Physical chemistry
Chemical equilibrium, equilibrium constant
How the reaction quotient compares with the equilibrium constant, and how K depends on the balanced equation and temperature.
IntuitionIntuition: balance between two opposing flows
When a reversible reaction is left alone, the mixture does not stop reacting — it reaches a state where the forward reaction supplies products at exactly the rate the reverse reaction consumes them. From outside, concentrations freeze; inside, molecules keep interconverting. The equilibrium constant is the fingerprint of where this dynamic stalemate lands.
SchoolSchool: writing and reading K
Definition: Equilibrium constant
For the reaction aA + bB ⇌ cC + dD at equilibrium, Kc = [C]^c[D]^d/([A]^a[B]^b). Pure solids and liquids do not appear; a value of K ≫ 1 means products dominate at equilibrium, K ≪ 1 means reactants dominate. K is fixed by the reaction equation and temperature, not by how the mixture was prepared.
Two simple rules turn a measured composition into a value of K. First, combine steps: adding reactions multiplies constants, reversing a reaction inverts its K, doubling all coefficients squares K. Second, compare the reaction quotient Q — the same algebraic form evaluated at the current state — with K: Q < K drives forward, Q > K drives backward.
| Operation on equations | Effect on K |
|---|---|
| reverse the reaction | K → 1/K |
| multiply by a factor n | K → Kⁿ |
| add two reactions | K → K₁K₂ |
Example: ICE table
For A ⇌ B with Kc = 4.0, a vessel is filled with 1.00 mol L⁻¹ of A only. Find the equilibrium concentrations.
Solution
Let x be the amount converted: [A] = 1.00 − x, [B] = x. Then x/(1.00 − x) = 4.0 gives x = 0.80. Hence [A] = 0.20 and [B] = 0.80 mol L⁻¹ — exactly the plateau shown in the K = 4-like curves of the simulation.
UndergraduateUniversity: thermodynamic basis
Equilibrium is the composition that minimises G at fixed T and p. The driving force vanishes when ΔrG = ΔrG° + RT ln Q = 0, which is possible only at Q = K = exp(−ΔrG°/RT). This single relation explains three facts at once: K is a temperature-only constant, large |ΔrG°| gives extreme K, and any disturbance that changes Q (addition, removal, compression) shifts composition toward restoring Q = K.
The van ’t Hoff equation makes Le Chatelier quantitative. For an endothermic reaction (ΔrH° > 0), ln K grows with T; for an exothermic one it shrinks. Over a narrow range where ΔrH° is nearly constant, a plot of ln K versus 1/T is a straight line with slope −ΔrH°/R — the standard laboratory route to reaction enthalpies.
Example: Ammonia synthesis
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔrH° ≈ −92 kJ mol⁻¹ (exothermic). Should K increase or decrease when the reactor is heated? Which observation do you expect for the equilibrium yield?
Solution
Exothermic reactions have d ln K/dT < 0, so K falls as T rises and the equilibrium yield of NH₃ decreases. The industrial Haber–Bosch process therefore compromises: moderate-high temperature for acceptable kinetics, high pressure for yield, and a catalyst to reach equilibrium faster — the catalyst speeds arrival but does not move the equilibrium.
AdvancedAdvanced: activities and real mixtures
In concentrated solutions and high-pressure gases, interactions make effective concentrations differ from nominal ones. The rigorous constant is written in activities aᵢ: K = ∏ aᵢ^νᵢ. For ions in water, Debye–Hückel theory relates the activity coefficient γᵢ to ionic strength; in gases the fugacity coefficient φᵢ corrects the pressure. These corrections explain why the “constant” measured in concentrations drifts at high ionic strength or high pressure even though the thermodynamic K has not changed.
References
- Studies in Chemical Dynamics · J. H. van ’t Hoff, 1896
- Thermodynamics · G. N. Lewis, M. Randall, 1923