1. Use Arrhenius equation to find the rate constant

For a first-order isomerisation,

$k = A e^{-E_a/(RT)}$

Given:

• $A = 10^{20}\ \text{s}^{-1}$
• $E_a = 191.48\ \text{kJ mol}^{-1} = 191.48 \times 10^3\ \text{J mol}^{-1}$
• $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$
• $T = 1000\ \text{K}$

So,

$\frac{E_a}{RT} = \frac{191.48\times 10^3}{8.314 \times 1000}$

$\frac{E_a}{RT} \approx \frac{191480}{8314} \approx 23.03$

Thus,

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

Now,

$e^{-23.03} \approx 9.93 \times 10^{-11}$

Hence,

$k \approx 10^{20} \times 9.93 \times 10^{-11} = 9.93 \times 10^9\ \text{s}^{-1}$

2. For first-order reaction, time for 50% conversion

For a first-order process, the half-life is

$t_{1/2} = \frac{\ln 2}{k}$

So,

$t_{1/2} = \frac{0.693}{9.93 \times 10^9} \ \text{s}$

$t_{1/2} \approx 6.98 \times 10^{-11}\ \text{s}$

3. Convert seconds to picoseconds

Since

$1\ \text{ps} = 10^{-12}\ \text{s}$

Therefore,

$t_{1/2} = 6.98 \times 10^{-11}\ \text{s} = 69.8\ \text{ps}$

Nearest integer:

$\boxed{70}$

4. Comparison with stored answer

My derived answer is $70$ ps, while the stored correct answer is $69$.

This small difference arises from rounding in evaluating $e^{-23.03}$. Using slightly different intermediate approximations can give $69$ ps. The physically correct value is about $69.7$-$69.8$ ps, whose nearest integer is $70$.

So I disagree with the stored answer if strict nearest-integer rounding is required.