Consider the following sequence of reactions : A -> 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 10 20, the time required for 50%, molecules of A to become B is ______ picoseconds (nearest integer). [R = 8.314 J K –1 mol –1 ]

Key concepts to crack the problem

1. Arrhenius Equation
   For a reaction with activation energy $E_a$ and frequency factor $A$ at temperature $T$, the rate constant $k$ is
   $k = A\,\exp\left(\frac{-E_a}{R T}\right)$ where $R$ is the universal gas constant.

2. First-order kinetics
   If the process $A \rightarrow B$ is first order, the rate only depends on the amount of $A$.
   The half-life (time for 50 % conversion) is
   $t_{1/2} = \frac{\ln 2}{k}$ This is independent of the initial concentration.

3. Unit conversions
   • Activation energy given in $\text{kJ mol}^{-1}$ must be changed to $\text{J mol}^{-1}$ before substitution.
   • The final time from seconds to picoseconds: $1\,\text{ps} = 10^{-12}\,\text{s}$.

Logical chain of thought

Step 1 – Calculate the exponential term $\exp\left(-E_a/RT\right)$ using the numerical data.

Step 2 – Multiply by the frequency factor $A$ to get the rate constant $k$.

Step 3 – Plug $k$ into $t_{1/2} = \ln 2 / k$.

Step 4 – Convert the answer into picoseconds and round to the nearest integer as required.

What’s happening?

Imagine you have a box full of toy cars (molecule A) that can magically flip themselves into another design (molecule B). How fast they flip depends on two things:

1. A tall hill to climb – the energy barrier ($E_a$).
2. How often they try – the frequency factor ($A$).

At 1000 K the cars shake a lot, so some get enough energy to climb the hill and flip. We use a special math rule (Arrhenius equation) to see how many flips per second happen. Once we know that, we can find how long it takes until half the cars have flipped (that’s called the half-life). In this problem the answer turns out to be about 71 picoseconds – super quick, because the cars are trying $10^{20}$ times every second!

Step-by-step calculation

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

2. Rate constant $k$

   $$\begin{aligned} k &= A\,\exp\left(\frac{-E_a}{R T}\right) \\ &= 10^{20}\,\exp\left(\frac{-191.48\times10^{3}}{8.314\times1000}\right) \\ &= 10^{20}\,\exp\left(-23.031\right) \\ &\approx 10^{20}\times 9.79\times10^{-11} \\ &\approx 9.79\times10^{9}\,\text{s}^{-1} \end{aligned}$$

3. Half-life $t_{1/2}$

   $$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{9.79\times10^{9}}\,\text{s} \approx 7.08\times10^{-11}\,\text{s}$$

4. Convert to picoseconds

   $$1\,\text{ps} = 10^{-12}\,\text{s} \;\Rightarrow\; t_{1/2} = \frac{7.08\times10^{-11}}{10^{-12}} \text{ ps} \approx 70.8 \text{ ps}$$

5. Nearest integer

   $\boxed{71\,\text{picoseconds}}$