Chemical Kinetics and Equilibrium
This domain tests how fast reactions go and how far they go: rate measurement, collision theory, rate laws, and the quantitative treatment of dynamic equilibrium.
Rate and rate laws
Rate is the change in concentration per unit time, measured by gas volume, mass loss, colour change or titration. For rate = k[A]m[B]n, orders are found experimentally, not from the equation. Overall order = m + n, and the units of k follow from the overall order: mol1−p dm3p−3 s−1 for overall order p.
Collision theory and catalysts
Molecules react only if they collide with at least the activation energy and correct orientation. Raising temperature increases both collision frequency and the fraction of molecules exceeding Ea, so the effect is large. A catalyst provides an alternative route with lower Ea. On a Maxwell–Boltzmann plot, raising T flattens and broadens the curve to the right.
Equilibrium
At equilibrium the forward and reverse rates are equal and concentrations are constant, not necessarily equal. For aA + bB ⇌ cC + dD, Kc = [C]c[D]d/([A]a[B]b); pure solids and liquids are omitted. Kc depends only on temperature. Le Chatelier's principle predicts the direction of shift for changes in concentration, pressure (only for gaseous species) and temperature; a catalyst changes neither K nor the position of equilibrium.
Worked example
1.00 mol H2 and 1.00 mol I2 are sealed in a 1.00 dm³ vessel and reach equilibrium with [HI] = 1.60 mol dm−3. Since 2HI forms, 0.800 mol of each reactant reacted, leaving [H2] = [I2] = 0.200. Kc = 1.60²/(0.200 × 0.200) = 2.56/0.0400 = 64.0.
Traps that cost marks
- Reading orders from the stoichiometric equation instead of experimental data.
- Saying a catalyst increases yield; it only shortens the time to reach equilibrium.
- Including solids such as CaCO3 in Kc expressions.
- Linking K to pressure or concentration changes; only temperature alters K.
- Confusing rate with position of equilibrium - a fast reaction can still have a tiny K.