The water-gas-shift reaction, an essential step in the production of hydrogen from natural gas, CO(g) + H2 O(g) → CO2 (g) + H2 (g) comes to equilibrium under several sets of conditions enumerated below. Calculate the fraction of steam reacted in each case. Assume the mixture behaves as an

Question

The water-gas-shift reaction, an essential step in the production of hydrogen from natural gas,

[latex]CO(g) + H_2 O(g) → CO_2 (g) + H_2 (g)[/latex]

comes to equilibrium under several sets of conditions enumerated below. Calculate the fraction of steam reacted in each case. Assume the mixture behaves as an ideal gas.

(a) The reactants consist of 1 mol of [latex]H_2O[/latex] vapor and 1 mol of CO. The temperature is 1100 K and the pressure is 1 bar.

(b) Same as part (a) except that the pressure is 10 bar.

(c) Same as part (a) except that 2 mol of [latex]N_2[/latex] is included in the reactants.

(d) The reactants are 2 mol of [latex]H_2O[/latex] and 1 mol of CO. Other conditions are the same as in part (a).

(e) The reactants are 1 mol of [latex]H_2O[/latex] and 2 mol of CO. Other conditions are the same as in part (a).

(f ) The initial mixture consists of 1 mol of [latex] H_2O, 1 mol of CO, and 1 mol of CO_2[/latex] . Other conditions are the same as in part (a).

(g) Same as part (a) except that the temperature is 1650 K.

Accepted Answer

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

[latex] \prod_i\left(y_i ight)^{ u_i}=\left(\frac{P}{P^{\circ}} ight)^{- u} K [/latex] (14.28)

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

By Eq. (14.5),

[latex] y_i=\frac{n_i}{n}=\frac{n_{i_0}+ u_i \varepsilon}{n_0+ u \varepsilon} [/latex] (14.5)

[latex] 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} [/latex]

Substituting these values into Eq. (A) gives:

[latex] \frac{\varepsilon_e^2}{\left(1-\varepsilon_e ight)^2}=1 [/latex]

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 [latex]ε_e[/latex] is still 0.5.

(c) The [latex]N_2[/latex] does not take part in the reaction and serves only as a diluent. It does increase the initial number of moles [latex]n_0[/latex] 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, [latex]ε_e[/latex] is again 0.5.

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

[latex] 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} [/latex]

and Eq. (A) becomes:

[latex] \frac{\varepsilon_e^2}{\left(1-\varepsilon_e ight)\left(2-\varepsilon_e ight)}=1 [/latex] from which [latex]ε_e  = 0.667 [/latex]

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

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

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

[latex] \frac{\varepsilon_e\left(1+\varepsilon_e ight)}{\left(1-\varepsilon_e ight)^2}=1 [/latex] from which [latex] ε_e  = 0.333[/latex]

The fraction of steam reacted is 0.333.

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

[latex] \frac{\varepsilon_e^2}{\left(1-\varepsilon_e ight)^2}=0.316 \quad ext { from which } \quad \varepsilon_e=0.36 [/latex]

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

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

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