JEE Main 2026 · 8 Apr · Shift 2 · Chemistry Q57

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ChemistrySingle correct+4 / -1Question 57 of 75

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 plot of $t_{1/2}$ against $[\mathrm A]_o$ is as shown. Here $[\mathrm A]_o$ is the initial concentration of A and $t_{1/2}$ is half life of reaction.

In the light of the above statements, choose the correct answer from the options given below :

Statement II's graph: t½ (s) against [A]₀ (mol L⁻¹), a straight line rising from the origin.

Both Statement I and Statement II are true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Answer and worked solution

Official answer: C

Statement I.
$\ln\frac{k_2}{k_1}=\frac{E_a}R\left(\frac1{298}-\frac1{308}\right)=\frac{12600\times4.2}{8.314}\times1.09\times10^{-4}\approx0.69,$
so the rate constant doubles. True.

Statement II. For a first-order reaction $t_{1/2}=\dfrac{0.693}k$ , independent of $[\mathrm A]_0$ . The plot should be a horizontal line, not the rising line shown. False.

(C)

Topics:Arrhenius equation Half-life

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JEE Main 2026 · 8 Apr · Shift 2, Chemistry question 57