JEE MAIN - Chemistry (2024 - 30th January Morning Shift - No. 28)
The rate of First order reaction is $$0.04 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$ at 10 minutes and $$0.03 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$ at 20 minutes after initiation. Half life of the reaction is _______ minutes.
(Given $$\log 2=0.3010, \log 3=0.4771$$)
(Given $$\log 2=0.3010, \log 3=0.4771$$)
Answer
24
Explanation
$$\begin{aligned} & 0.04=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 10 \times 60} \quad \text{..... (1)}\\ & 0.03=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 20 \times 60} \quad \text{..... (2)} \end{aligned}$$
$$\begin{aligned} \frac{4}{3} & =\mathrm{e}^{600 \mathrm{k}(2-1)} /(2) \\ \frac{4}{3} & =\mathrm{e}^{600 \mathrm{k}} \\ \ln \frac{4}{3} & =600 \mathrm{k} \\ \ln \frac{4}{3} & =600 \times \frac{\ln 2}{\mathrm{t}_{1 / 2}} \\ \mathrm{t}_{1 / 2} & =600 \frac{\ln 2}{\ln \frac{4}{3}} \sec \\ \mathrm{t}_{1 / 2} & =600 \times \frac{\log 2}{\log 4-\log 3} \mathrm{sec} .=10 \times \frac{0.3010}{0.6020-0.477} \mathrm{~min} \\ \mathrm{t}_{1 / 2} & =24.08 \mathrm{~min} \end{aligned}$$
Ans. 24
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