Equilibrium and Le Chatelier's principle

MDCAT Chemistry · Chemistry · Chemical equilibrium. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A reversible reaction in a closed system reaches dynamic equilibrium. The forward and backward reactions are still going on, but at equal rates, so the concentrations stay constant. Constant does not mean equal.

The equilibrium constant, K_c or K_p, tells how far the reaction goes. A very large K means mostly products; a very small K means mostly reactants. For a given reaction, only a change of temperature changes K.

Le Chatelier's principle: if a system at equilibrium is disturbed, the position of equilibrium shifts in the direction that reduces the disturbance. It predicts the direction of a shift, not how fast it happens.

Rules to remember

For aA + bB ⇌ cC + dD: K_c = [C]^c[D]^d ÷ [A]^a[B]^b; K_p uses partial pressures in the same way; pure solids and liquids are left out
K_p = K_c(RT)^Δn, where Δn = moles of gaseous products − moles of gaseous reactants; K_p = K_c when Δn = 0
Reaction quotient Q: if Q < K the reaction moves forward; if Q > K it moves backward
Adding a reactant or removing a product shifts the equilibrium forward; K does not change
Higher pressure shifts towards the side with fewer moles of gas; higher temperature shifts in the endothermic direction
A catalyst speeds up both directions equally: equilibrium is reached sooner, with no change in position or in K
Haber process: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ mol⁻¹; high pressure (about 200 atm), about 400 to 450 °C as a compromise, iron catalyst

Common mistake

Students forget to raise each concentration to the power of its coefficient, or use starting amounts instead of equilibrium amounts. Work out what is present at equilibrium first, then substitute.

Worked example

1.0 mol of H₂ and 1.0 mol of I₂ are heated in a 1.0 dm³ flask: H₂(g) + I₂(g) ⇌ 2HI(g). At equilibrium 1.6 mol of HI is present. What is K_c?

1.6 mol of HI forms from 0.80 mol of H₂ and 0.80 mol of I₂ (ratio 1 : 1 : 2).
At equilibrium: [H₂] = 1.0 − 0.80 = 0.20; [I₂] = 0.20; [HI] = 1.6 mol dm⁻³.
K_c = [HI]² ÷ ([H₂][I₂]) = (1.6)² ÷ (0.20 × 0.20) = 2.56 ÷ 0.040 = 64.

Answer: K_c = 64 (no units, because the numbers of moles on the two sides are equal)

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