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(B) For a reversible reaction $A \rightleftharpoons B$,the rate of forward reaction is $r_f = k_f[A]$ and the rate of reverse reaction is $r_b = k_b[B

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$2$

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(B) For a reversible reaction $A \rightleftharpoons B$,the rate of forward reaction is $r_f = k_f[A]$ and the rate of reverse reaction is $r_b = k_b[B]$.Given that at identical concentrations,the rate of forward reaction is two times the rate of reverse reaction,we have $r_f = 2r_b$.Substituting the rate expressions: $k_f[A] = 2k_b[B]$.Since the concentrations are identical,$[A] = [B]$,which implies $k_f = 2k_b$.The equilibrium constant $K_{eq}$ is defined as the ratio of the rate constants: $K_{eq} = \frac{k_f}{k_b} = \frac{2}{1} = 2$.

The rate of forward reaction is two times that of reverse reaction at a given temperature and identical concentration. $K_{equilibrium}$ is

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