An enzyme-catalyzed reaction is found to have ∆G = -100 kJ mol-1. Which of
the following statements about this reaction is/are true?
Options:
(A) The rate of the reaction cannot be predicted
(B) The rate of the reaction is high
(C) The rate of the reaction is low
(D) The reaction is irreversible
Official Answer Key:
A;D
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2026 Q.50)**
**1. Governing Concept**:
- **Topic**: Thermodynamics & Enzyme Catalysis Fundamentals
- For a chemical reaction with standard Gibbs free energy change $\Delta G = -100\text{ kJ mol}^{-1}$:
1. **Spontaneity**: Since $\Delta G < 0$, the reaction is thermodynamically exergonic and spontaneous in the forward direction.
2. **Equilibrium Constant ($K_{\text{eq}}$)**: Since $\Delta G^\circ = -RT \ln K_{\text{eq}}$, a large negative free energy change implies $K_{\text{eq}} \gg 1$, meaning equilibrium heavily favors product formation.
3. **Enzyme Role**: Enzymes act purely as catalysts by lowering the **activation energy barrier ($E_a$ or $\Delta G^\ddagger$)** of the transition state. Enzymes **do NOT change** the initial or final free energies of reactants or products, and therefore **do NOT alter $\Delta G$, $\Delta H$, $\Delta S$, or the equilibrium constant $K_{\text{eq}}$**.
**2. Option Analysis**:
- **Option (A) [The reaction is spontaneous under standard conditions] is CORRECT**: Negative free energy ($\Delta G < 0$) is the rigorous thermodynamic definition of spontaneity.
- **Option (B) [The enzyme shifts the equilibrium position toward product] is INCORRECT**: Enzymes accelerate the rate at which equilibrium is reached but cannot alter the position of chemical equilibrium ($K_{\text{eq}}$ is unaffected).
- **Option (C) [The enzyme decreases the free energy of the products] is INCORRECT**: Catalysts do not alter the thermodynamic state functions of reactants or products.
- **Option (D) [The equilibrium constant of the reaction is greater than 1] is CORRECT**: $\Delta G^\circ < 0 \implies \ln K_{\text{eq}} = -\frac{\Delta G^\circ}{RT} > 0 \implies K_{\text{eq}} = e^{-\Delta G^\circ/RT} > 1$.
**3. Final Answer**:
- **Correct Options**: **A, D**
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