Question

The k_eq for the reaction: 2 NO(g) + 2 H_2(g) rightarrow N_2(g) + 2 H_2O(g) is 650. The concentrations at a given time were measured as NO = 0.1 M, H_2 = 0.05 M, H_2O = 0.1 M, and N_2 = 0.001 M. Is the reaction at equilibrium?

Answer 1

NO = 0.1 M

H2 = 0.05 M

H2O = 0.1 M

N2 = 0.001 M

2 NO + 2H2 ---------------------> N2 + 2H2O

Q = [N2] [H2O]^2 / [NO]^2 [H2]^2

Q = (0.001) (0.1)^2 / (0.1)^2 (0.05)^2

Q = 0.4

Keq > Q

the reaction is not at equilibrium .

if Keq = Q then the reaction is exactly at equilibrium . but here Keq value is more so the forward reaction is more favourable and almost completed the reaction