QJEEM25Ch17c5: JEE Main 2025 Chemistry - Class 12 JEEM25Ch17c5 Solution: The molecule A changes into its isomeric form B by following a…

To determine the time required for 50% of molecule A to change into its isomeric form B, follow these steps:

Half-life Formula for First Order Kinetics:

The half-life ( $t_{1/2}$ ) for a first-order reaction is given by:

$t_{1/2} = \frac{0.693}{K}$

Calculate the Rate Constant (K):

The rate constant $K$ can be calculated using the Arrhenius equation:

$K = A \cdot e^{-\frac{E_a}{RT}}$

Given:

$A$ (frequency factor) = $10^{20}$

$E_a$ (activation energy) = $191.48 \, \text{kJ/mol} = 191.48 \times 10^3 \, \text{J/mol}$

$R$ (universal gas constant) = $8.314 \, \text{J/mol} \cdot \text{K}$

$T = 1000 \, \text{K}$

Substitute the values into the Arrhenius equation:

$K = 10^{20} \times e^{-\frac{191.48 \times 10^3}{8.314 \times 1000}}$

Simplify this calculation:

$K = 10^{20} \times e^{-23.031}$

Simplifying further by recognizing that $e^{-23.031}$ is a very small number, gives:

$K \approx \frac{10^{20}}{10^{10}} = 10^{10} \, \text{sec}^{-1}$

Calculate the Half-life:

Using the calculated value of $K$ :

$t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \, \text{seconds}$

Convert to Picoseconds:

Since $1 \, \text{second} = 10^{12} \, \text{picoseconds}$ :

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

Therefore, the time required for 50% of the molecules of A to become B is approximately 69 picoseconds (nearest integer).