Question 14.10: Equilibrium Composition One kmol of hydrogen and one kmol of...

The elementary step H _{2}+0.5 O _{2} \leftrightarrows H _{2} O is not available in Table G-3, but the inverse reaction is listed with ln K = -5.13 (interpolated) at 2500 K. For the chosen elementary step, therefore, ln K = 5.13. This has also been verified in the previous example. Using this equilibrium constant, the unknown stoichiometric coefficients of an overall reaction can be evaluated.

Assumptions
The IG mixture model is applicable for the equilibrium mixture.

Analysis
The overall reaction, with the initial mixture on the LHS and the equilibrium mixture on the RHS, can be written in terms of three unknown coefficients:

H _{2}+ O _{2} \rightarrow a H _{2}+b O _{2}+c H _{2} O

Using the atom balance equations, b and c can be expressed in terms of a:

H _{2}+ O _{2} \rightarrow a H _{2}+\frac{1+a}{2} O _{2}+(1-a) H _{2} O

The total mole in the equilibrium mixture is \sum\limits_{p} \nu_{p}=\frac{(3+a)}{2} ; therefore, the mole fractions of the components in the equilibrium mixture are:

y_{ H _{2}}=\frac{2 a}{3+a} ; \quad y_{ O _{2}}=\frac{1+a}{3+a} ; \quad \text { and } \quad y_{ H _{2} O }=\frac{2(1-a)}{3+a}

Substituting these expressions in the equilibrium reaction equation, Eq. (14.50), we obtain:

K \equiv \frac{\prod\limits_{p} y_{p}^{\nu_{p}}}{\prod\limits_{r} y_{r}^{\nu_{r}}}\left(\frac{p}{p_{0}}\right)^{\sum\limits_{p} v_{p}-\sum\limits_{r} v_{r}}                   (14.50)

\begin{aligned}K &=\frac{y_{ H _{2} O }}{y_{ H _{2}} y_{ O _{2}}^{0.5}}\left(\frac{p}{p_{0}}\right)^{1-1-0.5} \\&=\frac{2(1-a)}{3+a}\left(\frac{2 a}{3+a}\right)^{-1}\left[\frac{1+a}{3+a}\right]^{-0.5}\left(\frac{p}{p_{0}}\right)^{-0.5}\end{aligned}                        (14.54)

In the previous example, ln K at 2500 K was calculated as 5.13. At p = 1 atm \left(p=p_{0}\right) , the equilibrium relation reduces to:

\begin{aligned}&e^{5.13}=\frac{2(1-a)}{3+a}\left(\frac{2 a}{3+a}\right)^{-1}\left(\frac{1+a}{3+a}\right)^{-0.5} \\&\Rightarrow \quad 170.91=\frac{1-a}{a} \sqrt{\frac{3+a}{1+a}}\end{aligned}

Iteration with a calculator produces a = 0.01.
The molar composition can now be evaluated as:

y_{ H _{2}}=0.0067 ; \quad y_{ H _{2} O }=0.6578 ; \quad y_{ O _{2}}=0.3356

TEST Analysis
Launch the IGE system-state TESTcalc. In the composition panel, select H2 and O2 as the reactants and enter their moles. Also select H2, H2O, and O2 as the possible products components. Switch to the state panel and evaluate State-1 for the products (make sure that the Products radio button is checked) at p1=100 kPa and T1=2500 K. In the I/O panel, the mole fractions can be found as 0.0067, 0.6578, and 0.3356, respectively, verifying the manual Solution.

What-if scenario
(b) Change p1 to 10 atm and click Calculate to obtain y_{ H _{2} O }=0.664 . (c) Change p1 to 1 atm and T1 to 1000 K. Click Calculate to obtain y_{ H _{2} O }=0.667 .

Discussion
At very high temperatures, water starts dissociating into hydrogen and oxygen and the elementary step shifts to the left. On the other hand, as pressure is increased, the reaction shifts to the right, producing more water. A shift to the right accompanies a decrease in mole (from 1.5 to 1), mitigating the rise in pressure. A shift in equilibrium due to an external factor, generally, tends to offset the effect of the external factor. This is a well-known principle called the Le Chatelier principle.