Chapter 16.6, Problem 20P

( a)

To determine

The equilibrium constant for the reported reaction at 537 R.

Answer to Problem 20P

The equilibrium constant for the reported reaction at 537 R is $1.12\times {10}^{40}$.

Explanation of Solution

Write the stoichiometric reaction for the reported process.

${\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}+1/2{\text{O}}_{\text{2}}$

From the stoichiometric reaction, infer that the stoichiometric coefficients for hydrogen $\left(v_{{\text{H}}_{\text{2}}}\right)$ is 1, for oxygen $\left(v_{{\text{O}}_{\text{2}}}\right)$ is 0.5, and for water $\left(v_{{\text{H}}_{\text{2}}\text{O}}\right)$ is 1 respectively.

Write the formula for equilibrium constant $\left(k_p\right)$.

$\begin{array}{l}{k}_p=e^{-\Delta G\left(T\right)/R_{u}T}\\ \ln k_p=-\Delta G^{\ast }\left(T\right)/R_{u}T\end{array}$ (I)

Here, temperature is T, Gibbs function is $\Delta G^{\ast }\left(T\right)$, and universal gas constant is $R_u$.

Write the formula for Gibbs energy $\left(\Delta G^{\ast }\left(T\right)\right)$.

$\Delta G^{\ast }\left(T\right)=v_{{\text{H}}_{\text{2}}}{\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }\left(T\right)+v_{{\text{O}}_{\text{2}}}{\overline{g}}_{{\text{O}}_{\text{2}}}^{\ast }\left(T\right)-v_{{\text{H}}_{\text{2}}\text{O}}{\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }\left(T\right)$ (II)

Here, Gibbs function of hydrogen is ${\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }$, Gibbs function of oxygen is ${\overline{g}}_{{\text{O}}_{\text{2}}}^{\ast }$, and Gibbs function of ${\text{H}}_{\text{2}}\text{O}$ is ${\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }$.

Conclusion:

Refer Table A-26, “Enthalpy formation table”, obtain the enthalpy of ${\text{H}}_{\text{2}}$$\left({\overline{h}}_{{\text{H}}_{\text{2}}}\right)$ as $0$.

Refer Table A-26, “Enthalpy formation table”, obtain the enthalpy of ${\text{O}}_{\text{2}}$$\left({\overline{h}}_{{\text{O}}_{\text{2}}}\right)$ as $0$.

Refer Table A-26, “Enthalpy formation table”, obtain the enthalpy formation of water vapor $\left({\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}\right)$ as $-241,820\text{kJ}/\text{kmol}$.

Substitute 1 for $v_{H_2}$, 0.5 for $v_{O_2}$, 1 for $v_{H_{2}O}$, 0 for ${\overline{g}}_{H_2}^{\ast }\left(T\right)$, 0 for ${\overline{g}}_{O_2}^{\ast }\left(T\right)$, and $-228,590\text{kJ}/\text{mol}$${\overline{g}}_{H_{2}O}^{\ast }\left(T\right)$ in Equation (II).

$\begin{array}{c}\Delta G^{\ast }\left(T\right)=\left(1\right)\left(-228,590\text{kJ}/\text{mol}\right)-\left(1\right)\left(0\right)-\left(0.5\right)\left(0\right)\\ =-228,590\text{kJ}/\text{mol}\frac{1\text{Btu}/\text{lbmol}}{2.236\text{kJ}/\text{mol}}\\ =-98,350\text{Btu}/\text{lbmol}\end{array}$

Substitute $-98,350\text{Btu}/\text{lbmol}$ for $\Delta G^{\ast }\left(T\right)$, $1.986\text{Btu}/\text{lbmol}$ for $R_u$, and 537 R for T in Equation (I).

$\begin{array}{c}\ln k_p=\frac{-\left(-98,350\text{Btu}/\text{lbmol}\right)}{\left(1.986\text{Btu}/\text{lbmol}\right)\left(537\text{R}\right)}\\ =92.22\end{array}$

$\begin{array}{c}{k}_p=e^{92.22}\\ =1.12\times {10}^{40}\end{array}$

Thus, the equilibrium constant for the reported reaction at 537 R is $1.12\times {10}^{40}$.

Refer table A-28, “Natural logarithm of equilibrium constants”, obtain the value of $\ln k_p$ for water as $92.22$ at $537\text{R}$.

(b)

To determine

The equilibrium constant for the reported reaction at 3240 R.

Answer to Problem 20P

The equilibrium constant for the reported reaction at 3240 R is $1.9\times {10}^4$.

Explanation of Solution

Write the Gibbs function of water $\left({\overline{g}}_{H_{2}O}^{\ast }\right)$ at 3240 R.

${\overline{g}}_{H_{2}O}^{\ast }={\overline{h}}_{H_{2}O}+{\left[{\overline{h}}_{4320}-{\overline{h}}_{537}-Ts\right]}_{H_{2}O}$ (IV)

Here, enthalpy of water at 537 R is ${\overline{h}}_{H_{2}O}$, enthalpy of water vapor at 4,320 R is ${\overline{h}}_{4320}$, enthalpy of water vapor at 537 R is ${\overline{h}}_{537}$, and entropy is s.

Write the Gibbs function of $H_2$$\left({\overline{g}}_{H_2}^{\ast }\right)$ at 3,240 R.

${\overline{g}}_{H_2}^{\ast }={\overline{h}}_{H_2}+{\left[{\overline{h}}_{4320}-{\overline{h}}_{298}-Ts\right]}_{H_2}$ (V)

Here, enthalpy of $H_2$ at 537 R is ${\overline{h}}_{H_2}$, enthalpy of $H_2$ gas at 4,320 R is ${\overline{h}}_{4320}$, and enthalpy of $H_2$ gas at 298 R is ${\overline{h}}_{298}$.

Write the Gibbs function of $O_2$$\left({\overline{g}}_{O_2}^{\ast }\right)$ at 3,240 R.

${\overline{g}}_{O_2}^{\ast }={\overline{h}}_{O_2}+{\left[{\overline{h}}_{4320}-{\overline{h}}_{298}-Ts\right]}_{O_2}$ (VI)

Here, enthalpy of $O_2$ at 537 R is ${\overline{h}}_{O_2}$, enthalpy of $O_2$ gas at 4,320 R is ${\overline{h}}_{4320}$, enthalpy of $O_2$ gas at 298 R is ${\overline{h}}_{298}$.

Conclusion:

Refer table A-26E, “Enthalpy formation table”,obtain the enthalpy formation of water vapor $\left({\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}\right)$ as $-104,040\text{Btu}/\text{lbmol}$.

Refer table A-23E, “Ideal gas properties of water vapor”, obtain the following properties of water vapor.

Enthalpy of water vapor at 4,320 R as $31,205\text{Btu}/\text{lbmol}$.

Enthalpyof water vapor at 537 R as $4,258\text{Btu}/\text{lbmol}$.

Entropy of water vapor at 3,240 R as $61.9\text{Btu}/\text{lbmol}\cdot \text{K}$.

Substitute $-104,040\text{Btu}/\text{lbmol}$ for ${\overline{h}}_{H_{2}O}$, $31,205\text{Btu}/\text{lbmol}$ for ${\overline{h}}_{4320}$, $4,258\text{Btu}/\text{lbmol}$ for ${\overline{h}}_{537}$, 3,240 K for T, and $61.9\text{Btu}/\text{lbmol}\cdot \text{K}$ for s in Equation (IV).

$\begin{array}{c}{\overline{g}}_{H_{2}O}^{\ast }=\left\{\begin{array}{l}-104,040\text{Btu}/\text{lbmol}\\ +\left[\begin{array}{l}31,205\text{Btu}/\text{lbmol}\\ -4,258\text{Btu}/\text{lbmol}\\ -\left(3,240\text{K}\right)\left(61.9\text{Btu}/\text{lbmol}\cdot \text{K}\right)\end{array}\right]\end{array}\right\}\\ =-277,805\text{Btu}/\text{lbmol}\end{array}$

Refer table A-26E, “Enthalpy formation table”, obtain the enthalpy of ${\text{H}}_{\text{2}}$$\left({\overline{h}}_{{\text{H}}_{\text{2}}}\right)$ as $0$.

Refer table A-22E, “Ideal gas properties of Hydrogen”, obtain the following properties of hydrogen gas.

Enthalpy of hydrogen gas at 4320 R as $23,485\text{Btu}/\text{lbmol}$.

Enthalpy of hydrogen gas at 298 R as $3,640\text{Btu}/\text{lbmol}$.

Entropy of hydrogen gas at 3,240 R as $44.125\text{Btu}/\text{lbmol}\cdot \text{K}$.

Substitute $0$ for ${\overline{h}}_{H_2}$, $23,485\text{Btu}/\text{lbmol}$ for ${\overline{h}}_{4320}$, $3,640\text{Btu}/\text{lbmol}$ for ${\overline{h}}_{298}$, 3,240 K for T, and $44.125\text{Btu}/\text{lbmol}\cdot \text{K}$ for s in Equation (V).

$\begin{array}{c}{\overline{g}}_{H_2}^{\ast }=\left\{0+\left[\begin{array}{l}23,485\text{Btu}/\text{lbmol}\\ -3,640\text{Btu}/\text{lbmol}\\ -\left(3,240\text{K}\right)\left(44.125\text{Btu}/\text{lbmol}\cdot \text{K}\right)\end{array}\right]\right\}\\ =-123,120\text{Btu}/\text{lbmol}\end{array}$

Refer table A-26E, “Enthalpy formation table”, obtain the enthalpy of ${\text{O}}_{\text{2}}$$\left({\overline{h}}_{{\text{O}}_{\text{2}}}\right)$ as $0$.

Refer table A-19E, “Ideal gas properties of Oxygen”, obtain the following properties of Oxygen gas.

Enthalpy of Oxygengas at 4,320 R as $25,972\text{Btu}/\text{lbmol}$.

Enthalpy of Oxygen gas at 298 R as $3,725\text{Btu}/\text{lbmol}$.

Entropy of Oxygen gas at 3240 R as $63.22\text{Btu}/\text{lbmol}\cdot \text{K}$.

Substitute $0$ for ${\overline{h}}_{O_2}$, $25,972\text{Btu}/\text{lbmol}$ for ${\overline{h}}_{4320}$, $3,725\text{Btu}/\text{lbmol}$ for ${\overline{h}}_{298}$, 3,240 K for T, and $63.22\text{Btu}/\text{lbmol}\cdot \text{K}$ for s in Equation (VI).

$\begin{array}{c}{\overline{g}}_{O_2}^{\ast }=\left\{0+\left[\begin{array}{l}25,972\text{Btu}/\text{lbmol}\\ -3,725\text{Btu}/\text{lbmol}\\ -\left(3,240\text{K}\right)\left(63.22\text{Btu}/\text{lbmol}\cdot \text{K}\right)\end{array}\right]\right\}\\ =-182,585\text{Btu}/\text{lbmol}\end{array}$

Substitute 1 for $v_{H_2}$, 0.5 for $v_{O_2}$, 1 for $v_{H_{2}O}$, $-277,805\text{Btu}/\text{lbmol}$ for ${\overline{g}}_{H_{2}O}^{\ast }$, $-123,120\text{Btu}/\text{lbmol}$ for ${\overline{g}}_{H_2}^{\ast }$, and $-182,585\text{Btu}/\text{lbmol}$ for ${\overline{g}}_{O_2}^{\ast }$ in Equation (II).

$\begin{array}{c}\Delta G^{\ast }\left(T\right)=\left\{\begin{array}{l}\left(1\right)\left(-277,805\text{Btu}/\text{lbmol}\cdot \text{K}\right)\\ -\left(1\right)\left(-123,120\text{Btu}/\text{lbmol}\cdot \text{K}\right)\\ -\left(0.5\right)\left(-182,585\text{Btu}/\text{lbmol}\cdot \text{K}\right)\end{array}\right\}\\ =-63,392\text{Btu}/\text{lbmol}\end{array}$

Substitute $-63,392\text{Btu}/\text{lbmol}$ for $\Delta G^{\ast }\left(T\right)$, $1.986\text{Btu}/\text{lbmol}$ for $R_u$, and 3,240 R for T in Equation (I).

$\begin{array}{c}\ln k_p=\frac{-\left(-63,392\text{Btu}/\text{lbmol}\right)}{\left(1.986\text{Btu}/\text{lbmol}\right)\left(3,240\text{R}\right)}\\ =9.85\end{array}$

$\begin{array}{c}{k}_p=e^{9.85}\\ =1.9\times {10}^4\end{array}$

Thus, the equilibrium constant for the reported reaction at 3,240 R is $1.9\times {10}^4$.

Refer table A-28, “Natural logarithm of equilibrium constants”, obtain the value of $\ln k_p$ for water as $9.826$ at $3,240\text{R}$.

Thus, both Gibbs function data and equilibrium constants table provide the same data.