Chapter 16.6, Problem 23P

(a)

To determine

The natural Logarithm equilibrium constant for the reaction at $298\text{K}$.

Compare the results for the values of $K_P$ obtained from the equation of equilibrium constants and the equilibrium constant Table A-28.

Answer to Problem 23P

The natural Logarithm equilibrium constant for the reaction at $298\text{K}$ is $\underset{\_}{103.81}$.

The natural equilibrium constant obtained from the equilibrium constants of Table A-28 at 2$298\text{K}$ is $\underset{\_}{103.76}$.

Explanation of Solution

Express the standard-state Gibbs function change.

$\Delta G^{*}\left(T\right)=v_{{\text{CO}}_2}{\overline{g}}^{*}{}_{{\text{CO}}_2}\left(T\right)-v_{\text{CO}}{\overline{g}}^{*}{}_{\text{CO}}\left(T\right)-v_{{\text{O}}_2}{\overline{g}}^{*}{}_{{\text{O}}_2}\left(T\right)$ (I)

Here, the Gibbs function of components ${\text{CO}}_{\text{2}}\text{,}\text{CO,and}{\text{O}}_{\text{2}}$ at 1 atm pressure and temperature T are ${\overline{g}}^{*}{}_{{\text{CO}}_{\text{2}}}\left(T\right),{\overline{g}}^{*}{}_{\text{CO}}\left(T\right),\text{and}{\overline{g}}^{*}{}_{{\text{O}}_2}\left(T\right)$ respectively, temperature of ${\text{CO}}_{\text{2}}\text{,}\text{CO,and}{\text{O}}_{\text{2}}$ are $T_{{\text{CO}}_2},T_{\text{CO}},\text{and}{T}_{{\text{O}}_{\text{2}}}$,and stoichiometric coefficients of components ${\text{CO}}_{\text{2}}\text{,}\text{CO,and}{\text{O}}_{\text{2}}$ are $v_{{\text{CO}}_{\text{2}}},v_{\text{CO}},\text{and}{v}_{{\text{O}}_2}$ respectively.

Write the equation to calculate the natural logarithms of equilibrium constant for the chemical equilibrium of ideal-gas mixtures.

$\ln K_P=\frac{-\Delta G^{*}\left(T\right)}{R_{u}T}$ (II)

Here, universal gas constant is $R_u$ and temperature of Gibbs function of formation is T.

Conclusion:

From the equilibrium reaction, the values of $v_{{\text{CO}}_2},v_{\text{CO}},\text{and}{v}_{{\text{O}}_{\text{2}}}$ are 1, 1, and 0.5 respectively.

Refer Table A-26, obtain the values of ${\overline{g}}^{*}{}_{{\text{H}}_{\text{2}}\text{O}},{\overline{g}}^{*}{}_{{\text{H}}_{\text{2}}},\text{and}{\overline{g}}^{*}{}_{{\text{O}}_2}$ as below:

$\begin{array}{l}{\overline{g}}^{*}{}_{{\text{CO}}_2}=-\text{394,360 }\text{kJ}/\text{kmol}\\ {\overline{g}}^{*}{}_{\text{CO}}=-137,150\text{kJ}/\text{kmol}\\ {\overline{g}}^{*}{}_{O_2}=0\text{kJ}/\text{kmol}\end{array}$

Substitute 1 for $v_{{\text{CO}}_2}$, 1 for $v_{\text{CO}}$, 0.5 for $v_{{\text{O}}_{\text{2}}}$, $\text{-394,360 }\text{kJ}/\text{kmol}$ for ${\overline{g}}^{*}{}_{{\text{CO}}_2}$, $-137,150\text{kJ}/\text{kmol}$ for ${\overline{g}}^{*}{}_{\text{CO}}$, and 0 for ${\overline{g}}^{*}{}_{O_2}$ in Equation (I).

$\begin{array}{c}\Delta G*\left(T\right)=1\left(-\text{394,360 }\text{kJ}/\text{kmol}/\text{kmol}\right)-1\left(-137,150\text{kJ}/\text{kmol}\right)-0.5\left(0\right)\\ =-257,210\text{kJ}/\text{kmol}\end{array}$

Substitute $-257,210\text{kJ}/\text{kmol}$ for $\Delta G^{*}\left(T\right)$, $298\text{ K}$ for T, and $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$ in Equation (II).

$\begin{array}{c}\ln K_P=-\frac{-257,210\text{kJ}/\text{kmol}}{\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\times 298\text{ K}}\\ =103.81\end{array}$

Thus, the equilibrium constant obtained from the equilibrium reaction at 298 K is $\underset{\_}{103.81}$.

From table A-28, “Natural logarithms of the equilibrium constant” obtain the equilibrium constant for the reaction at the temperature of 298 K as $\underset{\_}{103.76}$.

The value obtained for equilibrium constant at 298 K from the definition of the equilibrium constant is $103.81$ which is equal to the value obtained for equilibrium constant as $103.76$ from the Table A-28.

(b)

To determine

The natural logarithm equilibrium constant for the reaction at 2000 K.

Compare the results for the values of $K_P$ obtained from the equilibrium constants of Table A-28.

Answer to Problem 23P

The natural logarithm equilibrium constant for the reaction at 2000 K is $\underset{\_}{6.64}$.

The natural Logarithm equilibrium constant obtained from the equilibrium constants of Table A-28 at 2000K is $\underset{\_}{6.635}$

Explanation of Solution

Express the standard-state Gibbs function change.

$\begin{array}{c}\Delta G^{*}\left(T\right)=v_{{\text{CO}}_2}{\overline{g}}^{*}{}_{{\text{CO}}_2}\left(T\right)-v_{\text{CO}}{\overline{g}}^{*}{}_{\text{CO}}\left(T\right)-v_{{\text{O}}_2}{\overline{g}}^{*}{}_{{\text{O}}_2}\left(T\right)\\ =v_{{\text{CO}}_2}{\left(\overline{h}-T\overline{s}\right)}_{{\text{CO}}_2}-v_{\text{CO}}{\left(\overline{h}-T\overline{s}\right)}_{\text{CO}}-v_{{\text{O}}_2}{\left(\overline{h}-T\overline{s}\right)}_{{\text{O}}_2}\\ =\left[\begin{array}{l}{v}_{{\text{CO}}_2}\left({\overline{h}}_f^o{,}_{{\text{CO}}_2}+\left({\overline{h}}_{{\text{CO}}_2}-{\overline{h}}^o{}_{{\text{CO}}_2}\right)-T{\overline{s}}_{{\text{CO}}_2}\right)-\\ v_{{\text{H}}_{\text{2}}}\left({\overline{h}}_{f,\text{CO}}^o+\left({\overline{h}}_{H_2}-{\overline{h}}^o{}_{\text{CO}}\right)-T{\overline{s}}_{\text{CO}}\right)-\\ v_{{\text{O}}_2}\left({\overline{h}}_f^o{,}_{{\text{O}}_2}+\left({\overline{h}}_{{}_{{\text{O}}_2}}-{\overline{h}}^o{}_{{}_{{\text{O}}_2}}\right)-T{\overline{s}}_{{}_{{\text{O}}_{\text{2}}}}\right)\end{array}\right]\end{array}$ (III)

Here, the Gibbs function of components ${\text{CO}}_2\text{,}\text{CO},\text{and}{\text{O}}_{\text{2}}$ at 1 atm pressure and temperature T are ${\overline{g}}^{*}{}_{{\text{CO}}_2}\left(T\right),{\overline{g}}^{*}{}_{\text{CO}}\left(T\right),\text{and}{\overline{g}}^{*}{}_{{\text{O}}_2}\left(T\right)$ respectively, enthalpy on the unit mole basis of ${\text{H}}_{\text{2}}{\text{,O}}_{\text{2}},\text{and}{\text{H}}_{\text{2}}\text{O}$ are ${\overline{h}}_{{\text{H}}_{\text{2}}},{\overline{h}}_{{\text{O}}_{\text{2}}},\text{and}{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}$ respectively, absolute entropy of ${\text{CO}}_2\text{,}\text{CO},\text{and}{\text{O}}_{\text{2}}$ are ${\overline{s}}_{{\text{CO}}_2},{\overline{s}}_{\text{CO}},\text{and}{\overline{s}}_{{\text{O}}_{\text{2}}}$, temperature of ${\text{CO}}_2\text{,}\text{CO},\text{and}{\text{O}}_{\text{2}}$ are $T_{\text{CO}},T_{\text{CO}},\text{and}{T}_{{\text{O}}_{\text{2}}}$, sensible enthalpy at the specified state of ${\text{CO}}_2\text{,}\text{CO},\text{and}{\text{O}}_{\text{2}}$ are ${\overline{h}}_{{\text{CO}}_2}{\overline{h}}_{\text{CO}},\text{and}{\overline{h}}_{{\text{O}}_{\text{2}}}$, the sensible enthalpy at the standard reference state of $25°\text{C}$ and 1 atm of ${\text{CO}}_2\text{,}\text{CO},\text{and}{\text{O}}_{\text{2}}$ are ${\overline{h}}_{{\text{CO}}_{\text{2}}}^o,{\overline{h}}_{\text{CO}}^o,\text{and}{\overline{h}}_{{\text{O}}_{\text{2}}}^o$, the enthalpy of formation at a specified states of $25°\text{C}$ and 1 atm for the components ${\text{CO}}_2\text{,}\text{CO},\text{and}{\text{O}}_{\text{2}}$ are ${\overline{h}}_{f,{\text{CO}}_{\text{2}}}^o,{\overline{h}}_{f,\text{CO}}^o,\text{and}{\overline{h}}_{f,{\text{O}}_{\text{2}}}^o$, stoichiometric coefficients of components ${\text{CO}}_2\text{,}\text{CO},\text{and}{\text{O}}_{\text{2}}$ are $v_{\text{CO}}{}_{{}_2},v_{\text{CO}},\text{and}{v}_{{\text{O}}_{\text{2}}}$ respectively.

Write the equation to calculate the natural logarithms of equilibrium constant for the chemical equilibrium of ideal-gas mixtures.

$\ln K_P=\frac{-\Delta G^{*}\left(T\right)}{R_{u}T}$ (IV)

Here, universal gas constant is $R_u$ and temperature of Gibbs function of formation is T.

Conclusion:

From the equilibrium reaction, the values of $v_{\text{CO}}{}_{{}_2},v_{\text{CO}},\text{and}{v}_{{\text{O}}_{\text{2}}}$ are 1, 1, and 0.5 respectively.

Refer Table A-26, obtain the values of ${\overline{h}}_{f,{\text{H}}_{\text{2}}}^o,{\overline{h}}_{f,{\text{O}}_{\text{2}}}^o,\text{and}{\overline{h}}_{f,{\text{H}}_{\text{2}}\text{O}}^o$ as below:

$\begin{array}{l}{\overline{h}}_{f,{\text{CO}}_2}^o=-393,520\text{ kJ/kmol}\\ {\overline{h}}_{f,\text{CO}}^o=-110,530\text{kJ}/\text{kmol}\\ {\overline{h}}_{f,{\text{O}}_2}^o=\text{0}\text{kJ/kmol}\end{array}$

Refer Table A-20, obtain the value of ${\overline{h}}_{{\text{CO}}_{\text{2}}}^o$ at temperature of 298K.

${\overline{h}}_{{\text{CO}}_{\text{2}}}^o=9634\text{kJ/kmol}$

Refer Table A-20, obtain the value of ${\overline{h}}_{{\text{CO}}_{\text{2}}}$ and ${\overline{s}}_{{\text{CO}}_{\text{2}}}$ at temperature of 2000K.

$\begin{array}{l}{\overline{h}}_{{\text{CO}}_{\text{2}}}=\text{100,804 kJ/kmol}\\ {\overline{s}}_{{\text{CO}}_{\text{2}}}=309.210\text{kJ/kmol}\cdot \text{K}\end{array}$

Refer Table A-21, obtain the value of ${\overline{h}}_{CO}^o$ at temperature of 298K.

${\overline{h}}_{CO}^o=\text{8669}\text{kJ/kmol}$

Refer Table A-21, obtain the value of ${\overline{h}}_{CO}$ and ${\overline{s}}_{CO}$ at temperature of 2000K.

$\begin{array}{l}{\overline{h}}_{CO}=\text{65,408}\text{kJ/kmol}\\ {\overline{s}}_{CO}=258.600\text{kJ/kmol}\cdot \text{K}\end{array}$

Refer Table A-19, obtain the value of ${\overline{h}}_{{\text{O}}_2}^o$ at temperature of 298K.

${\overline{h}}_{{\text{O}}_{\text{2}}}^o=8682\text{ kJ/kmol}$

Refer to Table A-19, obtain the value of ${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}$ and ${\overline{s}}_{{\text{H}}_{\text{2}}\text{O}}$ at temperature of 2000 K.

$\begin{array}{l}{\overline{h}}_{{\text{O}}_{\text{2}}}^o=67,881\text{ kJ/kmol}\\ {\overline{s}}_{{\text{O}}_2}=268.655\text{ kJ/kmol}\cdot \text{K}\end{array}$

Substitute 1 for $v_{{\text{H}}_{\text{2}}\text{O}}$, $-241,820\text{kJ/kmol}$ for ${\overline{h}}_f^o{,}_{{\text{H}}_{\text{2}}\text{O}}$, $82593\text{ kJ/kmol}$ for ${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}$, $9904\text{kJ/kmol}$ for ${\overline{h}}^o{}_{{\text{H}}_{\text{2}}\text{O}}$, $2000\text{ K}$ for T, $264.571\text{ kJ/kmol}\cdot \text{K}$ for ${\overline{s}}_{{\text{H}}_{\text{2}}\text{O}}$, 1 for $v_{{\text{H}}_{\text{2}}}$, 0 for ${\overline{h}}_{f,H_2}^o$, $61,400\text{ kJ/kmol}$ for ${\overline{h}}_{H_2}$, $\text{8468kJ/kmol}$ for ${\overline{h}}^o{}_{H_2}$, $188.297\text{ kJ/kmol}\cdot \text{K}$ for $T{\overline{s}}_{H_2}$, $0.5$ for $v_{{\text{O}}_2}$, $0$ for ${\overline{h}}_f^o{,}_{{\text{O}}_2}$, $67881\text{ kJ/kmol}$ for ${\overline{h}}_{{}_{{\text{O}}_2}}$, $8682\text{ kJ/kmol}$ for ${\overline{h}}^o{}_{{}_{{\text{O}}_2}}$, $268.655\text{ kJ/kmol}\cdot \text{K}$ for ${\overline{s}}_{{}_{{\text{O}}_2}}$ in Equation (III).

$\begin{array}{c}\Delta G^{*}\left(T\right)=\left[\begin{array}{l}1{\left(\begin{array}{l}-393,520\text{ kJ/kmol}+\text{100,804 kJ/kmol}\\ -9634\text{kJ/kmol}-2000\text{ K}\times 309.210\text{kJ/kmol}\cdot \text{K}\end{array}\right)}_{{\text{CO}}_{\text{2}}}-\\ 1{\left(\begin{array}{l}-110,530\text{kJ}/\text{kmol}+\text{65,408}\text{kJ/kmol}\\ -\text{8669}\text{kJ/kmol}-2000\text{K}\times 258.600\text{kJ/kmol}\cdot \text{K}\end{array}\right)}_{{\text{H}}_{\text{2}}}-\\ 0.5{\left(\begin{array}{l}0+\left(67881\text{ kJ/kmol}-8682\text{ kJ/kmol}\right)\\ -2000\text{K}\times 268.655\text{ kJ/kmol}\cdot \text{K}\end{array}\right)}_{{\text{O}}_{\text{2}}}\end{array}\right]\\ =-110.409\text{ kJ/kmol}\end{array}$

Substitute $-\text{110}\text{.409}\text{kJ/kmol}$ for $\Delta G^{*}\left(T\right)$, $2000\text{ K}$ for T, and $1.986\text{Btu/lbmol}\cdot \text{R}$ for $R_u$ in Equation (IV).

$\begin{array}{c}\ln K_P=\frac{-\text{110}\text{.409}\text{kJ/kmol}}{\left(1.986\text{kJ/kmol}\cdot \text{K}\right)\times 2000\text{K}}\\ =6.64\end{array}$

Thus, the natural logarithm equilibrium constant obtained from the equilibrium reaction at 2000K is $\underset{\_}{6.64}$.

Refer Table A-28, “Natural logarithms of the equilibrium constant” obtain the equilibrium constant for the reaction by interpolating for the temperature of 2000 K as $6.635$.

The value obtained for equilibrium constant at 2000K from the definition of the equilibrium constant is $6.64$ which is almost equal to the value obtained for equilibrium constant as $6.635$ from the Table A-28.