Question

An enzyme-catalyzed reaction is found to have \(\Delta G = -100\) kJ mol\(^{-1}\). Which of the following statements about this reaction is/are true?

• 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

Hint:

\(\Delta G\) fixes only the equilibrium position (via \(K\)), not the reaction rate; a very large negative \(\Delta G\) gives a huge \(K\), making the reaction effectively one-directional.

Answer 1

The Correct Option is A, D

This question checks whether you can tell apart two different ideas in reaction engineering: how favorable a reaction is (thermodynamics, captured by $\Delta G$) and how fast a reaction goes (kinetics, captured by the rate constant and activation energy). A big negative $\Delta G$ only tells us where the reaction ends up, not how quickly it gets there.

1. The rate of the reaction cannot be predicted: True. $\Delta G$ is a state function that depends only on the initial and final states of the reaction, so it fixes the equilibrium position, not the path taken to reach it. The rate is set separately by the activation energy and the enzyme's catalytic efficiency, so knowing $\Delta G = -100$ kJ mol$^{-1}$ alone gives no information about whether the reaction is fast or slow.
2. The rate of the reaction is high: False. This assumes a large negative $\Delta G$ automatically means a fast reaction, which is not true. Many thermodynamically favorable reactions (large negative $\Delta G$) are kinetically slow because of a high activation energy, and only proceed quickly once a catalyst such as an enzyme lowers that barrier.
3. The rate of the reaction is low: False for the same reason as option B: rate and $\Delta G$ are independent, so we cannot fix the rate as "low" just from the size of $\Delta G$.
4. The reaction is irreversible: True. Using $\Delta G^{\circ} = -RT\ln K$, at $T = 298$ K,
   $$ \ln K = \frac{100000}{8.314 \times 298} \approx 40.4 $$
   so $K \approx 3.5 \times 10^{17}$. An equilibrium constant this large means almost no reactant is left at equilibrium, so for all practical purposes the reverse reaction does not occur and the reaction is irreversible.

Checking each option this way shows the correct choices are A and D: the huge negative $\Delta G$ fixes the reaction as effectively one-directional (irreversible), but it tells us nothing about the actual speed of the reaction.

Let's summarize:

• $\Delta G$ is a thermodynamic quantity that decides the equilibrium position, not the rate.
• Rate depends on activation energy and enzyme kinetics, which are independent of $\Delta G$.
• A very large negative $\Delta G$ gives a huge equilibrium constant $K$, making the reaction effectively irreversible.

So the correct statements are A (rate cannot be predicted) and D (the reaction is irreversible).