The correct option is (A) The equilibrium of the reaction will not change.​

Introduction

In biochemistry and enzymology, a frequently tested concept is the effect of enzyme on chemical reaction equilibrium. Understanding this principle is essential for solving MCQs on enzyme catalysis, Gibbs free energy, and reversible reactions in solution.​

The given question:

“In a chemical reaction where the substrate and product are in equilibrium in solution, what will occur if an enzyme is added?”

Options:
(A) The equilibrium of the reaction will not change.
(B) There will be a decrease in product formed.
(C) Additional substrate will be formed.
(D) The free energy of the system will change.

Core concept: enzymes and equilibrium

• Enzymes are biological catalysts that increase the rate of both forward and reverse reactions equally by lowering activation energy, not by altering equilibrium.​

• At equilibrium, the forward and reverse reaction rates are already equal, so adding an enzyme cannot shift the equilibrium position or change the equilibrium constant KeqKeq.​

Therefore, when substrate and product are already at equilibrium, adding the enzyme will only allow rapid interconversion between substrate and product, but their equilibrium concentrations remain the same.​

Detailed analysis of each option

Option (A): The equilibrium of the reaction will not change

• The equilibrium constant depends on the difference in standard free energy ΔG∘ΔG∘ between substrate and product, which is a thermodynamic property independent of the presence of a catalyst.​

• Enzymes lower activation energy for both directions but do not change ΔG∘ΔG∘, so the equilibrium ratio [P]eq/[S]eq[P]eq/[S]eq remains unchanged.​

Hence, option (A) is correct.

Option (B): There will be a decrease in product formed

• A decrease in product at equilibrium would imply a shift of equilibrium toward substrate, meaning the equilibrium constant changed, which enzymes cannot do.​

• Enzymes accelerate attainment of equilibrium but do not favor product or substrate intrinsically at equilibrium; they catalyze both directions.​

Therefore, option (B) is incorrect.

Option (C): Additional substrate will be formed

• Similar to option (B), this suggests that equilibrium shifts toward substrate, which would require a change in ΔG∘ΔG∘ or external perturbation such as changing concentrations, pressure, or temperature, not just adding a catalyst.​

• Enzymes form transient enzyme–substrate complexes, but net equilibrium concentrations of free substrate and product remain defined by thermodynamics, not by catalysis.​

Thus, option (C) is incorrect.

Option (D): The free energy of the system will change

• Enzymes lower the activation free energy ΔG‡ΔG‡ (kinetic barrier), not the overall free energy change $\Delta G$ between reactants and products.​

• Since $\Delta G$ and ΔG∘ΔG∘ determine equilibrium, and these are unchanged, the thermodynamic free energy of the system at equilibrium remains the same.​

So, option (D) is incorrect.

Key exam points about enzyme effect on equilibrium

• Enzymes:

  • Lower activation energy for forward and reverse reactions equally.​

  • Increase the rate at which equilibrium is reached, but do not change the position of equilibrium or KeqKeq.​

• Chemical equilibrium:

  • At equilibrium, forward rate = reverse rate, and adding a catalyst does not disturb this balance.​

  • To change equilibrium position, one must change conditions (concentrations, temperature, etc.), not merely add an enzyme.​

This reasoning directly supports that the correct answer to the question is (A) The equilibrium of the reaction will not change.