# DBT JRF 2022 — Question 21

Exam: DBT JRF 2022 · Section: B · Topic: Biochemistry › Chemical Equilibrium & Reaction Kinetics · Difficulty: Medium

## Question

A following reaction is occurring in a test tube: A + B ⇌ C 100 nM solution of A is incubated with an equal volume of 100 nM solution of B and left at 25°C till the equilibrium is reached. Finally, when the concentration of A is measured, it is found to be 10 nM. What is the approximate concentration of C when 2 nM solution of A is incubated with an equal volume of 100 nM solution of B under identical conditions?                                                                                                                                                                                                                       Choose the correct answer :

## Options

- **A.** ~0.95 nM
- **B.** ~0.8 nM
- **C.** ~0.1 nM
- **D.** 10 nM

## Answer

Option A

## Explanation

Chemical equilibrium: $A + B \rightleftharpoons C$ Step 1 — Calculate $K_{eq}$ from experiment 1: Initial: [A]₀ = 100 nM, [B]₀ = 100 nM, but mixed in equal volumes → each diluted to 50 nM. At equilibrium: [A] = 10 nM (given), so $\Delta = 50 - 10 = 40$ nM consumed. $[A]_{eq} = 10\,\text{nM}$ , $[B]_{eq} = 50 - 40 = 10\,\text{nM}$ , $[C]_{eq} = 40\,\text{nM}$ $K_{eq} = \frac{[C]}{[A][B]} = \frac{40}{10 \times 10} = \frac{40}{100} = 0.4\,\text{nM}^{-1}$ Step 2 — Experiment 2: [A]₀ = 2 nM, [B]₀ = 100 nM → after mixing: [A] = 1 nM, [B] = 50 nM. With large excess B and Keq = 0.4 nM⁻¹, nearly all A is converted: $x \approx 1 \times \frac{K_{eq}[B]}{1 + K_{eq}[B]} = \frac{0.4 \times 50}{1 + 20} = \frac{20}{21} \approx 0.952\,\text{nM}$ $[C] \approx 0.95\,\text{nM}$ Answer: ~0.95 nM (Option 1).

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