Chapter 16.6, Problem 24P

(a)

To determine

The equilibrium constant obtained from the equilibrium reaction at 298 K.

Answer to Problem 24P

The equilibrium constant obtained from the equilibrium reaction at 298 K is $8.24\times {10}^{-46}$.

Explanation of Solution

Express the standard-state Gibbs function change.

$\Delta G^{*}\left(T\right)=v_{\text{CO}}{\overline{g}}^{*}{}_{\text{CO}}\left(T\right)+v_{{\text{O}}_{\text{2}}}{\overline{g}}^{*}{}_{{\text{O}}_{\text{2}}}\left(T\right)-v_{{\text{CO}}_2}{\overline{g}}^{*}{}_{{\text{CO}}_2}\left(T\right)$                                                         (I)

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

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

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

Conclusion:

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

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

$\begin{array}{l}{\overline{g}}^{*}{}_{\text{CO}}=-\text{137150 }\text{kJ}/\text{kmol}\\ {\overline{g}}^{*}{}_{{\text{O}}_{\text{2}}}=0\text{kJ}/\text{kmol}\\ {\overline{g}}^{*}{}_{{\text{CO}}_{\text{2}}}=-394360\text{kJ}/\text{kmol}\end{array}$

Substitute 1 for $v_{\text{CO}}$, 1 for $v_{{\text{CO}}_2}$, 0.5 for $v_{{\text{O}}_{\text{2}}}$, $-\text{137150 }\text{kJ}/\text{kmol}$ for ${\overline{g}}^{*}{}_{\text{CO}}$, $0\text{kJ}/\text{kmol}$ for ${\overline{g}}^{*}{}_{{\text{O}}_2}$, and $-394360\text{kJ}/\text{kmol}$ for ${\overline{g}}^{*}{}_{{\text{CO}}_{\text{2}}}$ in equation (I).

$\begin{array}{c}\Delta G*\left(T\right)=1\left(-\text{137150 }\text{kJ}/\text{kmol}\right)+0.5\left(0\right)-1\left(-394360\text{kJ}/\text{kmol}\right)\\ =257210\text{kJ}/\text{kmol}\end{array}$

Substitute $257210\text{kJ}/\text{kmol}$ for $\Delta G^{*}\left(T\right)$, $25°\text{C}$ for T, and $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$ in Equation (II).

$\begin{array}{c}\ln K_P=-\frac{257210\text{kJ}/\text{kmol}}{\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\times 298\text{}\text{ K}}\\ =-103.81\end{array}$

Substitute $-103.81$ for $K_P$ in Equation (III).

$\begin{array}{c}{K}_P=e^{\left(-103.81\right)}\\ =8.24\times {10}^{-46}\end{array}$

Thus, the equilibrium constant obtained from the equilibrium reaction at 298 K is $8.24\times {10}^{-46}$.

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 Table A-28.

(b)

To determine

The equilibrium constant obtained from the equilibrium reaction at 1800K.

Answer to Problem 24P

The equilibrium constant obtained from the equilibrium reaction at 1800Kis $2.03\times {10}^{-4}$.

Explanation of Solution

Write the expression to obtain standard-state Gibbs function change.

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

Here, the Gibbs function of components $\text{CO,}{\text{O}}_{\text{2}},\text{and}{\text{CO}}_2$ at 1 atm pressure and temperature T are ${\overline{g}}^{*}{}_{\text{CO}}\left(T\right),{\overline{g}}^{*}{}_{{\text{O}}_{\text{2}}}\left(T\right),\text{and}{\overline{g}}^{*}{}_{{\text{CO}}_2}\left(T\right)$ respectively, enthalpy on the unit mole basis of $\text{CO,}{\text{O}}_{\text{2}},\text{and}{\text{CO}}_2$ are ${\overline{h}}_{\text{CO}},{\overline{h}}_{{\text{O}}_{\text{2}}},\text{and}{\overline{h}}_{{\text{CO}}_2}$ respectively, absolute entropy of $\text{CO,}{\text{O}}_{\text{2}},\text{and}{\text{CO}}_2$ are ${\overline{s}}_{\text{CO}},{\overline{s}}_{{\text{O}}_{\text{2}}},\text{and}{\overline{s}}_{{\text{CO}}_2}$, the sensible enthalpy at the standard reference state of $298\text{ K}$ and 1 atm of $\text{CO,}{\text{O}}_{\text{2}},\text{and}{\text{CO}}_2$ are ${\overline{h}}_{\text{CO}}^o,{\overline{h}}_{{\text{O}}_{\text{2}}}^o,\text{and}{\overline{h}}_{{\text{CO}}_2}^o$, the enthalpy of formation at a specified states of $298\text{ K}$ and 1 atm for the components $\text{CO,}{\text{O}}_{\text{2}},\text{and}{\text{CO}}_2$ are ${\overline{h}}_{f,\text{CO}}^o,{\overline{h}}_{f,{\text{O}}_{\text{2}}}^o,\text{and}{\overline{h}}_{f,{\text{CO}}_2}^o$, stoichiometric coefficients of components $\text{CO,}{\text{O}}_{\text{2}},\text{and}{\text{CO}}_2$ are $v_{\text{CO}},v_{{\text{O}}_{\text{2}}},\text{and}{v}_{{\text{CO}}_2}$ respectively.

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

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

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

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

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

Conclusion:

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

Refer Table A-26, obtain the values of ${\overline{h}}_{f,\text{CO}}^o,{\overline{h}}_{f,{\text{O}}_{\text{2}}}^o,\text{and}{\overline{h}}_{f,{\text{CO}}_2}^o$ as below:

$\begin{array}{l}{\overline{h}}_{f,\text{CO}}^o=-\text{110530}\text{kJ/kmol}\\ {\overline{h}}_{f,{\text{O}}_{\text{2}}}^o=0\\ {\overline{h}}_{f,{\text{CO}}_2}^o=-\text{393520}\text{kJ/kmol}\end{array}$

Refer Table A-22, obtain the value of ${\overline{h}}_{\text{CO}}^o$ at temperature of 298K.

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

Refer Table A-22, obtain the value of ${\overline{h}}_{\text{CO}}$ and ${\overline{s}}_{\text{CO}}$ at temperature of 1800K.

$\begin{array}{l}{\overline{h}}_{\text{CO}}=\text{58191 kJ/kmol}\\ {\overline{s}}_{\text{CO}}=254.797\text{kJ/kmol}\cdot \text{K}\end{array}$

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

${\overline{h}}_{{\text{O}}_2}^o=\text{8682}\text{kJ/kmol}$

Refer Table A-19, obtain the value of ${\overline{h}}_{{\text{O}}_2}$ and ${\overline{s}}_{{\text{O}}_2}$ at temperature of 1800K.

$\begin{array}{l}{\overline{h}}_{{\text{O}}_2}=\text{60371}\text{kJ/kmol}\\ {\overline{s}}_{{\text{O}}_2}=264.701\text{kJ/kmol}\cdot \text{K}\end{array}$

Refer Table A-23, obtain the value of ${\overline{h}}_{{\text{CO}}_2}^o$ at temperature of 298K.

${\overline{h}}_{{\text{CO}}_2}^o=\text{9364 kJ/kmol}$

Refer Table A-23, obtain the value of ${\overline{h}}_{{\text{CO}}_2}$ and ${\overline{s}}_{{\text{CO}}_2}$ at temperature of 1800 K.

$\begin{array}{l}{\overline{h}}_{{\text{CO}}_2}^o=88806\text{ kJ/kmol}\\ {\overline{s}}_{{\text{CO}}_2}=302.884\text{ kJ/kmol}\cdot \text{K}\end{array}$

Substitute 1 for $v_{{\text{CO}}_2}$, $-393520\text{kJ/kmol}$ for ${\overline{h}}_f^o{,}_{{\text{CO}}_2}$, $88806\text{ kJ/kmol}$ for ${\overline{h}}_{{\text{CO}}_2}$, $9364\text{kJ/kmol}$ for ${\overline{h}}^o{}_{{\text{CO}}_2}$, $1800\text{ K}$ for T, $302.884\text{ kJ/kmol}\cdot \text{K}$ for ${\overline{s}}_{{\text{CO}}_2}$, 1 for $v_{\text{CO}}$, $-110530\text{ kJ/kmol}$ for ${\overline{h}}_{f,\text{CO}}^o$, $58191\text{ kJ/kmol}$ for ${\overline{h}}_{\text{CO}}$, $\text{8669 kJ/kmol}$ for ${\overline{h}}^o{}_{\text{CO}}$, $254.797\text{ kJ/kmol}\cdot \text{K}$ for ${\overline{s}}_{\text{CO}}$, $0.5$ for $v_{{\text{O}}_2}$, $0$ for ${\overline{h}}_f^o{,}_{{\text{O}}_2}$, $60371\text{ kJ/kmol}$ for ${\overline{h}}_{{}_{{\text{O}}_2}}$, $8682\text{ kJ/kmol}$ for ${\overline{h}}^o{}_{{}_{{\text{O}}_2}}$, $264.701\text{ kJ/kmol}\cdot \text{K}$ for ${\overline{s}}_{{}_{{\text{O}}_2}}$ in Equation (IV).

$\begin{array}{c}\Delta G^{*}\left(T\right)=\left[\begin{array}{l}1\left(\begin{array}{l}-110530\text{kJ/kmol}+58191\text{ kJ/kmol}-8669\text{kJ/kmol}\\ -1800\text{ K}\times 254.797\text{ kJ/kmol}\cdot \text{K}\end{array}\right)\\ +0.5\left(\begin{array}{l}0+\left(60371\text{ kJ/kmol}-8682\text{ kJ/kmol}\right)\\ -1800\text{K}\times 264.701\text{ kJ/kmol}\cdot \text{K}\end{array}\right)\\ -1\left(\begin{array}{l}-393520+88806\text{ kJ/kmol}-\text{9364 kJ/kmol}\\ -1800\text{K}\times 302.884\text{ kJ/kmol}\cdot \text{K}\end{array}\right)\end{array}\right]\\ =127240.2\text{ kJ/kmol}\end{array}$

Substitute $127240.2\text{ kJ/kmol}$ for $\Delta G^{*}\left(T\right)$, $1800\text{ K}$ for T, and $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$ in Equation (V).

$\begin{array}{c}\ln K_P=\frac{-127240.2\text{ kJ/kmol}}{\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\times 1800\text{K}}\\ =-8.502\end{array}$

Substitute $-8.502$ for $K_P$ in Equation (VI).

$\begin{array}{c}{K}_P=e^{\left(-8.502\right)}\\ =2.03\times {10}^{-4}\end{array}$

Thus, the equilibrium constant obtained from the equilibrium reaction at 1800Kis $2.03\times {10}^{-4}$.

The value obtained for equilibrium constant at 1800K from the definition of the equilibrium constant is $-8.502$ which is almost equal to the value obtained for equilibrium constant as $-8.497$ from Table A-28.