Question 14.5: The water-gas-shift reaction, an essential step in the produ...

(a) For the given reaction at 1100  K,  10^4/T = 9.05, and from Fig. 14.2, ln K = 0 and K = 1. For this reaction \nu=\sum_i \nu_i=1+1-1-1=0 . Because the reaction mixture can be treated as an ideal gas, Eq. (14.28) applies, and here becomes:

\prod_i\left(y_i\right)^{\nu_i}=\left(\frac{P}{P^{\circ}}\right)^{-\nu} K      (14.28)

\frac{y_{\mathrm{H}_2} y_{\mathrm{CO}_2}}{y_{\mathrm{CO}} y_{\mathrm{H}_2 \mathrm{O}}}=K=1           (A)

By Eq. (14.5),

y_i=\frac{n_i}{n}=\frac{n_{i_0}+\nu_i \varepsilon}{n_0+\nu \varepsilon}       (14.5)

y_{\mathrm{CO}}=\frac{1-\varepsilon_e}{2} \quad y_{\mathrm{H}_2 \mathrm{O}}=\frac{1-\varepsilon_e}{2} \quad y_{\mathrm{CO}_2}=\frac{\varepsilon_e}{2} \quad y_{\mathrm{H}_2}=\frac{\varepsilon_e}{2}

Substituting these values into Eq. (A) gives:

\frac{\varepsilon_e^2}{\left(1-\varepsilon_e\right)^2}=1

Therefore the fraction of the steam that reacts is 0.5.

(b) Because ν = 0, the increase in pressure has no effect on the ideal-gas reaction, and ε_e is still 0.5.

(c) The N_2 does not take part in the reaction and serves only as a diluent. It does increase the initial number of moles n_0 from 2 to 4, and the mole fractions are all reduced by a factor of 2. However, Eq. (A) is unchanged and reduces to the same expression as before. Therefore, ε_e is again 0.5.

(d) In this case the mole fractions at equilibrium are :

y_{\mathrm{CO}}=\frac{1-\varepsilon_e}{3} \quad y_{\mathrm{H}_2 \mathrm{O}}=\frac{2-\varepsilon_e}{3} \quad y_{\mathrm{CO}_2}=\frac{\varepsilon_e}{3} \quad y_{\mathrm{H}_2}=\frac{\varepsilon_e}{3}

and Eq. (A) becomes:

\frac{\varepsilon_e^2}{\left(1-\varepsilon_e\right)\left(2-\varepsilon_e\right)}=1 from which ε_e  = 0.667

The fraction of steam that reacts is then 0.667/2 = 0.333.

(e) Here the expressions for yCO and y_{\mathrm{H}_2 \mathrm{O}} are interchanged, but this leaves the equilibrium equation the same as in (d). Therefore ε_e = 0.667, and the fraction of steam that reacts is 0.667.

( f ) In this case Eq. (A) becomes:

\frac{\varepsilon_e\left(1+\varepsilon_e\right)}{\left(1-\varepsilon_e\right)^2}=1        from which        ε_e  = 0.333

The fraction of steam reacted is 0.333.

(g) At 1650 K, 10^4/T = 6.06, and from Fig. 14.2, ln K = −1.15 or K = 0.316. Therefore Eq. (A) becomes:

\frac{\varepsilon_e^2}{\left(1-\varepsilon_e\right)^2}=0.316 \quad \text { from which } \quad \varepsilon_e=0.36

The reaction is exothermic, and the extent of reaction decreases with increasing temperature.