Q. A → B (first reaction)

C → D (second reaction)

Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is ______ × 10⁻¹ hour⁻¹ (nearest integer).

Correct Answer: 5

Explanation

For a first-order reaction, time for completion of 50% reaction (half-life) is given by:

t_1/2 = 0.693 / k

At 500 K, half-life = 2 hour.

k_1,500 = 0.693 / 2 = 0.3465 h⁻¹

Given that for the first reaction:

k_1,500 = 2k_1,300

So,

k_1,300 = 0.173 h⁻¹

According to Arrhenius equation:

ln(k₅₀₀/k₃₀₀) = E_a/R (1/300 − 1/500)

For the second reaction, activation energy is half of the first reaction. Hence:

ln(k_2,500/k_2,300) = 1/2 ln(k_1,500/k_1,300)

Given:

k_2,500 = 2k_1,500 = 0.693 h⁻¹

Thus,

ln(0.693 / k_2,300) = 1/2 ln(2)

k_2,300 ≈ 0.50 h⁻¹

Hence,

k_2,300 = 5 × 10⁻¹ h⁻¹

Therefore, the required value is 5.

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