Chemical equilibrium: equilibrium constant, Le Châtelier and solubility product
The law of mass action, Kc and Kp, the reaction quotient, an ICE-table problem, Le Châtelier's principle with ammonia synthesis and the solubility product.
Many reactions do not go to completion: they reach a point where reactants and products coexist and their concentrations stop changing. Not because the reaction stops, but because the forward and the reverse reactions run at the same rate. This is chemical equilibrium, a dynamic equilibrium.
The equilibrium constant
For the reaction aA + bB ⇌ cC + dD, at equilibrium the law of mass action holds:
Kc = [C]c[D]d / ([A]a[B]b)
K depends only on temperature. Pure solids and pure liquids (such as water as the solvent) do not appear in the expression. For gases Kp is also used, with partial pressures; the two values are related by Kp = Kc(RT)Δn, where Δn is the change in moles of gas.
- Very large K (above about 103): products dominate at equilibrium, the reaction is “nearly complete”.
- Very small K (below 10−3): reactants dominate.
- K close to 1: comparable concentrations.
The reaction quotient Q has the same form as K but uses the current concentrations. If Q < K the reaction proceeds towards products, if Q > K towards reactants, if Q = K it is at equilibrium. The link with thermodynamics is ΔG° = −RT ln K: a negative ΔG° corresponds to K > 1.
A problem with an ICE table
For H2 + I2 ⇌ 2 HI at 450 °C, Kc = 50.0. We start from 1.00 mol/L of H2 and 1.00 mol/L of I2. What are the equilibrium concentrations?
| H2 | I2 | HI | |
|---|---|---|---|
| Initial | 1.00 | 1.00 | 0 |
| Change | −x | −x | +2x |
| Equilibrium | 1.00 − x | 1.00 − x | 2x |
Kc = (2x)2 / (1.00 − x)2 = 50.0. Taking the square root: 2x / (1.00 − x) = 7.07, so x = 0.780. At equilibrium [H2] = [I2] = 0.220 mol/L and [HI] = 1.56 mol/L.
Le Châtelier’s principle
A system at equilibrium that is disturbed reacts so as to counteract the disturbance:
- Adding a reactant shifts the equilibrium towards the products; removing a product has the same effect.
- Increasing the pressure (by reducing the volume) favours the side with fewer moles of gas.
- Raising the temperature favours the endothermic direction, and in this case K changes too.
- A catalyst does not shift the equilibrium: it only gets there sooner.
The classic case is ammonia synthesis, N2 + 3 H2 ⇌ 2 NH3, which is exothermic: 4 moles of gas become 2, so high pressure helps (150-300 atm); a low temperature would favour the yield but slow the reaction too much, and the Haber-Bosch process runs at about 400-450 °C with an iron-based catalyst. It is a compromise between thermodynamics and kinetics.
Heterogeneous equilibria and solubility
For a sparingly soluble salt, such as AgCl(s) ⇌ Ag+ + Cl−, the constant is called the solubility product: Ksp = [Ag+][Cl−] = 1.8·10−10 at 25 °C. The solubility is its square root: about 1.3·10−5 mol/L. Adding chloride from another source lowers the solubility further (the common ion effect), a direct application of Le Châtelier.
