Question

Determine the value of the equilibrium constant, Kgoal, for the reaction C(s)+12O2(g)+H2(g)⇌12CH3OH(g)+12CO(g), Kgoal=? by making use of the following information:
1. CO2(g)+3H2(g) ⇌ CH3OH(g)+H2O(g), K1 = 1.40×102
2. CO(g)+H2O(g) ⇌ CO2(g)+H2(g), K2 = 1.00×105
3. 2C(s)+O2(g) ⇌ 2CO(g), K3 = 2.10×1047

Answer 1

Answer:

1.71x10²⁷

Explanation:

If we sum 1/2 of (3) + 1/2 of (1):

1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷  = 4.58x10²³

1/2 (1)   1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8

C(s) + 1/2O₂(g) + 1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g) + CO(g)

K' = 4.58x10²³ * 11.8 = 5.42x10²⁴

+1/2 (2):

1/2 CO(g) + 1/2H₂O(g) ⇌ 1/2CO₂(g) + 1/2H₂ (g), K = √1.00×10⁵ = 316.2

C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)

K'' = 5.42x10²⁴* 316.2 =

1.71x10²⁷