Chemical Equilibrium Formulae

Introduction

An equilibrium formula is built from a balanced reaction . Its quotient compares product and reactant activities with exponents equal to stoichiometric coefficients. The same algebraic expression is called Q at any current composition and K at equilibrium. Comparing Q and K predicts the thermodynamic direction of change, while Δ rG° = −RT ln K connects that balance to free energy.

Core explanation

For aA + bB ⇌ cC + dD, write Q = a C^c a D^d/(a A^a a B^b), where each a i is dimensionless activity. At equilibrium Q = K for the stated temperature and standard-state convention. In dilute classroom examples, dissolved-species activities are often approximated by concentration divided by a standard concentration, and ideal-gas activities by partial pressure divided by standard pressure. Writing bare concentration units into a logarithm is not thermodynamically rigorous.

Pure solids and pure liquids have activity approximately one in the usual standard-state treatment, so they do not appear as variable factors in a simple K expression. Their presence can still be necessary for a particular equilibrium to exist. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the ideal-gas quotient depends on CO₂ activity, while the solid activities are one. Removing a solid entirely can change which phases are available even though it was absent from the quotient.

Compare Q with K. If Q < K, the forward direction lowers reaction Gibbs energy toward equilibrium. If Q > K, the reverse direction is favored. At Q = K, no net thermodynamic driving force remains under the specified constraints, though forward and reverse molecular events continue. This direction test assumes the reaction model and species activities are appropriate.

The free-energy relation Δ rG = Δ rG° + RT ln Q becomes zero at equilibrium. Therefore Δ rG° = −RT ln K. K > 1 gives negative Δ rG° for the reaction as written, but does not mean every initial mixture moves forward; a product-rich mixture with Q > K moves backward. Reversing a reaction changes K to 1/K and reverses Δ rG°. Multiplying all reaction coefficients by m changes K to K^m and multiplies Δ rG° by m.

Temperature changes K. A van 't Hoff approximation for nearly constant reaction enthalpy is ln(K₂/K₁) ≈ −Δ rH°/R (1/T₂ − 1/T₁). An endothermic forward reaction tends to have K increase as temperature rises. This is a thermodynamic relationship, not a rule that catalysts shift K. Catalysts can speed approach to equilibrium but do not change the equilibrium constant at fixed temperature for the same chemical system.

An ICE table—initial, change, equilibrium—translates stoichiometry into compositions. For A ⇌ B, if x mol L⁻¹ A converts, [A] decreases by x and [B] increases by x. For 2A ⇌ B, A decreases by 2x. Solve the K equation with physically allowed x, then check nonnegative concentrations and the approximation used. Squared coefficients in Q arise from stoichiometry, not from squaring the numerical change term without justification.

Real mixtures may need activities or fugacities because electrostatic interactions and nonideal gas behavior alter effective chemical potential. A concentration-based K c is a useful approximation in dilute settings, but its numerical expression should not be mixed carelessly with a thermodynamic dimensionless K in logarithmic free-energy formulae.

Step-by-step reasoning

Balance the chemical equation and identify phases. Write an activity quotient with stoichiometric powers. Evaluate Q from the current composition and compare with K. If solving equilibrium amounts, use a stoichiometric change variable and reject physically impossible roots. Link to ΔG only after making Q and K dimensionless.

Visual explanation

Draw a seesaw with reactant activities on one side and product activities on the other. A marker Q moves as reaction proceeds toward fixed K at the stated T. Under it show ΔG changing sign as Q crosses K.

Real-world analogy

A crowded room can exchange people with an adjacent room until movement in both directions balances. The ratio of occupants at any moment resembles Q, while the stable ratio resembles K. Molecules keep exchanging at equilibrium, even though average counts stop changing.

Real-world example

For N₂O₄(g) ⇌ 2 NO₂(g), ideal-gas Q p = (p NO₂/p°)²/(p N₂O₄/p°). Heating can shift the equilibrium and change the brown NO₂ fraction. A pressure change changes current Q through partial pressures, then the composition adjusts toward the temperature-specific K.

Why?

Equilibrium formulae predict reaction direction and final composition from a clear stoichiometric model. They link measured concentrations and pressures to the thermodynamic energy scale and help distinguish kinetic effects from equilibrium changes.

Common misconception

“K > 1 means the reaction always moves forward” ignores the current Q. Another error puts a catalyst into the K expression or assumes it changes the equilibrium composition at unchanged temperature.

Worked example

For A ⇌ B, K = 4.0 under an ideal dilute approximation. Initially [A] = 0.50 M and [B] = 0.50 M, so Q = 1.0 < K and the forward direction is favored. Let x convert: at equilibrium [A] = 0.50 − x and [B] = 0.50 + x. Solve (0.50 + x)/(0.50 − x) = 4, giving x = 0.30 M, [A] = 0.20 M and [B] = 0.80 M. Both are nonnegative and their ratio is 4.

Quick check

1. What direction is thermodynamically favored when Q > K? Answer: The reverse direction, which tends to lower Q toward K.

Exam focus

Use balanced coefficients as exponents and omit pure-solid and pure-liquid activity factors. Compare Q and K for direction. Keep logarithm arguments dimensionless and distinguish equilibrium from reaction rate.

Advanced insight

For coupled reactions, standard Gibbs energies add when reactions add, so equilibrium constants multiply. This allows a difficult equilibrium to be built from known steps. The relationship follows from ΔG° = −RT ln K and the logarithm rule.

Summary

Q is the current activity quotient and K is its equilibrium value. Their comparison predicts direction, while ΔG° = −RT ln K connects equilibrium to energy. Correct expressions depend on reaction stoichiometry, phases and the meaning of activity.

Practice questions

1. Write Q for A + 2B ⇌ C using ideal dilute activities. Answer: Q = a C/(a A a B²). 2. What is K for a reversed reaction if forward K is 5? Answer: K reverse = 1/5 = 0.20. 3. Does a catalyst change K at fixed temperature? Answer: No. It changes rates of approach, not the equilibrium constant. 4. What does Q = K mean? Answer: The system is at equilibrium for that reaction under the stated conditions.

Sources

- IUPAC standard equilibrium constant. - OpenStax Chemistry 2e: Equilibrium Calculations.