Chapter 16.6, Problem 93RP

A constant-volume tank contains a mixture of 1 mol of H₂ and 0.5 mol of O₂ at 25°C and 1 atm. The contents of the tank are ignited, and the final temperature and pressure in the tank are 2800 K and 5 atm, respectively. If the combustion gases consist of H₂O, H₂, and O₂, determine (a) the equilibrium composition of the product gases and (b) the amount of heat transfer from the combustion chamber. Is it realistic to assume that no OH will be present in the equilibrium mixture?

To determine

The equilibrium composition of mixture of ${\text{H}}_{\text{2}}\text{O}$, ${\text{H}}_2$ and ${\text{O}}_{\text{2}}$ at 5 atm pressure and 2,800 K.

Answer to Problem 93RP

The equilibrium composition of mixture of ${\text{H}}_{\text{2}}\text{O}$, ${\text{H}}_2$ and ${\text{O}}_{\text{2}}$ at 5 atm pressure and 2,800 K is $0.944{\text{H}}_{\text{2}}\text{O}+0.056{\text{H}}_2+0.028{\text{O}}_{\text{2}}$.

Explanation of Solution

Write the stoichiometric reaction for dissociation of water.

${\text{H}}_2+1/2{\text{O}}_2\rightleftharpoons {\text{H}}_{\text{2}}\text{O}$ (I)

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

Write the actual reaction for the combustion process.

${\text{H}}_2+0.5{\text{O}}_2\to x{\text{H}}_{\text{2}}\text{O}+\left(1-x\right){\text{H}}_2+\left(0.5-0.5x\right){\text{O}}_{\text{2}}$ (II)

From Equation (II), infer that the equilibrium composition contains x amount of ${\text{H}}_{\text{2}}\text{O}$ $\left(N_{{\text{H}}_{\text{2}}\text{O}}\right)$, $\left(1-x\right)$ amount of ${\text{H}}_{\text{2}}$ $\left(N_{{\text{H}}_{\text{2}}}\right)$, and $\left(0.5-0.5x\right)$ amount of ${\text{O}}_{\text{2}}$ $\left(N_{{\text{O}}_{\text{2}}}\right)$.

Write the formula for total number of moles $\left(N_{total}\right)$.

$N_{total}=N_{{\text{H}}_{\text{2}}\text{O}}+N_{{\text{H}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}$ (III)

Substitute x for $N_{{\text{H}}_{\text{2}}\text{O}}$, $\left(1-x\right)$ for $N_{{\text{H}}_{\text{2}}}$, and $\left(0.5-0.5x\right)$ for $N_{{\text{O}}_{\text{2}}}$ in Equation (III).

$\begin{array}{c}{N}_{total}=x+\left(1-x\right)+\left(0.5-0.5x\right)\\ =1.5-0.5x\end{array}$

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

$k_p=\frac{N_{{\text{H}}_{\text{2}}}^{v_{{\text{H}}_{\text{2}}}}{N}_{{\text{O}}_{\text{2}}}^{v_{{\text{O}}_{\text{2}}}}}{N_{{\text{H}}_{\text{2}}\text{O}}^{v_{{\text{H}}_{\text{2}}\text{O}}}}{\left(\frac{P}{N_{total}}\right)}^{\left(v_{{\text{H}}_{\text{2}}}+v_{{\text{O}}_{\text{2}}}-v_{{\text{H}}_{\text{2}}\text{O}}\right)}$ (IV)

Here, pressure is P.

Conclusion:

Refer the table A-28, “Natural logarithm of equilibrium constants’, obtain the value of $\ln k_p$ as $-3.812$ for the reported reaction at 2,800 K. Then the equilibrium constant for the reaction is $0.0221$.

Substitute $0.0221$ for $k_p$, x for $N_{{\text{H}}_{\text{2}}\text{O}}$, $\left(1-x\right)$ for $N_{{\text{H}}_{\text{2}}}$, $\left(0.5-0.5x\right)$ for $N_{{\text{O}}_{\text{2}}}$, $1.5-0.5x$ for $N_{total}$,5 atm for P, 1 for both $v_{{\text{H}}_{\text{2}}}$, and $v_{{\text{H}}_2\text{O}}$, and 0.5 for $v_{{\text{O}}_{\text{2}}}$ in Equation (IV).

$\begin{array}{l}0.0221=\frac{\left(1-x\right){\left(0.5-0.5x\right)}^{0.5}}{x}{\left(\frac{5\text{atm}}{1.5-0.5x}\right)}^{\left(1+0.5-1\right)}\\ 0.0221x=\left(1-x\right){\left(0.5-0.5x\right)}^{0.5}{\left(\frac{5\text{atm}}{1.5-0.5x}\right)}^{\left(0.5\right)}\\ 0.000488x^2=\left(1+x^2-2x\right)\left(0.5-0.5x\right)\left(\frac{5\text{atm}}{3\left(0.5-0.5x\right)}\right)\end{array}$

$\begin{array}{l}4.998x^2-10x+5=0\\ x=0.944\end{array}$

Substitute 0.944 for x in Equation (II).

$\begin{array}{l}{\text{H}}_2+0.5{\text{O}}_2\to x{\text{H}}_{\text{2}}\text{O}+\left(1-x\right){\text{H}}_2+\left(0.5-0.5x\right){\text{O}}_{\text{2}}\\ {\text{H}}_2+0.5{\text{O}}_2\to 0.944{\text{H}}_{\text{2}}\text{O}+\left(1-0.944\right){\text{H}}_2+\left(0.5-0.50\left(.944\right)\right){\text{O}}_{\text{2}}\\ {\text{H}}_2+0.5{\text{O}}_2\to 0.944{\text{H}}_{\text{2}}\text{O}+0.056{\text{H}}_2+0.028{\text{O}}_{\text{2}}\end{array}$

Thus, the equilibrium composition of mixture of ${\text{H}}_{\text{2}}\text{O}$, ${\text{H}}_2$ and ${\text{O}}_{\text{2}}$ at 5 atm pressure and 2,800 K is $0.944{\text{H}}_{\text{2}}\text{O}+0.056{\text{H}}_2+0.028{\text{O}}_{\text{2}}$.

To determine

The amount of heat released per kg of hydrogen.

Answer to Problem 93RP

The amount of heat released per kg of hydrogen is $132,600\text{kJ}/\text{kmol}{\text{H}}_2$.

Explanation of Solution

Write the energy balance equation for the combustion process.

$\begin{array}{c}-Q_{out}={{\displaystyle \sum N_P\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }-R_{u}T\right)}}_P-{{\displaystyle \sum N_R\left({\overline{h}}_f^{\circ }-R_{u}T\right)}}_R\\ =\left\{\begin{array}{c}{N}_{{\text{H}}_{\text{2}}\text{O}}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }-R_{u}T\right)}_{{\text{H}}_{\text{2}}\text{O}}\\ +N_{{\text{H}}_{\text{2}}}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }-R_{u}T\right)}_{{\text{H}}_{\text{2}}}\\ +N_{{\text{O}}_{\text{2}}}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }-R_{u}T\right)}_{{\text{O}}_{\text{2}}}\\ -N_{{\text{H}}_{\text{2}}}{\left({\overline{h}}_f^{\circ }-R_{u}T\right)}_{{\text{H}}_{\text{2}}}\\ -N_{{\text{O}}_{\text{2}}}{\left({\overline{h}}_f^{\circ }-R_{u}T\right)}_{{\text{O}}_{\text{2}}}\end{array}\right\}\end{array}$ (V)

Here, heat released during combustion is $Q_{out}$, number of moles of product is $N_P$, number of moles of reactants is $N_R$, number of moles of ${\text{H}}_2\text{O}$ is $N_{{\text{H}}_2\text{O}}$, number of moles of ${\text{H}}_2$ is $N_{{\text{H}}_2}$, number of moles of ${\text{O}}_{\text{2}}$ is $N_{{\text{O}}_{\text{2}}}$, enthalpy at reference state is ${\overline{h}}_f^{\circ }$, sensible enthalpy at specified state is $\overline{h}$, sensible enthalpy at reference state is ${\overline{h}}^{\circ }$, universal gas constant is $R_u$, and temperature is T.

Conclusion:

Refer the table A-26, “Enthalpy of formation table”, obtain the enthalpy of ${\text{H}}_2\text{O}$ ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{H}}_2\text{O}}$, ${\text{O}}_2$ ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{O}}_2}$,and ${\text{H}}_{\text{2}}$ ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{H}}_{\text{2}}}$ as $-241,820\text{kJ}/\text{kmol}$, 0, and 0.

Refer the table A-18, “Ideal gas properties of hydrogen gas”, obtain the following properties of hydrogen gas at different temperature.

Enthalpy of hydrogen gas at 2800 K, ${\left(\overline{h}\right)}_{{\text{H}}_{\text{2}}}=89,838\text{kJ}/\text{kmol}$.

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

Refer the table A-19, “Ideal gas properties of oxygen gas”, obtain the following properties of oxygen gas at different temperature.

Enthalpy of oxygen gas at 2800 K, ${\left(\overline{h}\right)}_{{\text{O}}_{\text{2}}}=98,826\text{kJ}/\text{kmol}$.

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

Refer the table A-19, “Ideal gas properties of water vapor”, obtain the following properties of water vapor at different temperature

Enthalpy of water vapor at 2800 K, ${\left(\overline{h}\right)}_{{\text{H}}_2\text{O}}=125,198\text{kJ}/\text{kmol}$.

Enthalpy of water vapor at 298 K, ${\left({\overline{h}}^{\circ }\right)}_{{\text{H}}_2\text{O}}=9,904\text{kJ}/\text{kmol}$.

Substitute $0$ for ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{H}}_2}$, $89,838\text{kJ}/\text{kmol}$ for ${\left(\overline{h}\right)}_{{\text{H}}_2}$, $8,468\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}^{\circ }\right)}_{{\text{H}}_2}$, $0$ for ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{O}}_2}$, $98,826\text{kJ}/\text{kmol}$ for ${\left(\overline{h}\right)}_{{\text{O}}_{\text{2}}}$, $8,682\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}^{\circ }\right)}_{{\text{O}}_{\text{2}}}$, $-241,820\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{H}}_2\text{O}}$, $125,198\text{kJ}/\text{kmol}$ for ${\left(\overline{h}\right)}_{{\text{H}}_2\text{O}}$, $9,904\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}^{\circ }\right)}_{{\text{H}}_2\text{O}}$ 0.944 for $N_{{\text{H}}_{\text{2}}\text{O}}$, 0.056 for $N_{{\text{H}}_2}$, 0.028 for $N_{{\text{O}}_2}$, $8.314\text{kJ}/\text{kmol}\cdot \text{K}$ for $R_u$, and 2800 K for T in Equation (V).

$\begin{array}{c}-Q_{out}=\left\{\begin{array}{l}0.944\left(\begin{array}{l}-241,820\text{kJ}/\text{kmol}+125,198\text{kJ}/\text{kmol}\\ -9,904\text{kJ}/\text{kmol}-\left(8.314\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(2800\text{K}\right)\end{array}\right)\\ +0.056\left(\begin{array}{l}0+89,838\text{kJ}/\text{kmol}-8,468\text{kJ}/\text{kmol}\\ -\left(8.314\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(2800\text{K}\right)\end{array}\right)\\ +0.028\left(\begin{array}{l}0+98,826\text{kJ}/\text{kmol}-8,682\text{kJ}/\text{kmol}\\ -\left(8.314\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(2800\text{K}\right)\end{array}\right)\\ -1\left(0-\left(8.314\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(298\text{K}\right)\right)\\ -0.5\left(0-\left(8.314\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(298\text{K}\right)\right)\end{array}\right\}\\ =-132,600\text{kJ}/\text{kmol}{\text{H}}_2\end{array}$

$Q_{out}=132,600\text{kJ}/\text{kmol}{\text{H}}_2$

Thus, the amount of heat released per kg of hydrogen is $132,600\text{kJ}/\text{kmol}{\text{H}}_2$.