Kinetics and Equilibria
Kinetics and equilibria covers how the rate of a chemical reaction is measured and explained using collision theory and the Maxwell-Boltzmann distribution, how reaction orders and rate equations are found from experimental data, and how rate-determining steps relate to reaction mechanisms. It is examined throughout A Level Chemistry, often combined with calculations.
Method
- Recall that a successful (reactive) collision needs particles to collide with a combined energy greater than or equal to the activation energy, and with the correct orientation.
- To find the order of reaction with respect to a reactant from an initial-rates table, compare two experiments where only that reactant's concentration changes; if the rate stays the same when concentration doubles the order is zero, if the rate doubles the order is one, if the rate quadruples the order is two.
- Combine the individual orders to write the rate equation (rate = k[A]^m[B]^n), add the orders to find the overall order, then rearrange the rate equation to calculate k, remembering to state its units.
- From a concentration-time graph, identify a zero order reactant by a constant negative gradient, identify a first order reactant by a constant half-life, and use the graph's shape to distinguish between orders.
- In a reaction mechanism, identify the rate-determining step as the step (or steps) whose reactants appear in the experimentally determined rate equation; species that appear in the mechanism but not in the overall equation are intermediates.
- For calculations, monitor a physical property changing over time (such as gas volume in a gas syringe, or the time taken for a fixed colour change to appear), and use a tangent to the curve at t = 0 to find the initial rate.
Worked example
A reaction between reactants X and Y was investigated using the initial rates method. Experiment 1: [X] = 0.15 mol/dm3, [Y] = 0.15 mol/dm3, rate = 3.0 x 10^-3 mol dm-3 s-1. Experiment 2: [X] = 0.30 mol/dm3, [Y] = 0.15 mol/dm3, rate = 6.0 x 10^-3 mol dm-3 s-1. Experiment 3: [X] = 0.30 mol/dm3, [Y] = 0.30 mol/dm3, rate = 2.4 x 10^-2 mol dm-3 s-1. Deduce the rate equation for this reaction and calculate the rate constant, k, including its units, using the data from Experiment 1.
- Compare Experiments 1 and 2: [X] doubles (0.15 to 0.30) with [Y] constant, and the rate doubles (3.0 x 10^-3 to 6.0 x 10^-3), so the order with respect to X is 1.
- Compare Experiments 2 and 3: [Y] doubles (0.15 to 0.30) with [X] constant, and the rate quadruples (6.0 x 10^-3 to 2.4 x 10^-2), so the order with respect to Y is 2.
- Write the rate equation: rate = k[X][Y]^2, with overall order = 1 + 2 = 3.
- Rearrange for k using Experiment 1 data: k = rate/([X][Y]^2) = (3.0 x 10^-3)/(0.15 x 0.15^2).
- Evaluate: 0.15 x 0.15^2 = 0.15 x 0.0225 = 3.375 x 10^-3, so k = (3.0 x 10^-3)/(3.375 x 10^-3) = 0.889 (3 sf).
- Final answer: rate = k[X][Y]^2; k = 0.889 mol^-2 dm^6 s^-1
Practice questions
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Q1Define the term rate of reaction.Show answer
Answer: The change in concentration of a reactant or product per unit time
Q2State the units of the rate constant, k, for a reaction that is overall first order.Show answer
Answer: s^-1
Q3In an experiment, a graph of initial rate against [Z] is a straight line through the origin. State the order of reaction with respect to Z.Show answer
Answer: First order (rate is directly proportional to [Z])
Q4Define activation energy.Show answer
Answer: The minimum energy that colliding particles must possess for a collision to result in a reaction
Q5In a rate investigation, doubling the concentration of reactant Q causes the initial rate to increase by a factor of 8. Deduce the order of reaction with respect to Q.Show answer
Answer: Third order (2^3 = 8, so rate is proportional to [Q]^3)
Q6A reaction has rate equation rate = k[A]^2[B]. In one experiment [A] = 0.20 mol/dm3, [B] = 0.40 mol/dm3 and the rate is 5.6 x 10^-3 mol dm-3 s-1. Calculate k, including its units.Show answer
Answer: 0.35 mol^-2 dm^6 s^-1 (k = rate/([A]^2[B]) = 5.6 x 10^-3/(0.20^2 x 0.40) = 5.6 x 10^-3/0.016 = 0.35)
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
State the two conditions that must be met for a collision between reactant particles to result in a successful (reactive) collision, and explain how increasing the temperature affects the proportion of collisions that are successful.
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The concentration of reactant M was monitored during its decomposition at a constant temperature: at t = 0 min, [M] = 0.640 mol/dm3; at t = 15 min, [M] = 0.320 mol/dm3; at t = 30 min, [M] = 0.160 mol/dm3; at t = 45 min, [M] = 0.0800 mol/dm3. Show that these data are consistent with the reaction being first order with respect to M, and calculate the rate constant, k, in s^-1, giving your answer to 3 significant figures.
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The reaction between bromate(V) ions and bromide ions in acidic solution has the overall equation: BrO3- + 5Br- + 6H+ -> 3Br2 + 3H2O. The experimentally determined rate equation is rate = k[BrO3-][Br-][H+]^2. Explain what this rate equation shows about the mechanism of the reaction, and why the rate equation cannot be predicted directly from the balanced overall equation.
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