Question

For the elementary reaction $2 \mathrm{~A} \rightleftharpoons \mathrm{B}$, the forward and backward rate constants are $k_1$ and $k_2$ respectively. Then the rate of disappearance of A is equal to:

(A) $k_1[\mathrm{~A}]^{2}-k_2[B]$

(B) $2k_1[A]^{2}+2k_2[B]$

(C) $2k_1[A]^{2}-2k_2[B]$

(D) $\left[2k_1-k_2\right][A]$

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Answer

The correct answer is Option C: $$ 2k_1[A]^2 - 2k_2[B] $$

Explanation:

The rate of an elementary reaction $$2A \rightleftharpoons B$$ with forward and reverse rate constants $ k_1 $ and $ k_2 $ respectively, follows the formula for the rate of change of concentration of reactants and products based on their stoichiometric coefficients.

For reactant $ A $, which has a stoichiometric coefficient of 2 in the balanced chemical equation, the rate of disappearance can be given by:

$$ -\frac{1}{2} \frac{d[A]}{dt} = k_1[A]^2 - k_2[B] $$

Multiplying both sides of the equation by 2 to get the actual rate of disappearance of $ A $ gives:

$$ -\frac{d[A]}{dt} = 2k_1[A]^2 - 2k_2[B] $$

This is expressed as the reaction proceeds, thus reflecting the concentration changes in the reagents.

Therefore, the rate of disappearance of $ A $ is:

$$ 2k_1[A]^2 - 2k_2[B] $$

Thus, the correct choice is (C).


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