Thermodynamics sign conventions: If the standard Gibbs free energy change (ΔG°) for a reaction is positive, what is favored at equilibrium under standard conditions?

Introduction / Context:
Gibbs free energy links thermodynamics to chemical equilibrium. The sign and magnitude of ΔG° determine the direction of spontaneity under standard conditions and relate quantitatively to the equilibrium constant Keq via ΔG° = −RT ln Keq.

Given Data / Assumptions:

• ΔG° > 0 (positive standard free energy change).
• Standard temperature and pressure; concentrations at 1 M for solutes (activities idealized).

Concept / Approach:
If ΔG° = −RT ln Keq is positive, then ln Keq is negative, meaning Keq < 1. This implies that, at equilibrium under standard conditions, the mixture contains more reactants than products—reactants are favored.

Step-by-Step Solution:
Start: ΔG° = −RT ln Keq. Given ΔG° > 0 ⇒ −RT ln Keq > 0 ⇒ ln Keq < 0. Therefore Keq < 1 (reactants favored at equilibrium). Select the statement consistent with Keq ≪ 1: reactants favored.

Verification / Alternative check:
Graphing ΔG° vs ln Keq shows the monotonic inverse relationship: positive ΔG° corresponds to small Keq.

Why Other Options Are Wrong:
Products favored requires Keq > 1 (ΔG° < 0); Keq = 1 implies ΔG° = 0; complete conversion ignores equilibrium; rate is kinetic, not governed by ΔG° alone.

Common Pitfalls:
Confusing thermodynamic favorability with speed; overlooking the log relationship between ΔG° and Keq.

Final Answer:
Reactants will be favored (Keq << 1).