Given below are two statements: … J K⁻¹…

Single correct+4 / -1Answer matches NTA’s final key

Given below are two statements:

$R = 8.314$ J K⁻¹ mol⁻¹ and 1 cal = 4.2 J

Statement I: When $E_a = 12.6$ 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,

Graph of half-life t½ (in s) against initial concentration [A]₀ (in mol L⁻¹): a straight line rising from the origin

Here $[A]_0$ is the initial concentration of A and $t_0.5$ is half life of reaction.

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

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

Solution

Statement I. By the Arrhenius equation,

$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) = \frac{12.6 \times 10^3 \times 4.2}{8.314} \times \frac{10}{298 \times 308} \approx 6365 \times 1.09 \times 10^{-4} \approx 0.693.$

$\ln\frac{k_2}{k_1} = \ln 2$ , so the rate constant doubles. True.

Statement II. For a first-order reaction $t_{1/2} = \frac{0.693}{k}$ , independent of $[A]_0$ : the graph should be a horizontal line. A straight line through the origin is the zero-order behaviour. False.

Answer: Statement I is true but Statement II is false.

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Chemical Kinetics

Solution

Keep going