Gibbs Energy and Equilibrium Constant

Introduction

The equation Δ rG° = −RT ln K connects thermodynamics to equilibrium composition. A negative standard Gibbs energy corresponds to K > 1 for the reaction as written; a positive value corresponds to K < 1. At equilibrium the actual reaction Gibbs energy is zero, but the standard value is generally not.

Core explanation

At a given temperature, Δ rG = Δ rG° + RT ln Q. Equilibrium means there is no net Gibbs driving force for an infinitesimal advance of the reaction, so Δ rG = 0 and Q = K. Substitution gives 0 = Δ rG° + RT ln K, hence Δ rG° = −RT ln K. K is built from dimensionless activities at equilibrium, and R and T must be in matching units.

If K > 1, ln K > 0, so Δ rG° < 0. Products are favored relative to the standard-state reference arrangement under that reaction definition. If K < 1, ln K < 0 and Δ rG° > 0; reactants are favored in that standard comparison. K = 1 gives Δ rG° = 0. These statements do not claim that a reaction with K < 1 never produces products; equilibrium still generally contains some products.

The reaction direction matters. Reverse the equation and K becomes 1/K, while Δ rG° changes sign. Double all stoichiometric coefficients and the new K becomes K², while Δ rG° doubles. This follows because ln(K²) = 2 ln K. A quoted K without the corresponding balanced reaction is therefore incomplete.

Temperature affects K because Δ rG° and RT depend on T. A reaction that is exothermic may have K decrease as temperature rises under common conditions, but the precise relation involves the van 't Hoff equation and temperature-dependent enthalpy. The formula Δ rG° = −RT ln K applies at each stated temperature; do not use one K at another temperature without adjustment.

Most importantly, distinguish standard from actual values. Suppose K = 100. Then Δ rG° is negative, but a mixture already at Q = 100 has actual Δ rG = 0. A product-rich mixture with Q > 100 has positive forward Δ rG and tends backward. Conversely, a reaction with K = 0.01 can still tend forward if Q is even smaller than 0.01. Equilibrium is a balance at a particular composition, not a permanent property of the reaction direction independent of mixture.

Calculations require natural logarithms. For K = 10 at 298 K, Δ rG° ≈ −(8.314)(298)ln 10 J mol⁻¹ ≈ −5.71 kJ mol⁻¹. The base-10 logarithm cannot be substituted without a factor of ln 10. The sign and energy units should be checked before reporting.

K can be expressed through partial pressures or solution concentrations in elementary approximations, but a rigorous thermodynamic K is dimensionless and activity-based. Pure solids and liquids have unit activity under the usual reference convention. This matters when comparing equilibrium constants from different data presentations.

Step-by-step reasoning

1. Write the balanced reaction and its activity-based K expression. 2. Use absolute temperature and compatible R units. 3. Calculate ln K, not log₁₀K unless converted. 4. Compute Δ rG° = −RT ln K and interpret K size. 5. Separately use Q to determine actual direction at a current composition.

Visual explanation

Draw a number line for K with marks below 1, at 1 and above 1. Under it show the corresponding signs Δ rG° positive, zero and negative. In a second row, show Q moving toward K and actual Δ rG crossing zero at Q = K.

Real-world analogy

A destination may generally be favored from a reference starting point, but once a traveler has reached it there is no remaining drive to move farther in that direction. Standard Gibbs energy describes a reference comparison; actual Gibbs energy depends on the current reaction position.

Real-world example

For the Haber equilibrium, N₂ + 3H₂ ⇌ 2NH₃, changing temperature changes K. At a fixed temperature and pressure, changing gas composition changes Q and therefore the actual direction even though K remains fixed until temperature changes.

Why?

Why is Δ rG° not necessarily zero at equilibrium? The equilibrium mixture usually does not have every species at its standard-state activity. The RT ln K term offsets Δ rG° so the actual sum is zero.

Common misconception

“At equilibrium ΔG° = 0.” It is actual Δ rG that is zero. Δ rG° is zero only in the special case K = 1 at that temperature.

Worked example

At 300 K, let K = 100. Then ln K ≈ 4.605 and RT ≈ 2.494 kJ mol⁻¹, so Δ rG° ≈ −11.49 kJ mol⁻¹. At equilibrium Q = 100 and RT ln Q ≈ +11.49 kJ mol⁻¹. Their sum is actual Δ rG = 0. If Q = 1 instead, actual Δ rG = −11.49 kJ mol⁻¹ and the reaction tends forward.

Quick check

1. If K < 1, what is the sign of Δ rG° for the reaction as written? Answer: Positive, because ln K is negative and the formula has a leading minus sign.

Exam focus

Use natural log, dimensionless K and the correct reaction equation. State that actual Δ rG = 0 at Q = K, while Δ rG° = −RT ln K can be nonzero.

Advanced insight

Thermodynamic equilibrium is the minimum of total Gibbs energy under fixed temperature, pressure and material constraints. The condition Δ rG = 0 is the local slope condition with respect to reaction extent; the equilibrium constant expresses where that minimum occurs in activity space.

Summary

Standard Gibbs energy and equilibrium constant obey Δ rG° = −RT ln K. K > 1 corresponds to negative Δ rG°, K < 1 to positive Δ rG°. Actual reaction Gibbs energy depends on Q and becomes zero at equilibrium, regardless of whether the standard value vanishes.

Practice questions

1. If K = 1 at a stated temperature, what is Δ rG°? Answer: Zero, because ln 1 = 0. 2. If a reaction is reversed, what happens to K and Δ rG°? Answer: K becomes 1/K and Δ rG° changes sign. 3. At equilibrium with K = 20, is Δ rG° zero? Answer: No. Actual Δ rG is zero; Δ rG° = −RT ln 20 is negative.