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QJEEM25Chbf25: JEE Main 2025 Chemistry - Class 12 JEEM25Chbf25 Solution: For a given reaction is related to as given in table. Given:…

JEE Main 2025 (Online) 28th January Morning Shift Chemistry - Chemical Kinetics and Nuclear Chemistry Question 1 English

For a given reaction R→P,t1/2\mathrm{R} \rightarrow \mathrm{P}, \mathrm{t}_{1 / 2} is related to [A]0[\mathrm{A}]_0 as given in table.

Given: log⁡2=0.30\log 2=0.30

Which of the following is true?

A. The order of the reaction is 1/21 / 2 .

B. If [A]0[\mathrm{A}]_0 is 1 M , then t1/2\mathrm{t}_{1 / 2} is 20010 min200 \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. t1/2\mathrm{t}_{1 / 2} is 800 min for [A]0=1.6M[\mathrm{A}]_0=1.6 \mathrm{M}

Choose the correct answer from the options given below:

Multiple Choice Options

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

Step-by-Step Verified Solution

t1/2∝1 A0n−1(t1/2)1(t1/2)2=(A0)2n−1( A0)1n−1amp;200100=(0.0250.100)n−1amp;n−1=−122=(14)n−1amp;n=12( order )⇒t1/2∝ A0amp;200t1/2=(0.1)1/2(1)1/2amp; when A0=1Mt1/2=20010 min amp;\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 t1/2t_{1 / 2} independent of their concentration. So upon changing the concentration t1/2t_{1 / 2} should not change for first order reaction.

amp;200t1/2=(0.1)1/2(1.6)1/2 when A0=1.6Mamp;t1/2=800 min\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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