Reactivity 2 - How much, how fast and how far?
Reactivity 2 answers three quantitative questions: how much, how fast and how far. The exam tests stoichiometry, limiting reagents, solution calculations, rates and collision theory, and equilibrium.
Amounts and solutions
- c = n/V with V in dm³; dilution follows c₁V₁ = c₂V₂.
- Identify the limiting reagent by comparing mole ratios, then calculate theoretical yield. Percentage yield = actual/theoretical × 100.
- In titrations, use the balanced equation to convert moles of titrant to moles of analyte.
Rates
- Rate is the change in concentration per unit time; units mol dm⁻³ s⁻¹.
- Collision theory requires collisions with sufficient energy and correct orientation.
- Increasing concentration, pressure or surface area increases collision frequency. Temperature increases both collision frequency and the fraction with energy ≥ Ea. A catalyst lowers Ea.
- Maxwell–Boltzmann curves show the distribution of molecular energies; the area under the curve is constant.
Equilibrium
- Equilibrium is dynamic: forward and reverse rates are equal and concentrations are constant.
- Kc = [products]^coefficients/[reactants]^coefficients; only temperature changes Kc.
- Le Chatelier's principle predicts the direction of shift after a change in concentration, pressure or temperature.
Worked example
5.00 g CaCO₃ (M = 100.1) reacts with 100 cm³ of 0.500 M HCl. n(CaCO₃) = 0.0500 mol; n(HCl) = 0.0500 mol. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O needs 0.100 mol HCl, so HCl is limiting. n(CO₂) = 0.0250 mol, m = 1.10 g.
Traps
- Forgetting to convert cm³ to dm³.
- Treating a catalyst as changing equilibrium position.
- Assuming Kc changes when concentration changes.