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Chemistry Question 57 – JEE-MAIN 2026

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Given below are two statements : R=8.314 J K−1 mol−1 and 1 cal=4.2 J Statement I: When Ea=12.6 kcal/mol, the room temperature rate constant is doubled by a 10∘C increase in temperature (298 K to 308 K) Statement II: For a first order reactions A→B, the half-life t1/2 is plotted against initial concentration [A]0. The graph shows t1/2 increasing linearly with [A]0. Here [A]0 is the initial concentration of A and t1/2 is half life of reaction. In the light of the above statements, choose the correct answer from the options given below :

Recall the Arrhenius equation for temperature dependence of rate constants and the half-life expressions for different reaction orders.

Step 1: Evaluate Statement I using the Arrhenius Equation✦ Active

Convert activation energy Ea from kcal/mol to J/mol: Ea=12.6 kcal/mol=12.6×1000×4.2 J/mol=52920 J/mol. Apply the Arrhenius equation to find the ratio of rate constants:

ln⁡(k2k1)=EaR(1T1−1T2) ln⁡(k2k1)=52920 J/mol8.314 J K−1 mol−1(1298 K−1308 K) ln⁡(k2k1)=529208.314(308−298298×308)=529208.314(1091784)≈0.6933

Since ln⁡(2)≈0.693, the calculated value 0.6933 is approximately ln⁡(2). This implies k2k1≈2. Thus, Statement I is true.

Step 2: Evaluate Statement II based on reaction kinetics○ Expand

For a first-order reaction, the half-life t1/2 is given by the formula t1/2=ln⁡2k, where k is the rate constant. This equation shows that for a first-order reaction, t1/2 is independent of the initial concentration [A]0. The graph provided in the statement, however, shows t1/2 increasing linearly with [A]0. This behavior is characteristic of a zero-order reaction (t1/2=[A]02k), not a first-order reaction. Therefore, the graph contradicts the claim of a first-order reaction. Thus, Statement II is false.

Step 3: Determine the correct option○ Expand

Based on the analysis, Statement I is true and Statement II is false. This corresponds to option 3.

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