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

Answer 1

Lets start with the equation,

2C(s) + O2(g) + 2H2(g) ----> CH3OH + CO
K = [CH3OH][CO] / [C][H2]^2[O2]

Now write the equilibrium constant equations for 1., 2. and 3.,

K1 = [CH3OH][H2O]/ [CO2][H2]^3
K2 = [CO2][H2]/ [CO][H2O]
K3= [CO]^2 / [C]^2[O2]

Multiply: K1 x K2 x K3 = [CH3OH][H2O]/ [CO2][H2]^3 x [CO2][H2]/ [CO][H2O] x [CO]^2 / [C]^2[O2]

                                    = [CH3OH][CO] / [C]^2[H2]^2[O2]

Feeding the K values given above we get,

(1.40 x 10^2) x (1.00 x 10^5) x (2.1 x 10^47) = 2.94 x 10^54

Thus,
[CH3OH][CO] / [C)^2[H2]^2[O2] = 2.94 x 10^54

Take square root of both sides we get,

[CH3OH]^1/2[CO]^1/2 / [C)[H2][O2]1/2 = (2.94 x 10^54)^1/2

                                                              = 1.71 x 10^27
which is the equilibrium constant for:
C + H2 + 1/2O2 ===> 1/2CH3OH + 1/2CO [the final equation for the reaction]

Therefore, the equilibrium constant for the reaction is 1.71 x 10^27