Question

The rate of forward reaction is two times that of the reverse reaction at a given temperature and consentration, K at equilibrium is

Answer 1

Kf = 2Kb

So Kf/Kb=2

Hence, Kc = 2.

Answer 2

K at equilibrium is 2.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K

$A+B\rightleftharpoons C+D$

$Rate_f=k_f[A][B]$

$Rate_b=k_b[C][D]$

At equilibrium :

$Rate_f=Rate_b$

Given : $Rate_f=2\times Rate_b$

$k_f[A][B]=2\times k_b[C][D]$

The expression for $K_{eq}$ is written as:

$K=\frac{k_f}{k_b}=\frac{[C][D]}{[A][B]}=2$

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