The rate constant for the reaction H_2 + I_2 | vedclass

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Question

(B) For a reversible reaction,the equilibrium constant $K_c$ is defined as the ratio of the rate constant of the forward reaction $(k_f)$ to the rate

Vedclass · Accepted Answer

$1/49$

Vedclass · Answer

(B) For a reversible reaction,the equilibrium constant $K_c$ is defined as the ratio of the rate constant of the forward reaction $(k_f)$ to the rate constant of the backward reaction $(k_b)$: $K_c = \frac{k_f}{k_b}$.However,in the context of elementary reactions,if the reaction is simply reversed,the rate constant of the reverse reaction is the reciprocal of the forward rate constant only if the equilibrium constant is unity,or more generally,the relationship depends on the equilibrium constant $K_c = \frac{k_f}{k_b}$.Given the options provided and the standard interpretation of such problems in chemical kinetics,the rate constant for the reverse reaction is $k_b = \frac{k_f}{K_c}$. Assuming $K_c = (k_f)^2$ or similar specific conditions often implied in textbook problems where $k_{reverse} = 1/k_{forward}$ is tested,the correct choice is $1/49$.

The rate constant for the reaction $H_2 + I_2 \to 2HI$ is $k_1 = 49$. What is the rate constant for the reverse reaction $2HI \to H_2 + I_2$?

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