Interpret the elementary reaction

The options suggest the reaction is the elementary reversible dissociation-association: $A_2 \rightleftharpoons 2A$ with forward rate constant $K_1$ and backward rate constant $K_{-1}$.

Write rates of forward and backward steps

Since the reaction is elementary:

• Forward step: $A_2 \to 2A$ $r_f = K_1[A_2]$

• Backward step: $2A \to A_2$ $r_b = K_{-1}[A]^2$

Relate reaction rate to change in $[A]$

In the forward direction, 1 mole of $A_2$ produces 2 moles of $A$. So contribution of forward reaction to $\dfrac{d[A]}{dt}$ is: $+2K_1[A_2]$

In the backward direction, 2 moles of $A$ are consumed. So contribution of backward reaction to $\dfrac{d[A]}{dt}$ is: $-2K_{-1}[A]^2$

Net rate of change of $[A]$

Therefore, $\frac{d[A]}{dt} = 2K_1[A_2] - 2K_{-1}[A]^2$

Match with options

This matches: D: $2K_1[A_2] - 2K_{-1}[A]^2$

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

Hence, they agree.