Nothing will change overall; the reaction will remain at equilibrium, and there will be no net formation of either extra substrate or extra product.

Core idea

  • At equilibrium, the forward and reverse reaction rates are equal, so concentrations of substrate and product are not changing over time.
  • An enzyme lowers the activation energy for both directions equally, so it speeds up how fast equilibrium is reached, but it does not change the position of equilibrium or the overall free energy change ΔG\Delta GΔG.

So in the multiple‑choice framing of this classic question, the correct outcome is:

“Nothing; the reaction will stay at equilibrium.”

Why no net change?

  • The equilibrium constant KeqK_{eq}Keq​ depends only on the intrinsic energetics of reactants and products (and temperature), not on the presence of a catalyst.
  • Adding the enzyme increases both forward and reverse rates by the same factor, so the ratio [product][substrate]\frac{[\text{product}]}{[\text{substrate}]}[substrate][product]​ at equilibrium stays the same.

What does the enzyme do here?

Even though the macroscopic concentrations do not change:

  • Individual molecules of substrate and product will interconvert more frequently, cycling through the enzyme’s active site faster.
  • If the system were not at equilibrium to start with, the enzyme would simply help it reach that same equilibrium state more quickly, but still without altering ΔG\Delta GΔG or KeqK_{eq}Keq​.

Bottom line: adding an enzyme to a system already at equilibrium does not push the reaction toward more substrate or more product; it only affects the speed of reaching (or maintaining) that equilibrium.

Information gathered from public forums or data available on the internet and portrayed here.