2. Chemical Equilibrium

At equilibrium the forward and reverse reactions run at equal rates — concentrations stop changing, but the reaction has not stopped. The equilibrium constant $K$ quantifies the balance point:

$$K_{eq} = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$$
Example: For $\text{N}_2 + 3\,\text{H}_2 \rightleftharpoons 2\,\text{NH}_3$ with $[\text{N}_2] = 0.50$, $[\text{H}_2] = 0.30$, $[\text{NH}_3] = 0.20$ M: $$K = \frac{(0.20)^2}{(0.50)(0.30)^3} = \frac{0.040}{0.0135} \approx 2.96$$

Q vs K — Predicting Direction

The reaction quotient $Q$ uses the same expression with current (non-equilibrium) concentrations. If $Q < K$, the reaction shifts right toward products; if $Q > K$, it shifts left.

Le Châtelier's Principle

A system at equilibrium shifts to counteract a disturbance:

ICE Tables

Example: For $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\,\text{HI}$, $K = 50$, with 0.10 M of each reactant initially: let $x$ = amount reacted. $[\text{HI}] = 2x$, $[\text{H}_2] = [\text{I}_2] = 0.10 - x$. $$50 = \frac{(2x)^2}{(0.10 - x)^2} \;\Rightarrow\; \frac{2x}{0.10 - x} = 7.071 \;\Rightarrow\; x = 0.0779$$ Equilibrium: $[\text{HI}] = 0.156$ M, $[\text{H}_2] = [\text{I}_2] = 0.022$ M.

Practice Problems

  1. Write the Keq expression for 2 SO2 + O2 ⇌ 2 SO3.
  2. For N2O4 ⇌ 2 NO2 (K = 0.36), Q is currently 0.10. Which direction will the reaction shift?
  3. The Haber process N2 + 3 H2 ⇌ 2 NH3 is exothermic. How does raising the temperature affect the equilibrium yield of NH3?
  4. Removing NH3 as it forms shifts the equilibrium which way?

Answers: 1. K = [SO3]2/([SO2]2[O2])   2. Right (toward products), since Q < K   3. Lowers it — heat is a product, so heating shifts left   4. Right, toward products