Chapter 16.6, Problem 15P

(a)

To determine

The logarithm equilibrium constant for the reaction at 1440 R.

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

Answer to Problem 15P

The logarithm equilibrium constant for the reaction at 1440 R is $\underset{\_}{4.93\times {10}^{-14}}$.

The equilibrium constant obtained from the equilibrium constants of Table A-28 at 1440Ris $\underset{\_}{6.4712\times {10}^{-16}}$.

Explanation of Solution

Express the standard-state Gibbs function change.

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

$\ln K_P=\frac{-\Delta G^{*}\left(T\right)}{R_{u}T}$ (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 $v_{{\text{H}}_{\text{2}}\text{O}},v_{\text{CO}},v_{{\text{H}}_{\text{2}}},\text{and}{v}_{{\text{CO}}_{\text{2}}}$ 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:

$\begin{array}{l}{\overline{h}}_{f,{\text{H}}_{\text{2}}}^o=0\\ {\overline{h}}_{f,{\text{O}}_{\text{2}}}^o=0\\ {\overline{h}}_{f,{\text{H}}_{\text{2}}\text{O}}^o=-\text{104,040 Btu/lbmol}\end{array}$

Refer Table A-22, obtain the value of ${\overline{h}}_{{\text{H}}_{\text{2}}}^o$ at temperature of 537 R.

${\overline{h}}_{{\text{H}}_{\text{2}}}^o=\text{3640}\text{.}3\text{Btu/lbmol}$

Refer Table A-22, obtain the value of ${\overline{h}}_{{\text{H}}_{\text{2}}}$ and ${\overline{s}}_{{\text{H}}_{\text{2}}}$ at temperature of 1440R.

$\begin{array}{l}{\overline{h}}_{{\text{H}}_{\text{2}}}=\text{9956}\text{.9}\text{Btu/lbmol}\\ {\overline{s}}_{{\text{H}}_{\text{2}}}=38.079\text{Btu/lbmol}\cdot \text{R}\end{array}$

Refer Table A-19E, obtain the value of ${\overline{h}}_{{\text{O}}_2}^o$ at temperature of 537 R.

${\overline{h}}_{{\text{O}}_2}^o=\text{3725}\text{.1 Btu/lbmol}$

Refer Table A-19E, obtain the value of ${\overline{h}}_{{\text{O}}_2}$ and ${\overline{s}}_{{\text{O}}_2}$ at temperature of 1440 R.

$\begin{array}{l}{\overline{h}}_{{\text{O}}_2}=\text{10532}\text{.0}\text{Btu/lbmol}\\ {\overline{s}}_{{\text{O}}_2}=56.326\text{Btu/lbmol}\cdot \text{R}\end{array}$

Refer Table A-23, obtain the value of ${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}^o$ at temperature of 537 R.

${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}^o=\text{4258}\text{.0}\text{Btu/lbmol}$

Refer 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 2500 K.

$\begin{array}{l}{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}^o=\text{11993}\text{.4}\text{Btu/lbmol}\\ {\overline{s}}_{{\text{H}}_2\text{O}}=\text{53}\text{.428}\text{Btu/lbmol}\cdot \text{R}\end{array}$

Substitute 1 for $v_{{\text{H}}_{\text{2}}}$, 0 for ${\overline{h}}_f^o{}_{H_2}$, $9956.9\text{Btu/lbm}$ for ${\overline{h}}_{H_2}$ , $3640.3\text{ Btu/lbm}$ for ${\overline{h}}^o{}_{H_2}$, 1440 R for T, $38.079\text{Btu/lbm}\cdot \text{R}$ for ${\overline{s}}_{H_2}$, $0.5$ for $v_{{\text{O}}_2}$, 0 for ${\overline{h}}_f^o{,}_{{\text{O}}_2}$, $10532.0\text{Btu/lbm}$ for ${\overline{h}}_{{}_{{\text{O}}_2}}$, $3725.1\text{ Btu/lbm}$ for ${\overline{h}}^o{}_{{}_{{\text{O}}_2}}$, $56.325\text{Btu/lbm}\cdot \text{R}$ for ${\overline{s}}_{{}_{{\text{O}}_2}}$, 1 for $v_{{\text{H}}_{\text{2}}\text{O}}$, $-104,040\text{Btu/lbm}$ for ${\overline{h}}_f^o{,}_{{\text{H}}_{\text{2}}\text{O}}$, $11,933.4\text{Btu/lbm}$ for ${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}$, $4258\text{Btu/lbm}$ for ${\overline{h}}^o{}_{{\text{H}}_{\text{2}}\text{O}}$, and $53.428\text{Btu/lbm}\cdot \text{R}$ for ${\overline{s}}_{{\text{H}}_{\text{2}}\text{O}}$ in Equation (I).

$\begin{array}{c}\Delta G^{*}\left(T\right)=\left[\begin{array}{l}1{\left(0+\left(9956.9\text{Btu/lbm}-3640.3\text{ Btu/lbm}\right)-1440\text{R}\times 38.079\text{Btu/lbm}\cdot \text{R}\right)}_{{\text{H}}_{\text{2}}}+\\ 0.5{\left(0+\left(10532.0\text{Btu/lbm}-3725.1\text{ Btu/lbm}\right)-1440\text{R}\times 56.325\text{Btu/lbm}\cdot \text{R}\right)}_{{\text{O}}_{\text{2}}}-\\ 1{\left(\begin{array}{l}-104,040\text{Btu/lbm}+\left(11,933.4\text{Btu/lbm}-4258\text{Btu/lbm}\right)\\ -1440\text{R}\times 53.428\text{Btu/lbm}\cdot \text{R}\end{array}\right)}_{{\text{O}}_{\text{2}}}\end{array}\right]\\ =87,632\text{Btu}/\text{lbmol}\end{array}$

Substitute $87,632\text{Btu/lbmol}$ for $\Delta G^{*}\left(T\right)$, $1440\text{ R}$ for T, and $1.986\text{Btu/lbmol}\cdot \text{R}$ for $R_u$ in Equation (II).

$\begin{array}{c}{K}_P=\frac{-87,632\text{Btu/lbmol}}{\left(1.986\text{Btu/lbmol}\cdot \text{R}\right)\times 1440\text{R}}\\ =-30.64\end{array}$

Substitute $-30.64$ for $K_P$ in Equation (III).

$\begin{array}{c}{K}_P=e^{\left(-30.64\right)}\\ =4.93\times {10}^{-14}\end{array}$

Thus, the equilibrium constant obtained from the equilibrium reaction at 1440R is $\underset{\_}{4.93\times {10}^{-14}}$.

Convert the temperature from Rankine to Kelvin.

$\begin{array}{c}T=1440\text{R}\times \frac{1\text{ K}}{1.8\text{ R}}\\ =800\text{ K}\end{array}$

Refer Table A-28, “Natural logarithms of the equilibrium constant” obtain the equilibrium constant for the reaction by interpolating for the temperature of 800 K as $\underset{\_}{-34.974}$.

Substitute $-34.974$ for $K_P$ in Equation (III).

$\begin{array}{c}{K}_P=e^{\left(-34.974\right)}\\ =6.4712\times {10}^{-16}\end{array}$

Thus, the equilibrium constant obtained from the table A-28  at 1440 R is $\underset{\_}{6.4712\times {10}^{-16}}$.

Refer Table A-28 “Natural logarithms of the equilibrium constant”, obtain the equilibrium constant for the dissociation reaction ${\text{N}}_2\iff 2\text{N}$  at 3000 k is $\ln K_P=-22.359$

The value obtained for equilibrium constant at 1440R from the definition of the equilibrium constant is $-30.64$ which is smaller than the value obtained for equilibrium constant at $-34.974$ from the Table A-28.

(b)

To determine

The logarithm equilibrium constant for the reaction at 3960 R.

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

Answer to Problem 15P

The logarithm equilibrium constant for the reaction at 3960 R is $\underset{\_}{1.125\times {10}^{-3}}$.

The equilibrium constant obtained from the equilibrium constants of Table A-28 at 3960 R is.

$\underset{\_}{1.150\times {10}^{-3}}$

Explanation of Solution

Express the standard-state Gibbs function change.

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

$\begin{array}{l}{\overline{h}}_{f,{\text{H}}_{\text{2}}}^o=0\\ {\overline{h}}_{f,{\text{O}}_{\text{2}}}^o=0\\ {\overline{h}}_{f,{\text{H}}_{\text{2}}\text{O}}^o=-\text{104,040 Btu/lbmol}\end{array}$

Refer to Table A-22, obtain the value of ${\overline{h}}_{{\text{H}}_{\text{2}}}^o$ at temperature of 537 R.

${\overline{h}}_{{\text{H}}_{\text{2}}}^o=\text{3640}\text{.}3\text{Btu/lbmol}$

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 3960 R.

$\begin{array}{l}{\overline{h}}_{{\text{H}}_{\text{2}}}=\text{29,370}\text{.5}\text{Btu/lbmol}\\ {\overline{s}}_{{\text{H}}_{\text{2}}}=45.765\text{Btu/lbmol}\cdot \text{R}\end{array}$

Refer to Table A-19E, obtain the value of ${\overline{h}}_{{\text{O}}_2}^o$ at temperature of 537 R.

${\overline{h}}_{{\text{O}}_2}^o=\text{3725}\text{.1 Btu/lbmol}$

Refer to Table A-19E, obtain the value of ${\overline{h}}_{{\text{O}}_2}$ and ${\overline{s}}_{{\text{O}}_2}$ at temperature of 1440 R.

$\begin{array}{l}{\overline{h}}_{{\text{O}}_2}=\text{32,441}\text{.0}\text{Btu/lbmol}\\ {\overline{s}}_{{\text{O}}_2}=65.032\text{Btu/lbmol}\cdot \text{R}\end{array}$

Refer to Table A-23, obtain the value of ${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}^o$ at temperature of 537 R.

${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}^o=\text{4258}\text{.0}\text{Btu/lbmol}$

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 2500 K.

$\begin{array}{l}{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}^o=\text{39,989}\text{Btu/lbmol}\\ {\overline{s}}_{{\text{H}}_2\text{O}}=\text{64}\text{.402}\text{Btu/lbmol}\cdot \text{R}\end{array}$

Substitute 1 for $v_{{\text{H}}_{\text{2}}}$, 0 for ${\overline{h}}_f^o{}_{H_2}$, $29370.5\text{Btu/lbm}$ for ${\overline{h}}_{H_2}$ , $3640.3\text{ Btu/lbm}$ for ${\overline{h}}^o{}_{H_2}$, 3960 R for T, $45.765\text{Btu/lbm}\cdot \text{R}$ for ${\overline{s}}_{H_2}$, $0.5$ for $v_{{\text{O}}_2}$, 0 for ${\overline{h}}_f^o{,}_{{\text{O}}_2}$, $32,441\text{Btu/lbm}$ for ${\overline{h}}_{{}_{{\text{O}}_2}}$, $3725.1\text{ Btu/lbm}$ for ${\overline{h}}^o{}_{{}_{{\text{O}}_2}}$, $39650\text{Btu/lbm}\cdot \text{R}$ for ${\overline{s}}_{{}_{{\text{O}}_2}}$, 1 for $v_{{\text{H}}_{\text{2}}\text{O}}$, $-104,040\text{Btu/lbm}$ for ${\overline{h}}_f^o{,}_{{\text{H}}_{\text{2}}\text{O}}$, $39,989\text{Btu/lbm}$ for ${\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}$, $4258\text{Btu/lbm}$ for ${\overline{h}}^o{}_{{\text{H}}_{\text{2}}\text{O}}$, and $64.402\text{Btu/lbm}\cdot \text{R}$ for ${\overline{s}}_{{\text{H}}_{\text{2}}\text{O}}$ in equation (IV).

$\begin{array}{c}\Delta G^{*}\left(T\right)=\left[\begin{array}{l}1{\left(0+\left(29370.5\text{Btu/lbm}-3640.3\text{ Btu/lbm}\right)-3960\text{R}\times 45.765\text{Btu/lbm}\cdot \text{R}\right)}_{{\text{H}}_{\text{2}}}+\\ 0.5{\left(0+\left(32,441\text{Btu/lbm}-3725.1\text{ Btu/lbm}\right)-3960\text{R}\times 65.032\text{Btu/lbm}\cdot \text{R}\right)}_{{\text{O}}_{\text{2}}}-\\ 1{\left(-104,040\text{Btu/lbm}+\left(39,989\text{Btu/lbm}-4258\text{Btu/lbm}\right)-3960\text{R}\times 64.402\text{Btu/lbm}\cdot \text{R}\right)}_{{\text{O}}_{\text{2}}}\end{array}\right]\\ =53,436\text{Btu}/\text{lbmol}\end{array}$

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).

$\begin{array}{c}\ln K_P=\frac{-53,436\text{Btu/lbmol}}{\left(1.986\text{Btu/lbmol}\cdot \text{R}\right)\times 3960\text{R}}\\ =-6.79\end{array}$

Substitute $-6.79$ for $K_P$ in Equation (VI).

$\begin{array}{c}{K}_P=e^{\left(-6.79\right)}\\ =888.91\end{array}$

Thus, the equilibrium constant obtained from the equilibrium reaction at 3960 R is $\underset{\_}{1.120\times {10}^{-3}}$.

Convert the temperature from Rankine to Kelvin.

$\begin{array}{c}T=3960\text{R}\times \frac{1\text{ K}}{1.8\text{ R}}\\ =2200\text{ K}\end{array}$

Refer Table A-28, “Natural logarithms of the equilibrium constant” obtain the equilibrium constant for the reaction by interpolating for the temperature of 2200 K as $-6.768$.

Substitute $-6.768$ for $K_P$ in equation (VI).

$\begin{array}{c}{K}_P=e^{\left(-6.768\right)}\\ =1.150\times {10}^{-3}\end{array}$

Thus, the equilibrium constant obtained from the table A-28 at3960 R is $\underset{\_}{1.150\times {10}^{-3}}$.

The value obtained for equilibrium constant at 3960R from the definition of the equilibrium constant is $-6.79$ which is almost equal to the value obtained for equilibrium constant as $-6.768$ from the Table A-28.