For an elementary chemical reaction, $$A_2 \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} 2A$$, the expression for $$\frac{d[A]}{dt}$$ is:

We have the reversible elementary reaction $$A_2 \;\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}\; 2A.$$

For any elementary step, the rate law is directly proportional to the product of the concentrations of the reactant molecules raised to the power of their stoichiometric coefficients. Therefore:

1. Forward step $$A_2 \longrightarrow 2A$$ has rate

$$r_{\text{f}} = k_1[A_2].$$

2. Backward step $$2A \longrightarrow A_2$$ has rate

$$r_{\text{b}} = k_{-1}[A]^2$$

because two molecules of $$A$$ collide simultaneously in this elementary process.

Next, we translate these molecular events into the rate of change of concentration of $$A$$. The forward step creates two moles of $$A$$ per molecular event, while the backward step consumes two moles of $$A$$ per molecular event. Hence:

$$\frac{d[A]}{dt} = (+2)\,r_{\text{f}} \;-\; (2)\,r_{\text{b}}.$$

Substituting $$r_{\text{f}} = k_1[A_2]$$ and $$r_{\text{b}} = k_{-1}[A]^2$$ we obtain

$$\frac{d[A]}{dt} = 2\,k_1[A_2] \;-\; 2\,k_{-1}[A]^2.$$

So the required expression matches Option A.

Hence, the correct answer is Option A.