JEE MAIN - Chemistry (2025 - 28th January Morning Shift - No. 5)
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For a given reaction $\mathrm{R} \rightarrow \mathrm{P}, \mathrm{t}_{1 / 2}$ is related to $[\mathrm{A}]_0$ as given in table.
Given: $\log 2=0.30$
Which of the following is true?
A. The order of the reaction is $1 / 2$.
B. If $[\mathrm{A}]_0$ is 1 M , then $\mathrm{t}_{1 / 2}$ is $200 \sqrt{10} \mathrm{~min}$
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. $\mathrm{t}_{1 / 2}$ is 800 min for $[\mathrm{A}]_0=1.6 \mathrm{M}$
Choose the correct answer from the options given below:
Explanation
$$\begin{array}{ll} \mathrm{t}_{1 / 2} \propto \frac{1}{\mathrm{~A}_0^{\mathrm{n}-1}} \\ \frac{\left(\mathrm{t}_{1 / 2}\right)_1}{\left(\mathrm{t}_{1 / 2}\right)_2}=\frac{\left(\mathrm{A}_0\right)_2^{\mathrm{n}-1}}{\left(\mathrm{~A}_0\right)_1^{\mathrm{n}-1}} & \\ \frac{200}{100}=\left(\frac{0.025}{0.100}\right)^{\mathrm{n}-1} & \mathrm{n}-1=-\frac{1}{2} \\ 2=\left(\frac{1}{4}\right)^{\mathrm{n}-1} & \mathrm{n}=\frac{1}{2}(\text { order }) \\ \Rightarrow \mathrm{t}_{1 / 2} \propto \sqrt{\mathrm{~A}_0} & \\ \frac{200}{\mathrm{t}_{1 / 2}}=\frac{(0.1)^{1 / 2}}{(1)^{1 / 2}} & \text { when } \mathrm{A}_0=1 \mathrm{M} \\ \mathrm{t}_{1 / 2}=200 \sqrt{10} \text { min } & \end{array}$$
*Ist order kinetics have $t_{1 / 2}$ independent of their concentration. So upon changing the concentration $t_{1 / 2}$ should not change for first order reaction.
$$\begin{aligned} & \frac{200}{t_{1 / 2}}=\frac{(0.1)^{1 / 2}}{(1.6)^{1 / 2}} \quad \text { when } \mathrm{A}_0=1.6 \mathrm{M} \\ & \mathrm{t}_{1 / 2}=800 \mathrm{~min} \end{aligned}$$
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