Chapter 7, Problem 7.30E

Interpretation Introduction

Interpretation:

The values of ${\Delta }_{\text{mix}}G$ and ${\Delta }_{\text{mix}}S$ for an ideal solution are to be predicted.

Concept introduction:

According to Raoult’s law, the vapor pressure of the solution is equal to the product of vapor pressure of the solvent and its mole fraction in the solution. The ideal solution follows the Raoult’s law.

Answer to Problem 7.30E

The values of ${\Delta }_{\text{mix}}G$ and ${\Delta }_{\text{mix}}S$ for an ideal solution are $-4213.12\text{J}$ and $13.58\text{J}{\text{K}}^{-1}$ respectively.

Explanation of Solution

The $25.0\text{g}$ of pentane $\left({\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\right)$, $45.0\text{g}$ of hexane $\left({\text{C}}_6{\text{H}}_{\text{14}}\right)$, and $55.0\text{g}$ of cyclohexane $\left({\text{C}}_6{\text{H}}_{\text{12}}\right)$ are mixed to form an ideal solution.

The conversion of temperature in Celsius to Kelvin is given by the formula,

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.2$

Substitute the value of $T=37.0°\text{C}$ in the above formula.

$\begin{array}{rll}T\left(\text{K}\right)& =& 37+273.2\\ & =& 310.2\text{K}\end{array}$

Thus, the temperature of the ideal solution is $310.2\text{K}$.

The molar mass of pentane is calculated as,

$\begin{array}{rll}\text{Molar}\text{Mass}\text{of}{\text{C}}_{\text{5}}{\text{H}}_{\text{12}}& =& \left(5\times \text{Mass}\text{of}\text{carbon}\text{atom}\right)+\left(12\times \text{Mass}\text{of}\text{hydrogen}\text{atom}\right)\\ & =& \left(5\times 12\right)+\left(12\times 1\right)\\ & =& 60+12\\ & =& 72\text{g}\end{array}$

Thus, the molar mass of pentane is $72\text{g}$.

The molar mass of hexane is calculated as,

$\begin{array}{rll}\text{Molar}\text{Mass}\text{of}{\text{C}}_6{\text{H}}_{\text{14}}& =& \left(6\times \text{Mass}\text{of}\text{carbon}\text{atom}\right)+\left(14\times \text{Mass}\text{of}\text{hydrogen}\text{atom}\right)\\ & =& \left(6\times 12\right)+\left(14\times 1\right)\\ & =& 72+14\\ & =& 86\text{g}\end{array}$

Thus, the molar mass of hexane is $86\text{g}$.

The molar mass of cyclohexane is calculated as,

$\begin{array}{rll}\text{Molar}\text{Mass}\text{of}{\text{C}}_6{\text{H}}_{\text{12}}& =& \left(6\times \text{Mass}\text{of}\text{carbon}\text{atom}\right)+\left(12\times \text{Mass}\text{of}\text{hydrogen}\text{atom}\right)\\ & =& \left(6\times 12\right)+\left(12\times 1\right)\\ & =& 72+12\\ & =& 84\text{g}\end{array}$

Thus, the molar mass of cyclohexane is $84\text{g}$.

The number of moles is calculated by the formula,

$\text{No}\text{.}\text{of}\text{moles}=\frac{\text{Given}\text{Mass}}{\text{Molar}\text{Mass}}$

Substitute the given mass and molar mass of pentane in above formula.

$\begin{array}{rll}\text{No}\text{.}\text{of}\text{moles}& =& \frac{\text{25}}{\text{72}}\\ & =& 0.35\text{mol}\end{array}$

Thus, the moles of pentane is $0.35\text{mol}$.

Substitute the given mass and molar mass of hexane in above formula.

$\begin{array}{rll}\text{No}\text{.}\text{of}\text{moles}& =& \frac{\text{45}}{86}\\ & =& 0.52\text{mol}\end{array}$

Thus, the moles of hexane is $0.52\text{mol}$.

Substitute the given mass and molar mass of cyclohexane in above formula.

$\begin{array}{rll}\text{No}\text{.}\text{of}\text{moles}& =& \frac{\text{55}}{84}\\ & =& 0.66\text{mol}\end{array}$

Thus, the moles of cyclohexane is $0.66\text{mol}$.

Hence, the total number of moles is $1.53\text{mol}$ respectively.

The mole fraction is calculated by the formula,

$x_{\text{A}}=\frac{n_{\text{A}}}{n_{\text{A}}+n_{\text{B}}+n_{\text{C}}}$

Where,

• $n_{\text{A}}$ is the moles of A.

• $n_{\text{B}}$ is the moles of B.

• $n_{\text{C}}$ is the moles of C.

• $\left(n_{\text{A}}+n_{\text{B}}+n_{\text{C}}\right)$ represents the total moles of the solution.

Substitute the moles of pentane in above formula.

$\begin{array}{rll}{x}_{\text{pentane}}& =& \frac{n_{\text{pentane}}}{n_{\text{pentane}}+n_{\text{hexane}}+n_{\text{cyclohexane}}}\\ & =& \frac{0.35}{0.35+0.52+0.66}\\ & =& \frac{0.35}{1.53}\\ & =& 0.23\end{array}$

Thus, the mole fraction of pentane is $0.23$.

Substitute the moles of hexane in above formula.

$\begin{array}{rll}{x}_{\text{hexane}}& =& \frac{n_{\text{hexane}}}{n_{\text{pentane}}+n_{\text{hexane}}+n_{\text{cyclohexane}}}\\ & =& \frac{0.52}{0.35+0.52+0.66}\\ & =& \frac{0.52}{1.53}\\ & =& 0.34\end{array}$

Thus, the mole fraction of hexane is $0.34$.

Substitute the moles of cyclohexane in above formula.

$\begin{array}{rll}{x}_{\text{cyclohexane}}& =& \frac{n_{\text{cyclohexane}}}{n_{\text{pentane}}+n_{\text{hexane}}+n_{\text{cyclohexane}}}\\ & =& \frac{0.66}{0.35+0.52+0.66}\\ & =& \frac{0.66}{1.53}\\ & =& 0.43\end{array}$

Thus, the mole fraction of cyclohexane is $0.43$.

The formula used to calculate ${\Delta }_{\text{mix}}G$ is given by,

${\Delta }_{\text{mix}}G=RT{\displaystyle \sum _i{x}_i\ln }{x}_i$

Where,

• $R$ is the gas constant.

• $T$ is the temperature of the solution.

• $x_i$ is the mole fraction of i-th component of a solution.

Substitute the value of $x_{\text{pentane}},x_{\text{hexane}}$ and $x_{\text{cyclohexane}}$ in the above formula.

$\begin{array}{rll}{\Delta }_{\text{mix}}G& =& RT\left(x_{\text{pentane}}\ln x_{\text{pentane}}+x_{\text{hexane}}\ln x_{\text{hexane}}+x_{\text{cyclohexane}}\ln x_{\text{cyclohexane}}\right)\\ & =& \left(8.314\frac{\text{J}}{\text{mol}\text{.K}}\right)\left(310.2\text{K}\right)\left[\left(0.34\ln 0.34+0.23\ln 0.23+0.43\ln 0.43\right)\right]\\ & =& -2753.7\text{J}{\text{mol}}^{-1}\times 1.53\text{mol}\\ & =& -4213.12\text{J}\end{array}$

Therefore, the ${\Delta }_{\text{mix}}G$ of the ideal solution is $-4213.12\text{J}$.

The formula used to calculate ${\Delta }_{\text{mix}}S$ is given by,

${\Delta }_{\text{mix}}S=-R{\displaystyle \sum _i{x}_i\ln }{x}_i$

Where,

• $R$ is the gas constant.

• $x_i$ is the mole fraction of i-th component of a solution.

Substitute the value of $x_{\text{toluene}}$ and $x_{\text{benzene}}$ in the above formula.

$\begin{array}{rll}{\Delta }_{\text{mix}}S& =& -R\left(x_{\text{pentane}}\ln x_{\text{pentane}}+x_{\text{hexane}}\ln x_{\text{hexane}}+x_{\text{cyclohexane}}\ln x_{\text{cyclohexane}}\right)\\ & =& -\left(8.314\frac{\text{J}}{\text{mol}\text{.K}}\right)\left[\left(0.34\ln 0.34+0.23\ln 0.23+0.43\ln 0.43\right)\right]\\ & =& 8.878\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\times 1.53\text{mol}\\ & =& 13.58\text{J}{\text{K}}^{-1}\end{array}$

Therefore, the ${\Delta }_{\text{mix}}S$ of the ideal solution is $13.58\text{J}{\text{K}}^{-1}$.

Conclusion

The values of ${\Delta }_{\text{mix}}G$ and ${\Delta }_{\text{mix}}S$ for an ideal solution are $-4213.12\text{J}$ and $13.58\text{J}{\text{K}}^{-1}$ respectively.