The expression for the equilibrium constant for the reverse reaction is A - VCE - SSCE Chemistry - Question 1 - 2014 - Paper 1

To find the equilibrium constant expression for the reverse reaction, we first need to know the forward reaction. For a general reaction of the form:

$aA + bB \rightleftharpoons cC + dD$

the expression for the equilibrium constant (K) for the forward reaction is given by:

$K_{forward} = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

The equilibrium constant for the reverse reaction (K_reverse) is the inverse of the forward reaction:

$K_{reverse} = \frac{1}{K_{forward}} = \frac{[A]^a[B]^b}{[C]^c[D]^d}$

Assuming the forward reaction is:

$2H_2 + CO_2 \rightleftharpoons 2H_2O + CH_4$

the equilibrium constant expression can be written as:

$K_{forward} = \frac{[H_2O]^2[CH_4]}{[H_2]^2[CO_2]}$

Consequently, the expression for the reverse reaction would be:

$K_{reverse} = \frac{[H_2]^2[CO_2]}{[H_2O]^2[CH_4]}$

Hence, the correct choice is:

B. $K = \frac{[H_2]^1[CO_2]}{[H_2O]^2[CH_4]}$