For an elementary chemical reaction,

Question:

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:

1. $\mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-\mathrm{k}_{-1}[\mathrm{~A}]^{2}$

2. $2 \mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-\mathrm{k}_{-1}[\mathrm{~A}]^{2}$

3. $\mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]+\mathrm{k}_{-1}[\mathrm{~A}]^{2}$

4. $2 \mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-2 \mathrm{k}_{-1}[\mathrm{~A}]^{2}$

Correct Option: , 4

Solution:

Given: $\mathrm{A}_{2} \underset{\mathrm{k}_{-1}}{\stackrel{\mathrm{k}_{1}}{\rightleftharpoons}} 2 \mathrm{~A}$

Now, $-\frac{1}{2} \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}_{-1}[\mathrm{~A}]^{2}-\mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]$

$\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=-2 \mathrm{k}_{-1}[\mathrm{~A}]^{2}+2 \mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]$

$\Rightarrow \quad \frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=2 \mathrm{k}_{1}\left[\mathrm{~A}_{2}\right]-2 \mathrm{k}_{-1}[\mathrm{~A}]^{2}$