At 721 degree C, p^Theta, after a pure H_2 gas slowly passes through excessive C

Question

At 721 degree C, p^Theta, after a pure H_2 gas slowly passes through excessive CoO(s), part of the oxide is reduced into part Co(s). The flowing out equilibrium gas contains 2.5% (volume fraction) H_2. At the same temperature, if we use CO gas to reduce CoO(s), the flowing-out equilibrium gas contains 1.92% (volume fraction) CO. Then, if the same mole number of CO and water vapor are mixed up at 721 degree C, and they undergo chemical reaction after an appropriate catalyst is added, how much is the equilibrium conversion percentage?

Explanation / Answer

The reaction is

CoO(s) + H2(g) = Co(s) + H2O(g)

Basis : 1 mole of H2.

Given at equilibrium, the Volume fraction ( or mole fraction of H2) is 2.5%. Rest is H2O whose mole fraction is 97.5%.

Equilibrium constant K1= [H2O] /[H2] = 0.975/0.025 = 38.88

For the reaction, Co(s)+ H2O---àCoO+H2(g) (1), K1’ = 1/38.88= 0.026

CoO+CO(g) ---àCO2(g)+ Co (2)

Basis : 1 mole of CO

Given the volume fraction of CO is 1.92%. Rest is CO2 whose mole fraction is =100-1.92= 98.08%

Equilibrium constant, K2= 0.9808/0.0192 = 51.08

When we combine reaction 1 and 2

H2O+CO----àCO2+H2, K = K1’*K2= 51.08/0.026 = 1.33

Let 1 mole of H2O be mixed with 1 mole of CO

Let x= percentage decomposition of H2O

Hence at equilibrium, [CO] = [H2O] =1-x and [CO2] =[H2] =x

Hence x2/(1-x)2= 1.33 or x/(1-x)= 1.33 or x= 1.33-1.33x

2.33x= 1.33 or x= 1.33/2.33 =.578

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