Standard Gibbs energy when enthalpy change is zero: If ΔH° = 0 for a reaction, what is the correct expression for ΔG°?

Introduction / Context:Gibbs free energy (ΔG°) predicts the spontaneity of reactions at constant temperature and pressure. Understanding how ΔH° (enthalpy) and ΔS° (entropy) contribute helps interpret biochemical processes such as protein folding and ligand binding.

Given Data / Assumptions:

• ΔH° = 0 (no net enthalpy change under standard conditions).
• Temperature T is constant and in kelvin.
• Standard thermodynamic identity applies.

Concept / Approach:The fundamental relation is ΔG° = ΔH° − TΔS°. If ΔH° = 0, then ΔG° simplifies to −TΔS°. A positive ΔS° gives a negative ΔG° (favorable), while a negative ΔS° gives positive ΔG° (unfavorable) at constant T.

Step-by-Step Solution:Write identity: ΔG° = ΔH° − TΔS°.Substitute ΔH° = 0 → ΔG° = −TΔS°.Interpret sign: if ΔS° > 0, reaction tends to be spontaneous.

Verification / Alternative check:Statistical thermodynamics similarly predicts that increased microstates (higher entropy) lowers free energy when enthalpy is unchanged.

Why Other Options Are Wrong:

• TΔS°: Wrong sign.
• −ΔH°: Ignores entropy term.
• RT ln Keq: Correct general relation (ΔG° = −RT ln Keq) but the prompt asks the expression given ΔH° = 0; the simplified form is −TΔS°.
• ΔH° − T: Dimensionally incorrect.

Common Pitfalls:Dropping the negative sign or mixing ΔG with ΔG°; standard vs nonstandard conditions differ.

Final Answer:-TΔS°.