Chapter 16.6, Problem 48P

a)

To determine

The amount of heat transfer for the dissociation of water.

Answer to Problem 48P

The amount of heat transfer for the dissociation of water is $2056\text{kJ}/\min$.

Explanation of Solution

Write the expression for the stoichiometric reaction for reaction 1.

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

Here, the stoichiometric coefficient for ${\text{O}}_{\text{2}}$$\left(v_{{\text{O}}_{\text{2}}}\right)$ is 0.5, for ${\text{H}}_{\text{2}}\text{O}$ is 1, and for ${\text{H}}_{\text{2}}$$\left(v_{{\text{H}}_{\text{2}}}\right)$ is 1.

Write the expression for the stoichiometric reaction for reaction 1.

${\text{H}}_{\text{2}}\text{O}\rightleftharpoons \frac{1}{2}{\text{H}}_{\text{2}}+\text{OH}$ (II)

Here, the stoichiometric coefficient for ${\text{H}}_{\text{2}}$$\left(v_{{\text{H}}_{\text{2}}}\right)$ is 0.5, for ${\text{H}}_{\text{2}}\text{O}$ is 1, and for OH $\left(v_{\text{OH}}\right)$ is 1.

Write the expression for the actual reaction for the reported process.

${\text{H}}_{\text{2}}\text{O}\to x{\text{H}}_{\text{2}}\text{O}+y{\text{H}}_{\text{2}}+z{\text{O}}_{\text{2}}+w\text{OH}$ (III)

Here, the equilibrium composition contains x kmol of ${\text{H}}_{\text{2}}\text{O}$$\left(N_{{\text{H}}_{\text{2}}\text{O}}\right)$, y kmol of ${\text{H}}_{\text{2}}$$\left(N_{{\text{H}}_{\text{2}}}\right)$, z kmol of ${\text{O}}_{\text{2}}$$\left(N_{O_2}\right)$, and w kmol of OH $\left(N_{\text{OH}}\right)$.

Write the hydrogen balance equation From Equation (III).

$2=2x+2y+w$ (IV)

Write the oxygen balance equation From Equation (III).

$1=x+2z+w$ (V)

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

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

Here, number of moles of ${\text{H}}_{\text{2}}\text{O}$ is $N_{{\text{H}}_{\text{2}}\text{O}}$, number of moles of ${\text{H}}_{\text{2}}$ is $N_{{\text{H}}_{\text{2}}}$, number of moles of ${\text{O}}_{\text{2}}$ is $N_{{\text{O}}_{\text{2}}}$, and number of moles of $\text{OH}$ is $N_{\text{OH}}$.

Write the expression for the equilibrium constant for reaction 1 $\left(K_{p1}\right)$.

$K_{p1}=\frac{N_{{\text{H}}_2}^{v_{{\text{H}}_2}}{N}_{{\text{O}}_{\text{2}}}^{v_{{\text{O}}_{\text{2}}}}}{N_{{\text{H}}_{\text{2}}\text{O}}^{v_{{\text{H}}_{\text{2}}\text{O}}}}{\left(\frac{P}{N_{total}}\right)}^{\left(v_{{\text{H}}_{\text{2}}}+v_{{\text{O}}_{\text{2}}}-v_{{\text{H}}_{\text{2}}\text{O}}\right)}$ (VII)

Here, pressure is P.

Write the expression for the equilibrium constant for reaction 2 $\left(K_{p2}\right)$.

$K_{p2}=\frac{N_{{\text{H}}_2}^{v_{{\text{H}}_2}}{N}_{\text{OH}}^{v_{\text{OH}}}}{N_{{\text{H}}_{\text{2}}\text{O}}^{v_{{\text{H}}_{\text{2}}\text{O}}}}{\left(\frac{P}{N_{total}}\right)}^{\left(v_{{\text{H}}_{\text{2}}}+v_{\text{OH}}-v_{{\text{H}}_{\text{2}}\text{O}}\right)}$ (VIII)

Write the energy balance equation to determine the heat transfer $\left(Q_{\text{out}}\right)$ for the dissociation process.

$Q_{in}={\displaystyle \sum N_P{\left({\overline{h}}_f^o+\overline{h}-{\overline{h}}^o\right)}_P-N_R{\left({\overline{h}}_f^o+\overline{h}-{\overline{h}}^o\right)}_R}$ (IX)

Here, number of moles of products is $N_P$, number of moles of reactants is $N_R$, enthalpy of vaporization is ${\overline{h}}_f^o$, and the enthalpy of formation is $\overline{h}$.

Write the expression to calculate the molar flow rate of water $\left(\dot{N}\right)$.

$\dot{N}=\frac{\dot{m}}{M}$ (X)

Here, mass flow rate of water is $\dot{m}$ and the molar mass of water is $M$.

Write the expression to calculate the rate of heat transfer $\left({\dot{Q}}_{\text{out}}\right)$.

${\dot{Q}}_{in}=\dot{N}{Q}_{in}$ (XI)

Conclusion:

Refer Table A-28, “Natural logarithm of equilibrium constants”, select the value of $\ln k_p$ as $-3.086$ for reaction 1 $\left(\ln K_{p1}\right)$ at 3000 K. Hence, the equilibrium constant for reaction 1 is 0.04568.

Refer Table A-28, “Natural logarithm of equilibrium constants”, select the value of $\ln k_p$ as $-2.937$ for reaction 2 $\left(\ln K_{p2}\right)$ at 3000 K. Hence, the equilibrium constant for reaction 2 is 0.05302.

Substitute x for $N_{{\text{H}}_{\text{2}}\text{O}}$, y for $N_{{\text{H}}_{\text{2}}}$, $z$ for $N_{{\text{O}}_{\text{2}}}$, and w for $N_{\text{OH}}$ in Equation (VI).

$N_{total}=x+y+z+w$

Substitute $0.04568$ for $k_{p1}$, x for $N_{{\text{H}}_{\text{2}}\text{O}}$, y for $N_{{\text{H}}_{\text{2}}}$, $z$ for $N_{{\text{O}}_{\text{2}}}$, $x+y+z+w$ for $N_{total}$, 1 atm for P, 1 for $v_{{\text{H}}_{\text{2}}}$, and 0.5 for $v_{{\text{O}}_{\text{2}}}$ and 1 for $v_{{\text{H}}_2\text{O}}$ in Equation (VII).

$\begin{array}{l}0.04568=\frac{y{z}^{0.5}}{x}{\left(\frac{1\text{ atm}}{x+y+z+w}\right)}^{\left(1+0.5-1\right)}\\ 0.04568=\frac{y{z}^{0.5}}{x}{\left(\frac{1}{x+y+z+w}\right)}^{\left(0.5\right)}\end{array}$ (XII)

Substitute $0.05302$ for $k_{p2}$, x for $N_{{\text{H}}_{\text{2}}\text{O}}$, y for $N_{{\text{H}}_{\text{2}}}$, w for $N_{\text{OH}}$$x+y+z+w$ for $N_{total}$, 1 atm for P, 0.5 for $v_{{\text{H}}_{\text{2}}}$, 1 for $v_{{\text{O}}_{\text{2}}}$ and 1 for $v_{{\text{H}}_2\text{O}}$ in Equation (VIII).

$\begin{array}{l}0.05302=\frac{w{y}^2}{x}{\left(\frac{1\text{atm}}{x+y+z+w}\right)}^{\left(1+0.5-1\right)}\\ 0.05302=\frac{w{y}^2}{x}{\left(\frac{1}{x+y+z+w}\right)}^{0.5}\end{array}$ (XIII)

Solve Equations (IV), (V), (XII), and (XIII) simultaneously and find the values of x, y, z, and w, as 0.7835, 0.1622, 0.05396, and 0.1086 respectively.

Substitute 0.7835 for x, 0.1622 for y, 0.05396 for z, and 0.1086 for w in Equation (III).

$\begin{array}{l}{\text{H}}_{\text{2}}\text{O}\to x{\text{H}}_{\text{2}}\text{O}+y{\text{H}}_{\text{2}}+z{\text{O}}_{\text{2}}+w\text{OH}\\ {\text{H}}_{\text{2}}\text{O}\to 0.7835{\text{H}}_{\text{2}}\text{O}+0.1622{\text{H}}_{\text{2}}+0.05396{\text{O}}_{\text{2}}+0.1086\text{OH}\end{array}$

Refer the table A-26, “Enthalpy of formation table”, obtain the enthalpy of ${\text{H}}_2\text{O}$ ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{H}}_2\text{O}}$, ${\text{O}}_2$ ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{O}}_2}$,$\text{OH}$ ${\left({\overline{h}}_f^{\circ }\right)}_{\text{OH}}$, and ${\text{H}}_{\text{2}}$ ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{H}}_{\text{2}}}$ as $-241,820\text{kJ}/\text{kmol}$, 0, 0, and $39,460\text{kJ}/\text{kmol}$.

Refer the table A-23, “Ideal gas properties of water vapor”, obtain the enthalpy of water vapor at 3000 K ${\left(\overline{h}\right)}_{{\text{H}}_2\text{O}}$ and 298 K ${\left({\overline{h}}^{\circ }\right)}_{{\text{H}}_2\text{O}}$ as $136,264\text{kJ}/\text{kmol}$ and $9,904\text{kJ}/\text{kmol}$

Refer the table A-22, “Ideal gas properties of ${\text{H}}_{\text{2}}$ gas”, obtain the enthalpy of ${\text{H}}_{\text{2}}$ gas at 3000 K ${\left(\overline{h}\right)}_{{\text{H}}_{\text{2}}}$ and 298 K as ${\left({\overline{h}}^{\circ }\right)}_{{\text{H}}_{\text{2}}}$ as $97,211\text{kJ}/\text{kmol}$ and $8,468\text{kJ}/\text{kmol}$.

Refer the table A-19, “Ideal gas properties of ${\text{O}}_2$ gas”, obtain the enthalpy of ${\text{O}}_2$ gas at 3000 K ${\left(\overline{h}\right)}_{{\text{O}}_2}$ and 298 K ${\left({\overline{h}}^{\circ }\right)}_{{\text{O}}_2}$ as $106,780\text{kJ}/\text{kmol}$ and $8,682\text{kJ}/\text{kmol}$.

Refer the table A-25, “Ideal gas properties of $\text{OH}$ gas”, obtain the enthalpy of $\text{OH}$ gas at 3000 K ${\left(\overline{h}\right)}_{\text{OH}}$ and 298 K ${\left({\overline{h}}^{\circ }\right)}_{\text{OH}}$ as $98,763\text{kJ}/\text{kmol}$ and $9,188\text{kJ}/\text{kmol}$.

Write the heat input equation $\left(Q_{in}\right)$ using Equation (X).

$Q_{in}=\left\{\begin{array}{l}{N}_{{\text{H}}_2\text{O}}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }\right)}_{{\text{H}}_2\text{O}}+N_{{\text{H}}_{\text{2}}}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }\right)}_{{\text{H}}_{\text{2}}}\\ +N_{{\text{O}}_2}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }\right)}_{{\text{O}}_2}+N_{\text{OH}}{\left({\overline{h}}_f^{\circ }+\overline{h}-{\overline{h}}^{\circ }\right)}_{\text{OH}}-{\left({\overline{h}}_f^{\circ }\right)}_{{\text{H}}_2\text{O}}\end{array}\right\}$ (XIV)

Here, 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 ${\text{H}}_{\text{2}}$ is $N_{{\text{H}}_{\text{2}}}$, number of moles of ${\text{O}}_{\text{2}}$ is $N_{{\text{O}}_{\text{2}}}$, number of moles of $\text{OH}$ is $N_{\text{OH}}$, enthalpy at reference state is ${\overline{h}}_f^{\circ }$, sensible enthalpy at specified state is $\overline{h}$, and sensible enthalpy at reference state is ${\overline{h}}^{\circ }$.

Substitute $-241,820\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{H}}_2\text{O}}$, $136,264\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{3000\text{K}}\right)}_{{\text{H}}_2\text{O}}$, $9,904\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}^{\circ }\right)}_{{\text{H}}_2\text{O}}$, $0$ for ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{H}}_{\text{2}}}$, $97,211\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{3000\text{K}}\right)}_{{\text{H}}_{\text{2}}}$, $8,468\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}^{\circ }{}_{298\text{K}}\right)}_{{\text{H}}_{\text{2}}}$, 0 for ${\left({\overline{h}}_f^{\circ }\right)}_{{\text{O}}_2}$, $106,780\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{3000\text{K}}\right)}_{{\text{O}}_2}$, $8,682\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}^{\circ }{}_{298\text{K}}\right)}_{{\text{O}}_2}$, $9,188\text{kJ}/\text{kmol}$ for , ${\left({\overline{h}}^{\circ }{}_{298\text{K}}\right)}_{\text{OH}}$, $98,763\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{3000\text{K}}\right)}_{\text{OH}}$,0.05396 kmol for $N_{{\text{O}}_{\text{2}}}$, and 0.7835 kmol for $N_{{\text{H}}_2\text{O}}$, 0.1622 for $N_{{\text{H}}_2}$ , and 0.1086 for $N_{\text{OH}}$ in Equation (XIV).

$\begin{array}{c}{\overline{h}}_R=\left\{\begin{array}{l}0.7835\text{kmol}\left(-241,820\text{kJ}/\text{kmol}+136,264\text{kJ}/\text{kmol}-9,904\text{kJ}/\text{kmol}\right)\\ +0.1622\text{kmol}\left(0+97,211\text{kJ}/\text{kmol}-8,468\text{kJ}/\text{kmol}\right)\\ +0.05396\text{kmol}\left(0+106,780\text{kJ}/\text{kmol}-8,682\text{kJ}/\text{kmol}\right)+0.1086\text{kmol}\\ \left(39,460\text{kJ}/\text{kmol}+98,763\text{kJ}/\text{kmol}-9,188\text{kJ}/\text{kmol}\right)-\left(-241,820\text{kJ}/\text{kmol}\right)\end{array}\right\}\\ =185,058\text{kJ}/\text{kmol}\end{array}$

Substitute $0.2\text{kg}/\min$ for $\dot{m}$ and $18\text{kg}/\text{kmol}$ for $M$ in Equation (X).

$\begin{array}{c}\dot{N}=\frac{0.2\text{kg}/\min }{18\text{kg}/\text{kmol}}\\ =0.0111\text{kmol}/\min \end{array}$

Substitute $0.0111\text{kmol}/\min$ for $\dot{N}$ and $185,058\text{kJ}/\text{kmol}$ for ${\dot{Q}}_{in}$ in Equation (XI)

$\begin{array}{c}{\dot{Q}}_{in}=\left(0.01111\text{kmol}/\min \right)\left(185,058\text{kJ}/\text{kmol}\right)\\ \simeq 2056\text{kJ}/\min \end{array}$

Thus, the amount of heat transfer for the dissociation of water is $2056\text{kJ}/\min$.

b)

To determine

The amount of heat transfer for the absence of dissociation of water.

Answer to Problem 48P

The amount of heat transfer for the absence of dissociation of water is $1553\text{kJ}/\min$.

Explanation of Solution

Write the expression to determine the heat transfer for the absence of dissociation of Water.

${\dot{Q}}_{in}=\dot{N}{\left({\overline{h}}_{3000\text{ K}}-{\overline{h}}_{298\text{K}}\right)}_{{\text{H}}_2\text{O}}$ (XV)

Conclusion:

Substitute $0.01111\text{kmol}/\min$ for $\dot{N}$, $136,264\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{3000\text{K}}\right)}_{{\text{H}}_2\text{O}}$, and $9904\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{298\text{K}}\right)}_{{\text{H}}_2\text{O}}$ in Equation (XV).

$\begin{array}{c}{\dot{Q}}_{in}=0.01111\text{kmol}/\min \left(136,264\text{kJ}/\text{kmol}-9904\text{kJ}/\text{kmol}\right)\\ =1404\text{kJ}/\min \end{array}$

Thus, the amount of heat transfer for the absence of dissociation of water is $1404\text{kJ}/\min$.