\(\mathrm{A} \rightarrow \mathrm{B}\) 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 \(191.48 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and the frequency factor is \(10^{20}\), the time required for \(50 \%\), molecules of A to become \(B\) is \_\_\_\_ picoseconds (nearest integer). \(\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]\)

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Chemical Kinetics

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}]

Answer: 69

Solution

Using the Arrhenius equation:

k = A e^{- E_{a} / R T} = 1 0^{20} \times e^{- \frac{191480}{8.314 \times 1000}} = 1 0^{20} \times e^{- 23.03}

Since

e^{- 23.03} \approx 1 0^{- 10}

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

For a first-order reaction, the half-life is:

t_{1/2} = \frac{0.693}{k} = \frac{0.693}{1 0 ^{10}} = 6.93 \times 1 0^{- 11} s = 69.3 picoseconds

The time for 50% conversion of A to B is 69 picoseconds.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Integer answer

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JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Chemical Kinetics

Solution

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