Q11 / 75 · JEE Main 2025 (28 Jan Shift 1) · Chemistry

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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: Options

  • A
    A and C Only
  • Correct answer
    B
    A, B and D Only
  • C
    C and D Only
  • D
    A and B Only

Solution

$\frac{t_1}{2} \propto\left(C_0\right)^{1-\eta}$ $\begin{aligned} & \frac{t_1}{t_2}=\left(\frac{C_1}{C_2}\right)^{1-\eta} \\ & \Rightarrow \frac{200}{100}=\left(\frac{0.100}{0.025}\right)^{1-\eta} \\ & \Rightarrow 2=(4)^{1-\eta} \\ & (1-\eta)=\frac{1}{2} \\ & \eta=\frac{1}{2} \\ & \text { For } \eta=\frac{1}{2} \\ & \frac{-d A}{d t}=k(A)^{\frac{1}{2}} \\ & \int_{C_0}^C \frac{d A}{(A)^{\frac{1}{2}}}=-\int_0^t k d t \\ & \Rightarrow 2 \mathrm{~A}^{\frac{1}{2}}=-\mathrm{kt} \\ & \Rightarrow \sqrt{\mathrm{C}}-\sqrt{\mathrm{C}_0}=\frac{-\mathrm{kt}}{2} \\ & \sqrt{\mathrm{C}}-\sqrt{\mathrm{C}_0}-\frac{\mathrm{kt}}{2} \\ & \text { For } \mathrm{C}_0=0.1 \Rightarrow \mathrm{t}_{\frac{1}{2}}=200 \mathrm{~min} \\ & \sqrt{\frac{c_0}{2}}=\sqrt{c_0}-\frac{k t}{2} \\ & \frac{\mathrm{kt}}{2}=\sqrt{\mathrm{c}_0}-\sqrt{\frac{\mathrm{c}_0}{2}} \\ & \frac{\mathrm{kt}}{2}=\sqrt{\mathrm{c}_0}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \\ & t_{\frac{1}{2}}=\frac{2 \sqrt{c_0}}{k}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \\ & 200=\frac{2 \sqrt{0.1}}{k}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \end{aligned}$ $\mathrm{k}=\frac{\sqrt{0.1}}{100}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$ For $\mathrm{C}_0=1 \mathrm{M}$ $t_{\frac{1}{2}}=\frac{2 \times 100(\sqrt{2})}{\sqrt{0.1}(\sqrt{2}-1)} \times \frac{(\sqrt{2}-1)}{\sqrt{2}}$ $\Rightarrow 200 \sqrt{10} \mathrm{~min}$. $\Rightarrow B$ is correct C is incorrect For $\mathrm{C}_0=1.6 \mathrm{M}$ $\begin{aligned} & \mathrm{t}_{\frac{1}{2}}=\frac{2 \sqrt{1.6}(\sqrt{2})(\sqrt{2}-1) \times 100}{\sqrt{0.1}(\sqrt{2}-1)(\sqrt{2})} \\ & \mathrm{t}_{\frac{1}{2}}=400 \times 2 \mathrm{~min} \\ & \mathrm{t}_{\frac{1}{2}}=800 \mathrm{~min}\end{aligned}$

Hence, the correct answer is option: 'B'.

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