Question

The following reaction is at equilibrium at 500 oC
CH4(g) + H2O(g)CO(g) + 3 H2(g)

Equilibrium concentrations were found to be:

[CH4(g)] = 2.46 x 10-3 M
[H2O(g)] = 4.97 x 10-3 M
[CO(g)] = 4.55 x 10-3 M
[H2(g)] = 2.20 x 10-3 M

Calculate the equilibrium constant for this reaction.

Answer 1

Find the expression for equilibrium constant for the reaction:

$K = [CO][H_2]^3/[CH_4][H_2O]$

Find the constant:

K = 4.55 * 10^-3 * (2.2 *10^-3)³ / 2.46 * 10^-3 * 4.97 * 10^-3 = 3.96 * 10^-6

Answer: 3.96 * 10^-6