Chemical Equilibrium and Reaction Kinetics
Equilibrium and kinetics questions separate two ideas that students often blur: how far a reaction goes (K) and how fast it goes (rate). The paper tests equilibrium expressions, ICE tables and Le Chatelier reasoning alongside experimental rate laws, integrated rate equations, half-lives and Arrhenius behaviour.
Equilibrium rules
- For aA + bB ⇌ cC + dD, K_c = [C]c[D]d/([A]a[B]b); pure solids and liquids are omitted.
- K_p = K_c(RT)Δn, where Δn counts gaseous moles only.
- Compare Q with K: Q < K means net forward reaction, Q > K means net reverse.
- K depends only on temperature. Catalysts never shift it, and neither does an inert gas added at constant volume.
- Le Chatelier: raising T favours the endothermic direction; compressing favours the side with fewer gas moles.
Kinetics rules
Rate laws come from experiment, not stoichiometry. Zero order: [A] = [A]0 - kt, t½ = [A]0/2k. First order: ln[A] = ln[A]0 - kt, t½ = 0.693/k, independent of concentration. Second order: 1/[A] = 1/[A]0 + kt. Arrhenius: k = A·exp(-Ea/RT), so ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). The rate-determining step is the slowest step, and the overall rate law must match experiment.
Worked example
H2 + I2 ⇌ 2HI with K_c = 49.0 at 700 K. Start with 1.00 mol dm-3 of each. ICE: H2 and I2 are each 1.00 - x, HI is 2x. 49.0 = (2x)2/(1.00 - x)2. Taking the positive root: 7.00 = 2x/(1.00 - x), so 9.00x = 7.00 and x = 0.778. Equilibrium: [HI] = 1.56 mol dm-3 and [H2] = [I2] = 0.22 mol dm-3. Check: 1.562/0.222 ≈ 49.
Traps
- Writing a rate law from stoichiometric coefficients.
- Forgetting that halving the volume multiplies every concentration term and also shifts the position of equilibrium.
- Using mol dm-3 units for K_p without converting.
- Believing t½ for a first-order reaction depends on starting concentration.
- Claiming a catalyst raises yield; it only shortens the time taken to reach equilibrium.