$$ A \rightarrow B $$ 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 $$ 191.48kJ \text{ }Mol^{-1}\text{ and the frequency factor is }10^{20},$$ the time required for 50% molecules of A to become B is
_________ picoseconds (nearest integer). $$[R=8.314 JK^{-1} mol^{-1}]$$

We need to find the time for 50% conversion using first order kinetics.

At T = 1000 K, the activation energy is 191.48 kJ/mol = 191480 J/mol, the frequency factor is A = 10^{20}, and R = 8.314 J K⁻¹ mol⁻¹.

We start by calculating the rate constant using the Arrhenius equation:

$$k = A \cdot e^{-E_a/RT}$$

Next, we compute the exponent:

$$\frac{E_a}{RT} = \frac{191480}{8.314 \times 1000} = \frac{191480}{8314} = 23.03$$

This gives:

$$k = 10^{20} \times e^{-23.03}$$

Since ln(10) = 2.303, we have:

$$e^{-23.03} = e^{-10 \times 2.303} = 10^{-10}$$

Substituting back yields:

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

For 50% conversion, the half-life in first order kinetics is given by:

$$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \text{ s}$$

Converting this to picoseconds using 1 ps = 10^{-12} s:

$$t_{1/2} = \frac{6.93 \times 10^{-11}}{10^{-12}} = 69.3 \text{ ps}$$

Rounding to the nearest integer gives a half-life of approximately 69 picoseconds.