Equilibrium and Gibbs Energy

Introduction

The reaction quotient predicts direction, while Gibbs energy expresses the thermodynamic driving force. These views are connected: current composition enters through Q, and the equilibrium constant enters when the reaction Gibbs energy becomes zero. A standard-state value is not generally zero at equilibrium.

Core explanation

At fixed temperature under the usual thermodynamic convention, ΔrG = ΔrG° + RT ln Q. Here ΔrG is the reaction Gibbs-energy change at the current composition, ΔrG° is the standard reaction Gibbs energy, R is the gas constant, T is absolute temperature, and Q is the dimensionless activity quotient for the written reaction. The expression requires consistent standard states.

At equilibrium, Q = K and the net reaction driving force vanishes: ΔrG = 0. Substituting gives 0 = ΔrG° + RT ln K, hence ΔrG° = −RT ln K. If K > 1, ln K > 0 and ΔrG° < 0 for the written direction. If K < 1, the standard reaction Gibbs energy is positive. These signs describe standard-state tendency, not a guarantee that every arbitrary initial mixture moves forward.

The current direction depends on Q/K. Combine the equations to get ΔrG = RT ln(Q/K). If Q < K, the logarithm is negative and forward reaction lowers Gibbs energy. If Q > K, the expression is positive and reverse change is favored. At Q = K, it is zero. This recovers the reaction-quotient direction test in thermodynamic language.

The distinction between ΔrG and ΔrG° is essential. A reaction may have ΔrG° < 0 yet be driven backward from a product-rich mixture with Q > K. At equilibrium, ΔrG = 0 but ΔrG° generally equals −RT ln K, which is zero only when K = 1 under the chosen standard states.

Temperature must be in kelvin. The logarithm takes a dimensionless Q, which is why activities and standard states matter. Reaction scaling also matters: double the balanced equation and ΔrG° doubles while K squares, preserving the identity. Reversing the equation changes both signs and reciprocates K.

Step-by-step reasoning

1. Write Q with dimensionless activities for the stated reaction. 2. Compare Q with K or calculate ΔrG = RT ln(Q/K). 3. Use the sign to identify forward, reverse or equilibrium behavior. 4. Keep current ΔrG distinct from standard ΔrG°.

Visual explanation

Draw a Gibbs-energy curve versus reaction extent with a minimum at equilibrium. Points left and right of the minimum have opposite downhill directions toward it.

Real-world analogy

A ball may roll downhill toward a valley from either side. The valley is equilibrium; the slope at the current position, not a reference slope elsewhere, determines its direction.

Real-world example

An industrial gas mixture may be product-rich enough to move net backward even when the standard-state reaction Gibbs energy favors products. Its current Q reveals this distinction.

Why?

Why is ΔrG zero at equilibrium? A small forward or reverse reaction change has no first-order Gibbs-energy advantage at the stable composition under the stated constraints.

Common misconception

“ΔrG° must be zero at equilibrium.” Current ΔrG is zero; ΔrG° is tied to the standard-state reference and equals −RT ln K.

Worked example

At a fixed temperature, let K = 10 and current Q = 1. Then ΔrG = RT ln(1/10), which is negative because ln 0.1 < 0. Net forward change is favored. At equilibrium Q becomes 10 and ΔrG = RT ln(10/10) = 0, while ΔrG° remains −RT ln 10, not zero.

Quick check

1. If Q = K, what is current ΔrG? Answer: Zero for the reaction extent at equilibrium under the stated thermodynamic conditions.

Exam focus

Use dimensionless Q and kelvin. Distinguish ΔrG from ΔrG° and remember that a negative standard value does not force forward motion for every current composition.

Advanced insight

At constant temperature and pressure, equilibrium minimizes total Gibbs energy subject to conserved elemental amounts. The reaction Gibbs energy is the derivative of that total with respect to reaction extent.

Summary

Current reaction Gibbs energy follows ΔrG = RT ln(Q/K). Its sign predicts net direction, and it vanishes at equilibrium, while standard ΔrG° equals −RT ln K.

Practice questions

1. What is the sign of ΔrG if Q < K? Answer: Negative, favoring net forward reaction. 2. When is ΔrG° zero under the chosen standard states? Answer: When K = 1 at that temperature. 3. Why must Q be dimensionless in the Gibbs-energy equation? Answer: It appears inside a logarithm and must be a pure number.