Chapter 16.6, Problem 21P

(a)

To determine

The logarithm equilibrium constant for the reaction at $25°\text{C}$.

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 21P

The logarithm equilibrium constant for the reaction at $25°\text{C}$ is $\underset{\_}{1.1695\times {10}^{40}}$.

The equilibrium constant obtained from the equilibrium constants of Table A-28 at 1440Ris $\underset{\_}{1.1125\times {10}^{40}}$.

Explanation of Solution

Express the standard-state Gibbs function change.

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

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

$\begin{array}{l}{\overline{g}}^{*}{}_{{\text{H}}_{\text{2}}\text{O}}=\text{228,590 }\text{kJ}/\text{kmol}\\ {\overline{g}}^{*}{}_{{\text{H}}_{\text{2}}}=0\text{kJ}/\text{kmol}\\ {\overline{g}}^{*}{}_{O_2}=0\text{kJ}/\text{kmol}\end{array}$

Substitute 1 for $v_{{\text{H}}_{\text{2}}\text{O}}$, 1 for $v_{{\text{H}}_{\text{2}}}$, 0.5 for $v_{{\text{O}}_{\text{2}}}$, $\text{228,590 }\text{kJ}/\text{kmol}$ for ${\overline{g}}^{*}{}_{{\text{H}}_{\text{2}}\text{O}}$, 0 for $0\text{kJ}/\text{kmol}$, and 0 for ${\overline{g}}^{*}{}_{O_2}$ in Equation (I).

$\begin{array}{c}\Delta G*\left(T\right)=1\left(-\text{228,590}\text{kJ}/\text{kmol}\right)-1\left(0\right)-0.5\left(0\right)\\ =-228,590\text{kJ}/\text{kmol}\end{array}$

Substitute $-228,590\text{kJ/kmol}$ 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).

$\begin{array}{c}\ln K_P=-\frac{-228,590\text{kJ/kmol}}{\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\times 25°\text{C}}\\ =-\frac{-228,590\text{kJ/kmol}}{\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\times \left(25+273\right)\text{K}}\\ =92.26\end{array}$

Substitute $92.26$ for $K_P$ in eEquation (III).

$\begin{array}{c}{K}_P=e^{\left(92.26\right)}\\ =1.1695\times {10}^{40}\end{array}$

Thus, the equilibrium constant obtained from the equilibrium reaction at 298 K is $\underset{\_}{1.1695\times {10}^{40}}$.

Refer Table A-28, “Natural logarithms of the equilibrium constant” obtain the equilibrium constant for the reaction at the temperature of 298 K as $\underset{\_}{92.21}$.

Substitute $92.21$ for $\ln K_P$ in Equation (III).

$\begin{array}{c}{K}_P=e^{\left(92.21\right)}\\ =1.1125\times {10}^{40}\end{array}$

Thus, the equilibrium constant obtained from the table A-28  at 1440 R is $\underset{\_}{1.1125\times {10}^{40}}$.

The value obtained for equilibrium constant at 298 K from the definition of the equilibrium constant is $92.26$ which is equal to the value obtained for equilibrium constant as $92.21$ from the Table A-28.

(b)

To determine

The 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 21P

The logarithm equilibrium constant for the reaction at 2000 K is $\underset{\_}{3471}$.

The equilibrium constant obtained from the equilibrium constants of Table A-28 at 2000K is.

$\underset{\_}{3446.1}$

Explanation of Solution

Express the standard-state Gibbs function change.

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

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

$\begin{array}{l}{\overline{h}}_{f,{\text{H}}_{\text{2}}}^o=0\\ {\overline{h}}_{f,{\text{O}}_{\text{2}}}^o=0\\ {\overline{h}}_{f,{\text{H}}_{\text{2}}\text{O}}^o=-\text{241,820}\text{kJ/kmol}\end{array}$

Refer to Table A-22, obtain the value of ${\overline{h}}_{{\text{H}}_{\text{2}}}^o$ at temperature of 298K.

${\overline{h}}_{{\text{H}}_{\text{2}}}^o=9904\text{kJ/kmol}$

Refer to Table A-22, obtain the value of ${\overline{h}}_{{\text{H}}_{\text{2}}}$ and ${\overline{s}}_{{\text{H}}_{\text{2}}}$ at temperature of 2000K.

$\begin{array}{l}{\overline{h}}_{{\text{H}}_{\text{2}}}=\text{61,400 kJ/kmol}\\ {\overline{s}}_{{\text{H}}_{\text{2}}}=188.297\text{kJ/kmol}\cdot \text{K}\end{array}$

Refer to Table A-19E, 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 to Table A-19E, obtain the value of ${\overline{h}}_{{\text{O}}_2}$ and ${\overline{s}}_{{\text{O}}_2}$ at temperature of 2000K.

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

Refer to Table A-23, obtain the value of ${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}^o$ at temperature of 298K.

${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}^o=\text{9904 kJ/kmol}$

Refer to Table A-23, 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.

$\begin{array}{l}{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}^o=82,593\text{ kJ/kmol}\\ {\overline{s}}_{{\text{H}}_2\text{O}}=264.571\text{ kJ/kmol}\cdot \text{K}\end{array}$

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 (IV).

$\begin{array}{c}\Delta G^{*}\left(T\right)=\left[\begin{array}{l}1{\left(\begin{array}{l}-241,820\text{kJ/kmol}+82593\text{ kJ/kmol}-\\ 9904\text{kJ/kmol}-2000\text{ K}\times 264.571\text{ kJ/kmol}\cdot \text{K}\end{array}\right)}_{{\text{H}}_{\text{2}}\text{O}}-\\ 1{\left(\begin{array}{l}0+61,400\text{ kJ/kmol}-\text{8468kJ/kmol}\\ -2000\text{K}\times 188.297\text{ kJ/kmol}\cdot \text{K}\end{array}\right)}_{{\text{H}}_{\text{2}}}-\\ 0.5{\left(\begin{array}{l}0+\left(67881\text{ kJ/kmol}-8682\text{ kJ/kmol}\right)\\ -2000\text{K}\times 268.655\text{ kJ/kmol}\cdot \text{K}\end{array}\right)}_{{\text{O}}_{\text{2}}}\end{array}\right]\\ =-135.556\text{ kJ/kmol}\end{array}$

Substitute $53,436\text{Btu/lbmol}$ for $\Delta G^{*}\left(T\right)$, $3960\text{ R}$ for T, and $1.986\text{Btu/lbmol}\cdot \text{R}$ for $R_u$ in Equation (V).

$\begin{array}{c}\ln K_P=\frac{-135,556\text{kJ/kmol}}{\left(1.986\text{kJ/kmol}\cdot \text{K}\right)\times 2000\text{K}}\\ =8.152\end{array}$

Substitute $8.152$ for $K_P$ in Equation (III).

$\begin{array}{c}{K}_P=e^{\left(8.152\right)}\\ =3471\end{array}$

Thus, the equilibrium constant obtained from the equilibrium reaction at 2000K is $\underset{\_}{3471}$.

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 $8.145$.

Substitute $8.145$ for $K_P$ in Equation (VI).

$\begin{array}{c}{K}_P=e^{\left(8.145\right)}\\ =3446.10\end{array}$

Thus, the equilibrium constant obtained from the table A-28 at2000K is $\underset{\_}{3446.10}$.

The value obtained for equilibrium constant at 2000K from the definition of the equilibrium constant is $8.152$ which is almost equal to the value obtained for equilibrium constant as $8.145$ from the Table A-28.