Chapter 16.6, Problem 15P

(a)

To determine

The equilibrium constant for the reported reaction at 1000 K.

Answer to Problem 15P

The equilibrium constant for the reported reaction at 1000 K is $8.66\times {10}^{-11}$.

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 ${\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }$.

Write the Gibbs function of ${\text{H}}_{\text{2}}$$\left({\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }\right)$ at 1000 K.

${\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }={\overline{h}}_{{\text{H}}_{\text{2}}}+{\left[{\overline{h}}_{1000}-{\overline{h}}_{298}-Ts\right]}_{{\text{H}}_{\text{2}}}$ (III)

Here, enthalpy of formation of ${\text{H}}_{\text{2}}$ is ${\overline{h}}_{{\text{H}}_{\text{2}}}$, enthalpy of ${\text{H}}_{\text{2}}$ gas at 1000 K is ${\overline{h}}_{1000}$, enthalpy of ${\text{H}}_{\text{2}}$ gas at 298 K is ${\overline{h}}_{298}$.

Write the Gibbs function of ${\text{O}}_{\text{2}}$$\left({\overline{g}}_{O_2}^{\ast }\right)$ at 1000 K.

${\overline{g}}_{{\text{O}}_{\text{2}}}^{\ast }={\overline{h}}_{{\text{O}}_{\text{2}}}+{\left[{\overline{h}}_{1000}-{\overline{h}}_{298}-Ts\right]}_{{\text{O}}_{\text{2}}}$ . (IV)

Here, enthalpy of formation of ${\text{O}}_{\text{2}}$ is ${\overline{h}}_{{\text{O}}_{\text{2}}}$, enthalpy of ${\text{O}}_{\text{2}}$ gas at 1000 K is ${\overline{h}}_{1000}$, enthalpy of ${\text{O}}_{\text{2}}$ gas at 298 K is ${\overline{h}}_{298}$.

Write the Gibbs function of water $\left({\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }\right)$ at 1000 K.

${\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }={\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}+{\left[{\overline{h}}_{1000}-{\overline{h}}_{298}-Ts\right]}_{{\text{H}}_{\text{2}}\text{O}}$ (V)

Here, enthalpy of formation of water vapor is ${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}$, enthalpy of water vapor at 1000 K is ${\overline{h}}_{1000}$, enthalpy of water vapor at 298 K is ${\overline{h}}_{298}$, and entropy is s.

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-22, “Ideal gas properties of Hydrogen”, obtain the following properties of hydrogen gas.

Enthalpy of hydrogen gas at 1000 K, ${\left(\overline{h}\right)}_{{\text{H}}_{\text{2}}}=29,154\text{kJ}/\text{kmol}$.

Enthalpy of hydrogen gas at 298 K, ${\left({\overline{h}}^{\circ }\right)}_{{\text{H}}_{\text{2}}}=8,468\text{kJ}/\text{kmol}$.

Entropy of hydrogen gas at 1000 K, $s=166.114\text{kJ}/\text{kmol}\cdot \text{K}$.

Substitute $0$ for ${\overline{h}}_{{\text{H}}_{\text{2}}}$, $29,154\text{kJ}/\text{kmol}$ for ${\overline{h}}_{1000}$, $8,468\text{kJ}/\text{kmol}$ for ${\overline{h}}_{298}$, 1000 K for T, and $166.114\text{kJ}/\text{kmol}\cdot \text{K}$ for s in Equation (III).

$\begin{array}{c}{\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }=\left\{0+\left[\begin{array}{l}29,154\text{kJ}/\text{kmol}\\ -8,468\text{kJ}/\text{kmol}\\ -\left(1,000\text{K}\right)\left(166.114\text{kJ}/\text{kmol}\cdot \text{K}\right)\end{array}\right]\right\}\\ =-145,428\text{kJ}/\text{kmol}\end{array}$

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-19, “Ideal gas properties of Oxygen”, obtain the following properties of Oxygen gas.

Enthalpy of oxygen gas at 1000 K, ${\left(\overline{h}\right)}_{{\text{O}}_{\text{2}}}=31,389\text{kJ}/\text{kmol}$.

Enthalpy of oxygen gas at 298 K, ${\left({\overline{h}}^{\circ }\right)}_{{\text{O}}_{\text{2}}}=8,682\text{kJ}/\text{kmol}$.

Entropy of oxygen gas at 1000 K, $s=243.471\text{kJ}/\text{kmol}\cdot \text{K}$.

Substitute $0$ for ${\overline{h}}_{{\text{O}}_{\text{2}}}$, $31,389\text{kJ}/\text{kmol}$ for ${\overline{h}}_{1000}$, $8,682\text{kJ}/\text{kmol}$ for ${\overline{h}}_{298}$, 1000 K for T, and $243.471\text{kJ}/\text{kmol}\cdot \text{K}$ for s in Equation (IV).

$\begin{array}{c}{\overline{g}}_{{\text{O}}_{\text{2}}}^{\ast }=\left\{0+\left[\begin{array}{l}31,389\text{kJ}/\text{kmol}\\ -8,682\text{kJ}/\text{kmol}\\ -\left(1,000\text{K}\right)\left(243.471\text{kJ}/\text{kmol}\cdot \text{K}\right)\end{array}\right]\right\}\\ =-220,764\text{kJ}/\text{kmol}\end{array}$

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 $-241,820\text{kJ}/\text{kmol}$.

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

Enthalpy of water vapor at 1000 K, ${\left(\overline{h}\right)}_{{\text{H}}_{\text{2}}\text{O}}=35,882\text{kJ}/\text{kmol}$.

Enthalpy of water vapor at 298 K, ${\left({\overline{h}}^{\circ }\right)}_{{\text{H}}_{\text{2}}\text{O}}=9,904\text{kJ}/\text{kmol}$.

Entropy of water vapor at 1000 K, $s=232.597\text{kJ}/\text{kmol}\cdot \text{K}$.

Substitute $-241,820\text{kJ}/\text{kmol}$ for ${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}$, $35,882\text{kJ}/\text{kmol}$ for ${\overline{h}}_{1000}$, $9,904\text{kJ}/\text{kmol}$ for ${\overline{h}}_{298}$, 1000 K for T, and $232.597\text{kJ}/\text{kmol}\cdot \text{K}$ for s in Equation (V).

$\begin{array}{c}{\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }=\left\{\begin{array}{l}-241,820\text{kJ}/\text{kmol}\\ +\left[\begin{array}{l}35,882\text{kJ}/\text{kmol}\\ -9,904\text{kJ}/\text{kmol}\\ -\left(1,000\text{K}\right)\left(232.597\text{kJ}/\text{kmol}\cdot \text{K}\right)\end{array}\right]\end{array}\right\}\\ =-448,439\text{kJ}/\text{kmol}\end{array}$

Substitute 1 for $v_{{\text{H}}_{\text{2}}}$, 0.5 for $v_{{\text{O}}_{\text{2}}}$, 1 for $v_{{\text{H}}_{\text{2}}\text{O}}$, $-448,439\text{kJ}/\text{kmol}$ for ${\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }$, $-145,428\text{kJ}/\text{kmol}$ for ${\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }$, and $-220,764\text{kJ}/\text{kmol}$ for ${\overline{g}}_{{\text{O}}_{\text{2}}}^{\ast }$ in Equation (II).

$\begin{array}{c}\Delta G^{\ast }\left(T\right)=v_{{\text{H}}_{\text{2}}}{\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }\left(T\right)+v_{{\text{O}}_{\text{2}}}{\overline{g}}_{{\text{O}}_{\text{2}}}^{\ast }\left(T\right)-v_{{\text{H}}_{\text{2}}\text{O}}{\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }\left(T\right)\\ =\left\{\begin{array}{l}\left(1\right)\left(-145,428\text{kJ}/\text{kmol}\right)\\ +\left(0.5\right)\left(-220,764\text{kJ}/\text{kmol}\right)\\ -\left(1\right)\left(-448,439\text{kJ}/\text{kmol}\right)\end{array}\right\}\\ =192,629\text{kJ}/\text{kmol}\end{array}$

Substitute $192,629\text{kJ}/\text{kmol}$ for $\Delta G^{\ast }\left(T\right)$, $8.314\text{kJ}/\text{kmol}$ for $R_u$, and 1000 K for T in Equation (I).

$\begin{array}{c}\ln k_p=-\Delta G^{\ast }\left(T\right)/R_{u}T\\ =\frac{-\left(192,629\text{kJ}/\text{kmol}\right)}{\left(8.314\text{kJ}/\text{kmol}\right)\left(1,000\text{K}\right)}\\ =-23.1692\end{array}$

$\begin{array}{c}{k}_p=e^{-23.1692}\\ =8.66\times {10}^{-11}\end{array}$

Thus, the equilibrium constant for the reported reaction at 1000 K is $8.66\times {10}^{-11}$.

Refer table A-28, “Natural logarithm of equilibrium constants”, obtain the value of $\ln k_p$ for water as $-23.163$ at 1000 K.

(b)

To determine

The equilibrium constant for the reported reaction at 2000 K.

Answer to Problem 15P

The equilibrium constant for the reported reaction at 2000 K is $2.88\times {10}^{-4}$.

Explanation of Solution

Write the Gibbs function of ${\text{H}}_{\text{2}}$$\left({\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }\right)$ at 2000 K.

${\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }={\overline{h}}_{{\text{H}}_{\text{2}}}+{\left[{\overline{h}}_{2000}-{\overline{h}}_{298}-Ts\right]}_{{\text{H}}_{\text{2}}}$ (VI)

Here, enthalpy of ${\text{H}}_{\text{2}}$ gas at 2000 K is ${\overline{h}}_{2000}$.

Write the Gibbs function of ${\text{O}}_{\text{2}}$$\left({\overline{g}}_{O_2}^{\ast }\right)$ at 2000 K.

${\overline{g}}_{{\text{O}}_{\text{2}}}^{\ast }={\overline{h}}_{{\text{O}}_{\text{2}}}+{\left[{\overline{h}}_{2000}-{\overline{h}}_{298}-Ts\right]}_{{\text{O}}_{\text{2}}}$ (VII)

Here, enthalpy of ${\text{O}}_{\text{2}}$ gas at 2000 K is ${\overline{h}}_{2000}$.

Write the Gibbs function of water $\left({\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }\right)$ at 2000 K.

${\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }={\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}+{\left[{\overline{h}}_{2000}-{\overline{h}}_{298}-Ts\right]}_{{\text{H}}_{\text{2}}\text{O}}$ (VIII)

Here, enthalpy of water vapor at 2000 K is ${\overline{h}}_{2000}$.

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-22, “Ideal gas properties of Hydrogen”, obtain the following properties of hydrogen gas.

Enthalpy of hydrogen gas at 2000 K, ${\left(\overline{h}\right)}_{{\text{H}}_{\text{2}}}=61,400\text{kJ}/\text{kmol}$.

Entropy of hydrogen gas at 2000 K, $s=188.297\text{kJ}/\text{kmol}\cdot \text{K}$.

Substitute $0$ for ${\overline{h}}_{{\text{H}}_{\text{2}}}$, $61,400\text{kJ}/\text{kmol}$ for ${\overline{h}}_{2000}$, $8,468\text{kJ}/\text{kmol}$ for ${\overline{h}}_{298}$, 2000 K for T, and $188.297\text{kJ}/\text{kmol}\cdot \text{K}$ for s in Equation (VI).

$\begin{array}{c}{\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }=\left\{0+\left[\begin{array}{l}61,400\text{kJ}/\text{kmol}\\ -8,468\text{kJ}/\text{kmol}\\ -\left(2000\text{K}\right)\left(188.297\text{kJ}/\text{kmol}\cdot \text{K}\right)\end{array}\right]\right\}\\ =-323,662\text{kJ}/\text{kmol}\end{array}$

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-19, “Ideal gas properties of Oxygen”, obtain the following properties of Oxygen gas.

Enthalpy of oxygen gas at 2000 K, ${\left(\overline{h}\right)}_{{\text{O}}_{\text{2}}}=67,881\text{kJ}/\text{kmol}$.

Entropy of oxygen gas at 2000 K, $s=268.655\text{kJ}/\text{kmol}\cdot \text{K}$.

Substitute $0$ for ${\overline{h}}_{{\text{O}}_{\text{2}}}$, $67,881\text{kJ}/\text{kmol}$ for ${\overline{h}}_{2000}$, $8,682\text{kJ}/\text{kmol}$ for ${\overline{h}}_{298}$, 2000 K for T, and $268.655\text{kJ}/\text{kmol}\cdot \text{K}$ for s in Equation (VII).

$\begin{array}{c}{\overline{g}}_{{\text{O}}_{\text{2}}}^{\ast }=\left\{0+\left[\begin{array}{l}67,881\text{kJ}/\text{kmol}\\ -8,682\text{kJ}/\text{kmol}\\ -\left(2000\text{K}\right)\left(268.655\text{kJ}/\text{kmol}\cdot \text{K}\right)\end{array}\right]\right\}\\ =-478,111\text{kJ}/\text{kmol}\end{array}$

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 $-241,820\text{kJ}/\text{kmol}$.

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

Enthalpy of water vapor at 2000 K, ${\left(\overline{h}\right)}_{{\text{H}}_{\text{2}}\text{O}}=82,593\text{kJ}/\text{kmol}$.

Entropy of water vapor at 2000 K, $s=264.571\text{kJ}/\text{kmol}\cdot \text{K}$.

Substitute $-241,820\text{kJ}/\text{kmol}$ for ${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}$, $82,593\text{kJ}/\text{kmol}$ for ${\overline{h}}_{2000}$, $9,904\text{kJ}/\text{kmol}$ for ${\overline{h}}_{298}$, 2000 K for T, and $264.571\text{kJ}/\text{kmol}\cdot \text{K}$ for s in Equation (VIII).

$\begin{array}{c}{\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }=\left\{\begin{array}{l}-241,820\text{kJ}/\text{kmol}\\ +\left[\begin{array}{l}82,593\text{kJ}/\text{kmol}\\ -9,904\text{kJ}/\text{kmol}\\ -\left(2000\text{K}\right)\left(264.571\text{kJ}/\text{kmol}\cdot \text{K}\right)\end{array}\right]\end{array}\right\}\\ =-698,273\text{kJ}/\text{kmol}\end{array}$

Substitute 1 for $v_{{\text{H}}_{\text{2}}}$, 0.5 for $v_{{\text{O}}_{\text{2}}}$, 1 for $v_{{\text{H}}_{\text{2}}\text{O}}$, $-698,273\text{kJ}/\text{kmol}$ for ${\overline{g}}_{{\text{H}}_{\text{2}}\text{O}}^{\ast }$, $-323,662\text{kJ}/\text{kmol}$ for ${\overline{g}}_{{\text{H}}_{\text{2}}}^{\ast }$, and $-478,111\text{kJ}/\text{kmol}$ for ${\overline{g}}_{{\text{O}}_{\text{2}}}^{\ast }$ in Equation (II).

$\begin{array}{c}\Delta G^{\ast }\left(T\right)=\left\{\begin{array}{l}\left(1\right)\left(-323,662\text{kJ}/\text{kmol}\right)\\ +\left(0.5\right)\left(-478,111\text{kJ}/\text{kmol}\right)\\ -\left(1\right)\left(-698,273\text{kJ}/\text{kmol}\right)\end{array}\right\}\\ =135,555.5\text{kJ}/\text{kmol}\end{array}$

Substitute $135,555.5\text{kJ}/\text{kmol}$ for $\Delta G^{\ast }\left(T\right)$, $8.314\text{kJ}/\text{kmol}$ for $R_u$, and 2000 K for T in Equation (I).

$\begin{array}{c}\ln k_p=-\Delta G^{\ast }\left(T\right)/R_{u}T\\ =\frac{-\left(135,555.5\text{kJ}/\text{kmol}\right)}{\left(8.314\text{kJ}/\text{kmol}\right)\left(2000\text{K}\right)}\\ =-8.1522\end{array}$

$\begin{array}{c}{k}_p=e^{-8.1522}\\ =2.88\times {10}^{-4}\end{array}$

Thus, the equilibrium constant for the reported reaction at 2000 K is $2.88\times {10}^{-4}$.

Refer table A-28, “Natural logarithm of equilibrium constants”, obtain the value of $\ln k_p$ for water as $-8.5$ at 2000 K.

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