Reactivity: Energetics, Rates, Equilibrium and Transformations
Reactivity is the second core area and covers energy changes, the speed and extent of reactions, and proton, electron and electron-pair transfer. HL questions test multi-step calculations, mechanisms, and explanations using collision theory, entropy and Le Chatelier's principle.
Core rules and concepts
- Enthalpy: exothermic reactions have negative ΔH. Hess's law allows indirect routes. Bond breaking is endothermic (positive), bond forming is exothermic (negative); mean bond enthalpies give estimates, not exact values.
- Rates: collision theory requires sufficient energy and correct orientation. Raising temperature increases both collision frequency and the fraction of collisions with E ≥ Ea; a catalyst lowers Ea.
- Equilibrium: Kc depends only on temperature. Compare Q with K to predict direction. Le Chatelier predicts shifts in concentration, pressure or temperature, but only temperature changes K.
- Acids and bases: pH = −log[H⁺]; weak acids require Ka because they are only partly ionised. Buffers resist pH change through a conjugate acid–base pair.
- Redox: assign oxidation numbers, balance half-equations, and relate cell potential to spontaneity.
- Organic mechanisms (SN1, SN2, electrophilic addition, elimination) depend on structure, solvent and conditions.
Worked example
At 700 K, Kc = 0.036 for H₂(g) + I₂(g) ⇌ 2HI(g). Starting with [H₂] = [I₂] = 0.100 mol dm⁻³, find [HI] at equilibrium. Let the change be x: [H₂] = [I₂] = 0.100 − x and [HI] = 2x. Then 0.036 = (2x)²/(0.100 − x)². Taking the square root: 0.1897 = 2x/(0.100 − x). Solving gives x = 0.00866, so [HI] = 0.0173 mol dm⁻³. Show the ICE table and the square-root step for full marks.
Traps that cost marks
Do not forget stoichiometric coefficients in Kc. Do not substitute initial concentrations into Kc. Do not confuse rate with equilibrium position. Do not change K for pressure or concentration changes-only temperature. Do not assume pH = −log(acid concentration) for a weak acid.