Medium INTEGER +4 / -1 PYQ · JEE Mains 2025

$\mathrm{A \rightarrow B}$

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

Answer (integer) 69

Solution

<p>To determine the time required for 50% of molecule A to change into its isomeric form B, follow these steps:</p> <p><p><strong>Half-life Formula for First Order Kinetics:</strong></p> <p>The half-life ($ t_{1/2} $) for a first-order reaction is given by:</p> <p>$ t_{1/2} = \frac{0.693}{K} $</p></p> <p><p><strong>Calculate the Rate Constant (K):</strong></p> <p>The rate constant $ K $ can be calculated using the Arrhenius equation:</p> <p>$ K = A \cdot e^{-\frac{E_a}{RT}} $</p> <p>Given:</p></p> <p><p>$ A $ (frequency factor) = $ 10^{20} $</p></p> <p><p>$ E_a $ (activation energy) = $ 191.48 \, \text{kJ/mol} = 191.48 \times 10^3 \, \text{J/mol} $</p></p> <p><p>$ R $ (universal gas constant) = $ 8.314 \, \text{J/mol} \cdot \text{K} $</p></p> <p><p>$ T = 1000 \, \text{K} $</p> <p>Substitute the values into the Arrhenius equation:</p> <p>$ K = 10^{20} \times e^{-\frac{191.48 \times 10^3}{8.314 \times 1000}} $</p> <p>Simplify this calculation:</p> <p>$ K = 10^{20} \times e^{-23.031} $</p> <p>Simplifying further by recognizing that $ e^{-23.031} $ is a very small number, gives:</p> <p>$ K \approx \frac{10^{20}}{10^{10}} = 10^{10} \, \text{sec}^{-1} $</p></p> <p><p><strong>Calculate the Half-life:</strong></p> <p>Using the calculated value of $ K $:</p> <p>$ t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \, \text{seconds} $</p></p> <p><p><strong>Convert to Picoseconds:</strong></p> <p>Since $ 1 \, \text{second} = 10^{12} \, \text{picoseconds} $:</p> <p>$ t_{1/2} = 6.93 \times 10^{-11} \times 10^{12} \, \text{picoseconds} = 69.3 \, \text{picoseconds} $</p></p> <p>Therefore, the time required for 50% of the molecules of A to become B is approximately 69 picoseconds (nearest integer).</p>

About this question

Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction

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