Chapter 18, Problem 18.45E

Determine $E,H,G$, and $S$ for ${\text{CH}}_{\text{4}}$ at standard pressure and $25°\text{C}$. $\sigma$ equals $12$ for methane and the atomization energy of ${\text{CH}}_{\text{4}}$ is $1163\text{kJ}/\text{mol}$. Compare your calculated value of $S$ with the tabulated (that is, experimentally determined) value in Appendix $2$.

Interpretation Introduction

Interpretation:

The values of $E,H,G$, and $S$ for ${\text{CH}}_{\text{4}}$ at standard pressure and $25°\text{C}$ are to be calculated. The calculated value of $S$ is to be compared with the tabulated experimentally determined value.

Concept introduction:

The $p$ is given by the formula,

$p=NkT{\left(\frac{\partial \ln {\text{Q}}_{\text{sys}}}{\partial V}\right)}_T$

The ${\text{Q}}_{\text{sys}}$ is given by the formula,

${\text{Q}}_{\text{sys}}=q_{\text{rot}}\cdot q_{\text{trans}}\cdot q_{\text{vib}}\cdot q_{\text{nuc}}\cdot q_{\text{ele}}$

Answer to Problem 18.45E

The values of $E,H,G$, and $S$ for ${\text{CH}}_{\text{4}}$ at standard pressure and $25°\text{C}$ are $-1154.32\text{kJ}/\text{mol}$, $-1151.85\text{kJ}/\text{mol}$, $-1246.83\text{kJ}/\text{mol}$ and $-1114.06\text{J}/\text{K}{\text{mol}}^{-1}$ respectively. The calculated value of $S$ is much less as compared with the tabulated experimentally determined value.

Explanation of Solution

The expression for $E$ is given below as,

$\begin{array}{rll}E& =& E_{\text{trans}}+E_{\text{rot}}+E_{\text{vib}}+E_{\text{nuc}}+E_{\text{elec}}\\ & =& \frac{3}{2}NkT+NkT+NkT\left(\frac{{\theta }_{\text{v}}}{2T}+\frac{{\theta }_{\text{v}}/T}{e^{{\theta }_{\text{v}}/T}-1}\right)+0+\left(-N{D}_0\right)\end{array}$ … (1)

Since the temperature $25°\text{C}$ is much lower than vibrational temperature, therefore, the vibration temperature term is taken as $1$. Substitute the values in equation (1) as follows.

$\begin{array}{rll}& =& \frac{3}{2}NkT+NkT+NkT\left(\frac{{\theta }_{\text{v}}}{2T}+\frac{{\theta }_{\text{v}}/T}{e^{{\theta }_{\text{v}}/T}-1}\right)+0+\left(-N{D}_0\right)\\ & =& \left(\begin{array}{l}\frac{3}{2}\times 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}+\\ 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}+\\ 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}-1\times 1163000\text{J}/\text{mol}\end{array}\right)\\ & =& \left(3717.42+2478.28+2478.28-1163000\right)\text{J}/\text{mol}\\ & =& -1154.32\text{kJ}/\text{mol}\end{array}$

The expression for $H$ is given below as,

$\begin{array}{rll}H& =& H_{\text{trans}}+H_{\text{rot}}+H_{\text{vib}}+H_{\text{nuc}}+H_{\text{elec}}\\ & =& \frac{5}{2}NkT+NkT+NkT\left(\frac{{\theta }_{\text{v}}}{2T}+\frac{{\theta }_{\text{v}}/T}{e^{{\theta }_{\text{v}}/T}-1}\right)+0+\left(-N{D}_0\right)\end{array}$ … (2)

Since the temperature $25°\text{C}$ is much lower than vibrational temperature, therefore, the vibration temperature term is taken as $1$. Substitute the values in equation (2) as follows.

$\begin{array}{rll}& =& \frac{5}{2}NkT+NkT+NkT\left(\frac{{\theta }_{\text{v}}}{2T}+\frac{{\theta }_{\text{v}}/T}{e^{{\theta }_{\text{v}}/T}-1}\right)+0+\left(-N{D}_0\right)\\ & =& \left(\begin{array}{l}\frac{5}{2}\times 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}+\\ 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}+\\ 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}-1\times 1163000\text{J}/\text{mol}\end{array}\right)\\ & =& \left(6195.70+2478.28+2478.28-1163000\right)\text{J}/\text{mol}\\ & =& -1151.85\text{kJ}/\text{mol}\end{array}$

The expression for $G$ is given below as,

$\begin{array}{rll}G& =& G_{\text{trans}}+G_{\text{rot}}+G_{\text{vib}}+G_{\text{nuc}}+G_{\text{elec}}\\ & =& -NkT\left(\ln {\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\frac{V}{N}\right)+NkT\left(1-\ln \frac{T}{\sigma {\theta }_{\text{r}}}\right)+NkT\left(\frac{{\theta }_{\text{v}}}{2T}+\ln \left(1-e^{{\theta }_{\text{v}}/T}\right)\right)+0+\left(-N{D}_0\right)\end{array}$

Since the temperature $25°\text{C}$ is much lower than vibrational temperature, therefore, the vibration temperature term is taken as $1$. Substitute the values in above equation as follows.

$\begin{array}{rll}& =& -NkT\left(\ln {\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\frac{kT}{p}\right)+NkT\left(1-\ln \frac{T}{\sigma {\theta }_{\text{r}}}\right)+NkT\left(\frac{{\theta }_{\text{v}}}{2T}+\ln \left(1-e^{{\theta }_{\text{v}}/T}\right)\right)+0+\left(-N{D}_0\right)\\ & =& \left(\begin{array}{l}-1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}\\ \left(\begin{array}{l}\ln {\left(\frac{2\times 3.14\times 1.53\times {10}^{-22}\text{kg}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}}{\left(6.626\times {10}^{-34}\text{Js}\right){}^2}\right)}^{3/2}\\ \frac{1.363\times {10}^{-25}\text{L}\text{atm}/\text{K}\times 298\text{K}}{1\text{atm}}\end{array}\right)\\ +1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}\left(1-\ln \frac{298\text{K}}{12\times 7.54\text{K}}\right)\\ +1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}-1\times 1163000\text{J}/\text{mol}\end{array}\right)\\ & =& -1246.83\text{kJ}/\text{mol}\end{array}$

The expression for $S$ is given below as,

$\begin{array}{rll}S& =& S_{\text{trans}}+S_{\text{rot}}+S_{\text{vib}}+S_{\text{nuc}}+S_{\text{elec}}\\ & =& -Nk\left(\ln {\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\frac{\left(kT/p\right)+5/2}{\left(kT/p\right)}\right)+Nk\ln \frac{T}{\sigma {\theta }_{\text{r}}}+Nk+Nk\left(\frac{{\theta }_{\text{v}}}{\left(e^{{\theta }_{\text{v}}/T}-1\right)}-\ln \left(1-e^{{\theta }_{\text{v}}/T}\right)\right)+0+\left(-Nk\ln g_1\right)\end{array}$

Since the temperature $25°\text{C}$ is much lower than vibrational temperature, therefore, the vibration temperature term is taken as $1$. Substitute the values in above equation as follows.

The value of $S$ given in Appendix $2$ is $188.66\text{J}/\text{mol}\text{K}$. This value is much higher from the calculated value of $S$ for ${\text{CH}}_{\text{4}}$. This is because of the assumptions taken in the calculation of $S$.

Conclusion

The values of $E,H,G$, and $S$ for ${\text{CH}}_{\text{4}}$ at standard pressure and $25°\text{C}$ are $-1154.32\text{kJ}/\text{mol}$, $-1151.85\text{kJ}/\text{mol}$, $-1246.83\text{kJ}/\text{mol}$ and $-1114.06\text{J}/\text{K}{\text{mol}}^{-1}$ respectively. The calculated value of $S$ is much less as compared with the tabulated experimentally determined value.