GATE 2026 BT Q.50: solution and answer

Official answer: A, DSource: Official answer key, GATE 2026 BT (IIT Guwahati)

1. $\Delta G$ is a thermodynamic quantity. Its sign says whether a reaction can proceed spontaneously and in which direction, and its size says how far the system is from equilibrium. It says nothing about how fast the reaction goes: speed is set by the activation energy, a kinetic property, and an enzyme only lowers that barrier.

2. (A) True. $\Delta G=-100{\text{kJ mol}}^{-1}$ gives no information about the activation barrier or the enzyme's turnover, so the rate cannot be predicted from it alone.

3. (B) False. A large negative $\Delta G$ does not imply a fast reaction. Many strongly exergonic reactions, such as the oxidation of glucose in air, are extremely slow without a catalyst.

4. (C) False. The same argument applies the other way. Nothing here shows the rate is low, because an enzyme may lower the barrier a great deal. Rate cannot be inferred either way.

5. (D) True. Write the free-energy change as $\Delta G=RT\ln \frac{Q}{K_{eq}}$, where $Q$ is the mass-action ratio under the actual conditions. At about $310\text{K}$, $RT\approx 2.58{\text{kJ mol}}^{-1}$, so $\ln \frac{K_{eq}}{Q}=\frac{100}{2.58}\approx 39$, which gives $\frac{K_{eq}}{Q}\sim 10^{17}$. The system is astronomically far from equilibrium on the product-forming side, and the reverse reaction would have $\Delta G=+100{\text{kJ mol}}^{-1}$, which is not spontaneous. The reaction therefore runs only forward in practice, i.e. it is irreversible. An enzyme does not change $\Delta G$ or the direction.

6. So the correct statements are (A) and (D).