$ is temperature.
• $\Delta G^{\text{ο}}$ is the standard free energy.
• $K_{eq}$ is the equilibrium constant.

Substitute the value of $R$ as $8.314\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}$, the value of $T$ as $\text{298}\text{K}$ and $\Delta G^{\text{ο}}$ as $6.86\text{kJ}{\text{mol}}^{-1}$ in above equation.

$\begin{array}{l}6.86\text{kJ}{\text{mol}}^{-1}=-2.3\times 8.314\times {10}^{-3}\times 298\times \log K_{eq}\\ 6.86\text{kJ}{\text{mol}}^{-1}=-5.69\times \log K_{eq}\end{array}$

Rearrange the above equation as shown below.

$\begin{array}{rll}-\frac{6.86\text{kJ}{\text{mol}}^{-1}}{5.69}& =& \log K_{eq}\\ -1.20& =& \log K_{eq}\\ K_{eq}& =& \text{antilog}\left(-1.20\right)\\ K_{eq}& =& 0.0630\end{array}$

The percentage of each compound in the mixture is calculated as shown below.

$K_{eq}=\frac{\left[\text{B}\right]}{\left[\text{A}\right]}$

Substitute the value of $K_{eq}$ as $0.0630$.

$\begin{array}{l}0.0630=\frac{\left[\text{B}\right]}{\left[\text{A}\right]}\\ 0.0630\left[\text{A}\right]=\left[\text{B}\right]\end{array}$

At equilibrium, consider total hydrocarbon $=1$. This is represented as shown below.

$\text{A}+\text{B}=1$

Substitute the value of B in the above equation.

$\begin{array}{rll}\text{A}+0.0630\left[\text{A}\right]& =& 1\\ 1.0630\left[\text{A}\right]& =& 1\\ \left[\text{A}\right]& =& \frac{1}{1.0630}\\ \left[\text{A}\right]& =& 0.941\end{array}$

The value of $\left[\text{A}\right]$ is converted into percentage as shown below.

$\begin{array}{rll}\%\text{of}\left[\text{A}\right]& =& 0.941\times 100\\ & =& 94.1\%\end{array}$

Therefore, the percentage of the compound $\left[\text{A}\right]$ is $94.1\%$ in the mixture.

The value of $\left[\text{B}\right]$ is calculated as shown below.

$\begin{array}{rll}\%\text{of}\left[\text{B}\right]& =& 100\%-94.1\%\\ & =& 5.9\%\end{array}$

Therefore, compound $\left[\text{B}\right]$ is present in $5.9\%$ in the mixture.

Conclusion

The percentage of $2,2-$ dimethylpropane is $94.1\%$ and the pentane is $5.9\%$ in the mixture.

Interpretation Introduction

(b)

Interpretation:

The percentage of anti-butane and gauche-butane in an equilibrium mixture at $25°\text{C}$ is to be stated.

Concept introduction:

The product of natural log of equilibrium constant, gas constant and specific temperature gives the value of free energy for the reaction. Free energy of a reaction is expressed in $\text{J}/\text{mol}$ or $\text{kJ}/\text{mol}$.

Answer to Problem 3.46AP

The percentage of anti-butane is $75.6\%$ and the gauche-butane is $24.4\%$ in the mixture.

Explanation of Solution

It is given that standard free energy $\left(\Delta G^{\text{ο}}\right)$ for gauche-butane and anti-butane is $2.8\text{kJ}{\text{mol}}^{-1}$.

Assume that anti-butane is A and gauche butane is B.

Then the equilibrium between A and B is shown below.

$\text{A}\rightleftarrows \text{B}$

The formula to calculate the $K_{eq}$ is shown below.

$\Delta G^{\text{ο}}=-2.3RT\log K_{eq}$

Where

• $R$ is gas constant.
• $T$ is temperature.
• $\Delta G^{\text{ο}}$ is the standard free energy.
• $K_{eq}$ is the equilibrium constant.

Substitute the value of $R$ as $8.314\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}$, the value of $T$ as $\text{298}\text{K}$ and $\Delta G^{\text{ο}}$ as $2.8\text{kJ}{\text{mol}}^{-1}$ in the above equation.

$\begin{array}{l}2.8\text{kJ}{\text{mol}}^{-1}=-2.3\times 8.314\times {10}^{-3}\times 298\times \log K_{eq}\\ 2.8\text{kJ}{\text{mol}}^{-1}=-5.69\times \log K_{eq}\end{array}$

Rearrange the above equation as shown below.

$\begin{array}{rll}-\frac{2.8\text{kJ}{\text{mol}}^{-1}}{5.69}& =& \log K_{eq}\\ -0.492& =& \log K_{eq}\\ K_{eq}& =& \text{antilog}\left(-0.492\right)\\ K_{eq}& =& 0.322\end{array}$

The percentage of each compound in the mixture is calculated as shown below.

$K_{eq}=\frac{\left[\text{B}\right]}{\left[\text{A}\right]}$

Substitute the value of $K_{eq}$ as $0.322$.

$\begin{array}{rll}0.322& =& \frac{\left[\text{B}\right]}{\left[\text{A}\right]}\\ 0.322\left[\text{A}\right]& =& \left[\text{B}\right]\end{array}$

At equilibrium, consider total hydrocarbon $=1$

$\text{A}+\text{B}=1$

Substitute the value of B in the above equation.

$\begin{array}{rll}\text{A}+0.322\left[\text{A}\right]& =& 1\\ 1.322\left[\text{A}\right]& =& 1\\ \left[\text{A}\right]& =& \frac{1}{1.322}\\ \left[\text{A}\right]& =& 0.756\end{array}$

The value of $\left[\text{A}\right]$ is converted into percentage as shown below.

$\begin{array}{rll}\%\text{of}\left[\text{A}\right]& =& 0.756\times 100\\ & =& 75.6\%\end{array}$

Therefore, the percentage of the compound $\left[\text{A}\right]$ is $75.6\%$ in the mixture.

The value of $\left[\text{B}\right]$ is calculated as shown below.

$\begin{array}{rll}\%\text{of}\left[\text{B}\right]& =& 100\%-75.6\%\\ & =& 24.4\%\end{array}$

Therefore, compound $\left[\text{B}\right]$ is present in $24.4\%$ in the mixture.

Conclusion

The percentage of anti-butane is $75.6\%$ and the gauche-pentane is $24.4\%$ in the mixture.