Question

The steam reforming reaction can be described by the following two reactions:

CH₄ +H₂O ↔ CO+3H₂

CH₄ +2H₂O ↔ CO2 +4H₂

Assume that both these reactions achieve equilibrium at 600 K. The equilibrium constants at this temperature for the two reactions are 0.41 and 1.09 respectively. Calculate the equilibrium composition if the starting composition is 5 moles of steam and 1 mole of methane at a pressure of 2 atm.

Answer 1

For the first reaction,

Kc = (CO)(H2)^3/(CH4)(H2O)

let x be the chainge in concentration at equilibrium

0.41 = (x)(3x)^3/(1 - x)(5 - x)

2.05 - 2.46x + 0.41x^2 = 27x^4

27x^4 - 0.41x^2 + 2.46x - 2.05 = 0

x = 0.443 mols

Equilibrium concentration from Ist reaction of,

CH4 = 1 - 0.443 = 0.56 mols

H2O = 5 - 0.443 = 4.57 mols

CO = 0.443 mols

H2 = 3 x 0.443 = 1.33 mols

For the second reaction,

Kc = (CO2)(H2)^4/(CH4)(H2O)^2

let x be the chainge in concentration at equilibrium

0.41 = (x)(4x)^4/(1 - x)(5 - x)^2

256x^5 - 0.41x^3 + 3.69x^2 - 6.15x - 10.25 = 0

x = 0.55 mols

Equilibrium concentration from IInd reaction of,

CH4 = 1 - 0.55 = 0.45 mols

H2O = 5 - 2 x 0.55 = 3.9 mols

CO2 = 0.55 mols

H2 = 4 x 0.55 = 2.2 mols