Thermodynamics of enzymatic reactions: If the equilibrium constant (Keq) for an enzyme-catalyzed reaction is greater than 1 under standard conditions, what can be concluded about its energetics?

Introduction / Context: Relating equilibrium constants to free energy changes is central in bioenergetics. Enzymes do not change equilibrium positions; they only accelerate reaching equilibrium. Understanding the sign of Delta G°′ from Keq helps predict spontaneity under standard biochemical conditions.

Given Data / Assumptions:

• Keq > 1 means products are favored at equilibrium.
• Standard relationship: Delta G°′ = -R * T * ln(Keq).
• Temperature is standard (usually 298 K) and conditions are biochemical standard state.

Concept / Approach: If Keq > 1, then ln(Keq) > 0, so Delta G°′ is negative. A negative Delta G°′ indicates the reaction is exergonic under standard conditions (spontaneous in the forward direction until equilibrium is reached). Therefore, statements claiming endergonic behavior or impossibility without energy input are incorrect.

Step-by-Step Solution: Start with Delta G°′ = -R*T*ln(Keq). For Keq > 1, ln(Keq) is positive. Compute sign: Delta G°′ becomes negative, indicating exergonic behavior. Conclude that options (a) and (b) are false; choose “None of these.”

Verification / Alternative check: Example: Keq = 10 → ln(10) ≈ 2.303; Delta G°′ ≈ -5.7 kJ/mol at 298 K (negative). The forward reaction is favored at standard state.

Why Other Options Are Wrong: (a) Endergonic implies Delta G°′ > 0; (b) also implies non-spontaneity at standard conditions; (c) compounds the same errors.

Common Pitfalls: Confusing instantaneous direction with equilibrium position; enzymes accelerate rate but do not alter Keq or Delta G°′.

Final Answer: None of these.