For a reaction A xrightarrow K 1 B xrightarrow K 2 C If the rate of formation of B is set to be zero then the concentration of B is given by :

Chemistry · Chemical Kinetics

JEE Main 2024 — 8 April, Shift 2 — Question 59

For a reaction $A \xrightarrow{\mathrm{~K}_{1}} \mathrm{~B} \xrightarrow{\mathrm{~K}_{2}} \mathrm{C}$ If the rate of formation of $B$ is set to be zero then the concentration of B is given by :

Option A:
$\mathrm{K}_{1} \mathrm{~K}_{2}[A]$
Option B:
$\left(\mathrm{K}_{1}-\mathrm{K}_{2}\right)[\mathrm{A}]$
Option C:
$\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right)[\mathrm{A}]$
Option D:
$\left(\mathrm{K}_{1} / \mathrm{K}_{2}\right)[\mathrm{A}]$
Correct

Answer: D

Step-by-step solution

Rate of formation of B is \begin{aligned} & \frac{\mathrm{d}[\mathrm{~B}]}{\mathrm{dt}}=\mathrm{k}{1}[\mathrm{~A}]-\mathrm{k}{2}[\mathrm{~B}] & 0=\mathrm{k}{1}[\mathrm{~A}]-\mathrm{k}{2}[\mathrm{~B}] & \left(\frac{\mathrm{k}{1}}{\mathrm{k}{2}}\right)[\mathrm{A}]=[\mathrm{B}] \end{aligned}$$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws