11. You are observing a chemical reaction (𝑅 → 𝑃) either in the absence or presence
of an enzyme catalyst, which has 𝐾M = 106 𝜇𝑀. At equilibrium 20 moles of reactant
and 2 moles of product are present in the un-catalysed reaction. What would the
reactant and product concentrations be at equilibrium in the presence of the enzyme?
a. 𝑅 = 0, 𝑃 = 22
b. 𝑅 = 2, 𝑃 = 20
c. 𝑅 = 20, 𝑃 = 2
d. This cannot be answered without knowing the reaction volume
Enzyme Equilibrium in Catalyzed Reactions
Enzymes accelerate reactions but do not alter equilibrium positions determined by thermodynamics. In this CSIR NET Life Sciences question, the un-catalysed equilibrium (20 moles R, 2 moles P) remains identical with the enzyme. The Km value (106 μM) indicates substrate affinity but holds no relevance to equilibrium ratios.
Core Principle
Equilibrium concentrations depend solely on the equilibrium constant \( K_{eq} = \frac{[P]}{[R]} \), unchanged by catalysts like enzymes. Here, \( K_{eq} = \frac{2}{20} = 0.1 \), so adding enzyme speeds attainment of this ratio without shifting it. Total moles (22) are conserved in the closed system.
Option Analysis
| Option | Concentrations | Explanation | Status |
|---|---|---|---|
| a. R = 0, P = 22 | R = 0, P = 22 | Wrong; implies complete conversion (\( K_{eq} = \infty \)), ignoring thermodynamics. | Incorrect |
| b. R = 2, P = 20 | R = 2, P = 20 | Wrong; reverses ratio (\( K_{eq} = 10 \)), contradicting un-catalysed data. | Incorrect |
| c. R = 20, P = 2 | R = 20, P = 2 | Correct; matches un-catalysed equilibrium as enzyme preserves \( K_{eq} \). | ✅ Correct Answer |
| d. Needs reaction volume | Requires volume info | Wrong; moles suffice for ratio-based equilibrium in fixed-volume systems. | Incorrect |
Km Irrelevance
Km governs kinetics (half Vmax at [S] = Km), not thermodynamics. High Km (106 μM) shows low substrate affinity, but equilibrium ignores kinetics. Enzyme boosts forward/reverse rates equally.
