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…
A following reaction is occurring in a test tube:A + B ⇌ C100 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 :
Select an option to attempt
Correct Answer: Option A
Explanation
Chemical equilibrium:Step 1 — Calculatefrom 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), sonM consumed.,,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:Answer: ~0.95 nM (Option 1).