Chapter 18, Problem 18.44E

Interpretation Introduction

Interpretation:

The values of $E,H,G$, and $S$ for $\text{HCl}$ at standard pressure and $25°\text{C}$ are to be calculated.

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 18.44E

The values of $E,H,G$, and $S$ for $\text{HCl}$ at standard pressure and $25°\text{C}$ are $-388.32\text{kJ}/\text{mol}$, $-385.84\text{kJ}/\text{mol}$, $-442.67\text{kJ}/\text{mol}$ and $-439.36\text{J}/\text{K}{\text{mol}}^{-1}$ respectively.

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.

$\begin{array}{rll}& =& \frac{3}{2}NkT+NkT+NkT\left(\frac{{\theta }_{\text{v}}}{2T}+\frac{{\theta }_{\text{v}}/T}{e^{{\theta }_{\text{v}}/T}-1}\right)+0+\left(-N{D}_0\right)\\ & =& \left(\begin{array}{l}\frac{3}{2}\times 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}+\\ 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}+\\ 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}-1\times 397000\text{J}/\text{mol}\end{array}\right)\\ & =& \left(3717.42+2478.28+2478.28-397000\right)\text{J}/\text{mol}\\ & =& -388.32\text{kJ}/\text{mol}\end{array}$

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.

$\begin{array}{rll}& =& \frac{5}{2}NkT+NkT+NkT\left(\frac{{\theta }_{\text{v}}}{2T}+\frac{{\theta }_{\text{v}}/T}{e^{{\theta }_{\text{v}}/T}-1}\right)+0+\left(-N{D}_0\right)\\ & =& \left(\begin{array}{l}\frac{5}{2}\times 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}+\\ 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}+\\ 1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}-1\times 397000\text{J}/\text{mol}\end{array}\right)\\ & =& \left(6195.70+2478.28+2478.28-397000\right)\text{J}/\text{mol}\\ & =& -385.84\text{kJ}/\text{mol}\end{array}$

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.

$\begin{array}{rll}& =& -NkT\left(\ln {\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\frac{kT}{p}\right)+NkT\left(1-\ln \frac{T}{\sigma {\theta }_{\text{r}}}\right)+NkT\left(\frac{{\theta }_{\text{v}}}{2T}+\ln \left(1-e^{{\theta }_{\text{v}}/T}\right)\right)+0+\left(-N{D}_0\right)\\ & =& \left(\begin{array}{l}-1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}\\ \left(\begin{array}{l}\ln {\left(\frac{2\times 3.14\times 1.627\times {10}^{-27}\text{kg}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}}{\left(6.626\times {10}^{-34}\text{Js}\right){}^2}\right)}^{3/2}\\ \frac{1.363\times {10}^{-25}\text{L}\text{atm}/\text{K}\times 298\text{K}}{1\text{atm}}\end{array}\right)\\ +1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}\left(1-\ln \frac{298\text{K}}{1\times 15.2\text{K}}\right)\\ +1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}-1\times 397000\text{J}/\text{mol}\end{array}\right)\\ & =& -442.67\text{kJ}/\text{mol}\end{array}$

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.

$\begin{array}{rll}& =& \left(\begin{array}{c}-Nk\left(\ln {\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\frac{\left(kT/p\right)+5/2}{\left(kT/p\right)}\right)+Nk\ln \frac{T}{\sigma {\theta }_{\text{r}}}+\\ Nk+Nk\left(\frac{{\theta }_{\text{v}}}{\left(e^{{\theta }_{\text{v}}/T}-1\right)}-\ln \left(1-e^{{\theta }_{\text{v}}/T}\right)\right)+0+\left(-Nk\ln g_1\right)\end{array}\right)\\ & =& \left(\begin{array}{l}-1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\\ \left(\begin{array}{l}\ln {\left(\frac{2\times 3.14\times 1.627\times {10}^{-27}\text{kg}\times 1.381\times {10}^{-23}\text{J}/\text{K}\times 298\text{K}}{\left(6.626\times {10}^{-34}\text{Js}\right){}^2}\right)}^{3/2}\\ \left(\frac{\frac{1.363\times {10}^{-25}\text{L}\text{atm}/\text{K}\times 298\text{K}}{1\text{atm}}+5/2}{\frac{1.363\times {10}^{-25}\text{L}\text{atm}/\text{K}\times 298\text{K}}{1\text{atm}}}\right)\end{array}\right)\\ +1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\left(\ln \frac{298\text{K}}{1\times 15.2\text{K}}\right)+\\ 2\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\\ +1\times 6.022\times {10}^{23}{\text{mol}}^{-1}\times 1.381\times {10}^{-23}\text{J}/\text{K}\end{array}\right)\\ & =& \left(\left(-489.055\right)+24.7478+24.9491\right)\text{J}/\text{K}{\text{mol}}^{-1}\\ & =& -439.36\text{J}/\text{K}{\text{mol}}^{-1}\end{array}$

Conclusion

The values of $E,H,G$, and $S$ for $\text{HCl}$ at standard pressure and $25°\text{C}$ are $-388.32\text{kJ}/\text{mol}$, $-385.84\text{kJ}/\text{mol}$, $-442.67\text{kJ}/\text{mol}$ and $-439.36\text{J}/\text{K}{\text{mol}}^{-1}$ respectively.