JEE MAIN - Chemistry (2024 - 4th April Evening Shift - No. 26)
Consider the following reaction, the rate expression of which is given below
$$\begin{aligned} & \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C} \\ & \text { rate }=\mathrm{k}[\mathrm{A}]^{1 / 2}[\mathrm{~B}]^{1 / 2} \end{aligned}$$
The reaction is initiated by taking $$1 \mathrm{~M}$$ concentration of $$\mathrm{A}$$ and $$\mathrm{B}$$ each. If the rate constant $$(\mathrm{k})$$ is $$4.6 \times 10^{-2} \mathrm{~s}^{-1}$$, then the time taken for $$\mathrm{A}$$ to become $$0.1 \mathrm{~M}$$ is _________ sec. (nearest integer)
Explanation
$$\begin{aligned} & A+B \rightarrow C \\ & \frac{-d[A]}{d t}=k[A]^{1 / 2}[B]^{1 / 2} \end{aligned}$$
Since, $$[A]=[B]$$
$$\begin{aligned} & \Rightarrow \quad \frac{-d[A]}{d t}=k[A] \\ & \Rightarrow \quad k t=\ln \frac{[A]_0}{[A]} \\ & \Rightarrow \quad t=\frac{1}{4.6 \times 10^{-2}} \times \ln \left(\frac{1}{0.1}\right) \\ & =\frac{2.303}{4.6} \times 100 \approx 50 \end{aligned}$$
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