Chemical Potential and Reaction Equilibrium

Introduction

A reaction changes several species amounts in fixed stoichiometric proportions. Chemical potential tells how Gibbs energy responds to each amount change; adding those responses gives the reaction Gibbs energy. Its sign predicts the locally favourable direction at constant temperature and pressure. Setting it to zero at equilibrium produces the familiar link between standard Gibbs energy and the equilibrium constant.

Core explanation

Write a reaction as Σ iν i A i = 0, with ν i negative for reactants and positive for products. Define extent ξ by dn i = ν i dξ. At fixed T and p, dG = Σ i μ i dn i = (Σ iν iμ i)dξ. Thus Δ rG = (∂G/∂ξ) T,p = Σ iν iμ i. If Δ rG < 0, a small forward change dξ > 0 lowers G; if Δ rG > 0, a small reverse change is favoured. Stable equilibrium with respect to the reaction requires Δ rG = 0 and a local G minimum under the allowed constraints.

For each component write μ i = μ i° + RT ln a i, where a i is dimensionless and based on a consistent standard state. Substitution yields Δ rG = Δ rG° + RT ln Q, with Δ rG° = Σν iμ i° and Q = Π i a i^ν i. At equilibrium Q = K and Δ rG = 0, so Δ rG° = −RT ln K. K is dimensionless in this thermodynamic formulation. A quoted concentration-style constant that carries units often reflects an unstated reference concentration; the thermodynamic activity expression avoids this ambiguity.

The standard Gibbs energy Δ rG° is not the current reaction Gibbs energy unless Q = 1. A reaction with Δ rG° > 0 can still proceed forward if Q is sufficiently small, such as when products are initially absent. Likewise, a negative Δ rG° does not guarantee net forward change under every mixture composition. Compare Q with K: if Q < K, then Δ rG = RT ln(Q/K) < 0 and forward reaction is favourable; if Q > K, reverse reaction is favourable. This result assumes the same temperature, standard states and activity definitions for Q and K.

For gases, ideal activities are p i/p°; for ideal-dilute solutes, an appropriate concentration ratio may be used, while real systems require fugacities or activities. Pure solids and pure liquids often have activity one in their standard states, so they do not appear explicitly in Q although their chemical potentials contribute to Δ rG°. Stoichiometric coefficients must be applied as exponents; doubling a reaction equation squares K and doubles Δ rG°.

Step-by-step reasoning

Balance the reaction and assign signed ν i. Express amount changes as dn i = ν i dξ, then sum μ iν i to obtain Δ rG. Insert μ i° + RT ln a i, collecting standard terms and the logarithm of a product to define Q. At equilibrium set Δ rG = 0 to find K or Δ rG°. For a non-equilibrium mixture compare Q with K before naming the favoured direction.

Visual explanation

Plot G against reaction extent ξ as a valley-shaped curve. At a point left of the minimum, the tangent slope Δ rG is negative; at the minimum it is zero; right of it the slope is positive. Under the plot, place a reaction arrow with Q < K on the left and Q > K on the right.

Real-world analogy

A ball on a shaped track tends to move toward a lower height, and its local slope determines which way it initially rolls. Reaction Gibbs energy is the local slope of G against extent. The analogy is limited because molecular reactions may face activation barriers and require catalysts even when the thermodynamic slope favours change.

Real-world example

For synthesis of ammonia, N₂ + 3H₂ ⇌ 2NH₃, the ideal-gas quotient is Q = (p NH3/p°)²/[(p N2/p°)(p H2/p°)³]. A mixture's composition and pressure set Q, so the reaction direction can change as ammonia accumulates. Industrial operating choices also depend on reaction rates and catalysts, not solely on the equilibrium sign.

Why?

Stoichiometry couples species amount changes, so their individual chemical-potential contributions must be weighted and added. Logarithmic activity terms then combine into a logarithm of a product. The equilibrium constant emerges as the value of that product where the Gibbs-energy slope with respect to reaction extent vanishes.

Common misconception

Δ rG° is not the same as Δ rG at every composition. The standard symbol means each species is evaluated in its standard state, a hypothetical collection that may not be the actual reacting mixture. Also, K is fixed by temperature and chosen standard states for a reaction, while Q changes as composition changes.

Worked example

At 298 K, let K = 10.0 for A ⇌ B and let current activities be a A = 0.50, a B = 1.0. Then Q = a B/a A = 2.0. Since Q < K, forward change is favourable. Quantitatively Δ rG = RT ln(Q/K) = (8.314)(298)ln(0.20) ≈ −3.99 kJ mol⁻¹ of reaction. The negative value is the local Gibbs-energy change per mole of forward extent, not a reaction rate.

Quick check

1. If Q equals K, what is Δ rG at fixed T and p? Answer: Δ rG = RT ln(Q/K) = 0. The mixture is at reaction equilibrium with respect to a small allowed extent change.

Exam focus

Use dimensionless activities and signed stoichiometric coefficients. Write Δ rG = Δ rG° + RT ln Q and Δ rG° = −RT ln K with consistent signs. Compare Q and K for direction. Do not infer rate from Δ rG, and check whether a pure phase has activity one under the stated standard convention.

Advanced insight

Changing the written reaction changes numerical thermodynamic descriptors systematically. Reversing the equation changes Δ rG° to its negative and K to 1/K. Multiplying all coefficients by two doubles Δ rG° and changes K to K². These transformations are powerful checks on algebra and on the definition of extent.

Summary

Reaction extent couples composition changes through dn i = ν i dξ. Therefore Δ rG = Σν iμ i is the Gibbs-energy slope along the reaction. Activity formulas give Δ rG = Δ rG° + RT ln Q; at equilibrium Q = K and Δ rG° = −RT ln K. Current direction depends on Q/K, not on Δ rG° alone.

Practice questions

1. For A + 2B ⇌ C, write the activity quotient. Answer: Q = a C/(a Aa B²), with each activity dimensionless and the exponent 2 inherited from B's stoichiometric coefficient. 2. At a temperature where K = 0.10, a mixture has Q = 1.0. Which direction lowers G? Answer: Q > K, so Δ rG = RT ln(Q/K) > 0 for the forward direction. A small reverse change lowers G. 3. If Δ rG° = −RT ln 4 for a reaction, what is the equilibrium constant for its reverse as written? Answer: The forward K is 4; reversing the reaction gives K reverse = 1/4 and changes the sign of Δ rG°.