For an elementary chemical reaction, $A_{2} \underset{\mathrm{k}_{-1}}{\stackrel{\mathrm{k}_{1}}{\rightleftharpoons}} 2 A$, the expression for $\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}$ is:
Correct Option: , 4
Given: $\mathrm{A}_{2} \underset{\mathrm{k}_{-1}}{\stackrel{\mathrm{k}_{1}}{\rightleftharpoons}} 2 \mathrm{~A}$
Now, $-\frac{1}{2} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}_{-1}[\mathrm{~A}]^{2}-\mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]$
$\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=-2 \mathrm{k}_{-1}[\mathrm{~A}]^{2}+2 \mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]$
$\Rightarrow \quad \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=2 \mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-2 \mathrm{k}_{-1}[\mathrm{~A}]^{2}$
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