JEE Main 2025 Jan 22 Shift 1, Chemistry Q24: Reactions of Various Orders

JEE Main2025Jan 22, Shift 1Chemistry

A $\to$ B

The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K. If the energy barrier with respect to reactant energy for such isomeric transformation is 191.48 kJ mol $^{- 1}$ and the frequency factor is $1 0^{20}$ , the time required for 50% molecules of A to become B is ______ picoseconds (nearest integer).

[R = 8.314 J K $^{- 1}$ mol $^{- 1}$ ]

Your answer

Solution

Arrhenius equation: $k = A e^{- E_{a} / R T}$ .

Exponent: $\frac{E _{a}}{R T} = \frac{191.48 \times 1 0 ^{3}}{8.314 \times 1000} \approx 23.03$ .

Note $23.03 = ln (1 0^{10})$ , so $e^{- 23.03} = 1 0^{- 10}$ .

$∴ k = 1 0^{20} \times 1 0^{- 10} = 1 0^{10} s^{- 1}$ .

For first-order kinetics, $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{1 0 ^{10}} = 6.93 \times 1 0^{- 11} s = 69.3$ ps.

Nearest integer $= 69$ ps.

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JEE Main 2025 Jan 22 Shift 1, Chemistry Q24: Reactions of Various Orders

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