Chemistry: Rate Equations and Equilibrium
Rate equations and equilibrium is the A-level Chemistry topic that extends the introductory kinetics and equilibrium ideas met earlier in the course into full quantitative treatments. It covers determining the order of reaction with respect to each reactant from experimental data, writing and using the rate equation and the rate constant k (including its units), using the Arrhenius equation to link the rate constant to activation energy and temperature, deducing a reaction mechanism from a rate equation, and calculating the equilibrium constant Kc for reactions in solution and Kp for reactions involving gases, including how catalysts and temperature (but not concentration or pressure) affect the value of K.
Method
- Find the order of reaction with respect to each reactant from a set of initial-rate experiments by comparing two experiments in which only one reactant's concentration changes, using the same doubling/no-change/quadrupling logic as for orders 0, 1 and 2; combine the individual orders into the rate equation, rate = k[A]^m[B]^n, and add them for the overall order.
- Rearrange the rate equation to find k from a set of data, and always state its units by working them out from the equation, rather than quoting a memorised unit: units of k = units of rate divided by units of concentration raised to the overall order.
- Use the Arrhenius equation, k = A x exp(-Ea / (R x T)), or its logarithmic form, ln k = ln A - Ea/(R x T), to link the rate constant to the activation energy Ea and the absolute temperature T; a graph of ln k (y-axis) against 1/T (x-axis) gives a straight line of gradient -Ea/R, letting Ea be found from experimental data at different temperatures.
- To deduce a reaction mechanism from an experimentally determined rate equation, identify the rate-determining (slowest) step as the one whose reactants, and only those reactants, appear in the rate equation; any species appearing in a proposed mechanism but not in the overall rate equation must react only in a faster step after the rate-determining step.
- Write the expression for Kc by putting the concentrations of the products (raised to their balancing numbers) over the concentrations of the reactants (raised to their balancing numbers), including only species in the same phase as the reaction; for a heterogeneous equilibrium, solids and pure liquids do not appear in the expression.
- For gas-phase equilibria, calculate the partial pressure of each gas as its mole fraction multiplied by the total pressure, then write Kp using partial pressures in the same way Kc uses concentrations.
- State the effect of a change on the value of K, not just the position of equilibrium: increasing temperature increases K for an endothermic forward reaction and decreases K for an exothermic forward reaction; a catalyst speeds up the approach to equilibrium but does not change the value of K, because it speeds up the forward and reverse reactions equally.
Worked example
The reaction between X and Y was investigated at constant temperature, giving the following initial rate data: Experiment 1: [X] = 0.10 mol/dm3, [Y] = 0.10 mol/dm3, initial rate = 2.0 x 10^-3 mol dm-3 s-1. Experiment 2: [X] = 0.20 mol/dm3, [Y] = 0.10 mol/dm3, initial rate = 8.0 x 10^-3 mol dm-3 s-1. Experiment 3: [X] = 0.20 mol/dm3, [Y] = 0.20 mol/dm3, initial rate = 1.6 x 10^-2 mol dm-3 s-1. Deduce the order of reaction with respect to X and with respect to Y, write the rate equation, and calculate the rate constant k, including its units.
- Compare Experiment 1 and Experiment 2: [Y] is constant, [X] doubles (0.10 to 0.20 mol/dm3), and the rate increases from 2.0 x 10^-3 to 8.0 x 10^-3, a factor of 4. Since 4 = 2^2, the reaction is second order with respect to X.
- Compare Experiment 2 and Experiment 3: [X] is constant, [Y] doubles (0.10 to 0.20 mol/dm3), and the rate increases from 8.0 x 10^-3 to 1.6 x 10^-2, a factor of 2. Since 2 = 2^1, the reaction is first order with respect to Y.
- Write the rate equation using these orders: rate = k[X]^2[Y]. The overall order is 2 + 1 = 3.
- Substitute the data from Experiment 1 to find k: 2.0 x 10^-3 = k x (0.10)^2 x (0.10) = k x 1.0 x 10^-3, so k = 2.0.
- Find the units of k by rearranging the rate equation: units of k = units of rate / (units of [X]^2 x units of [Y]) = (mol dm^-3 s^-1) / (mol^2 dm^-6 x mol dm^-3) = mol^-2 dm^6 s^-1.
- Final answer: the reaction is second order with respect to X, first order with respect to Y (third order overall); rate = k[X]^2[Y], and k = 2.0 mol^-2 dm^6 s^-1. Check using Experiment 3: k x (0.20)^2 x (0.20) = 2.0 x 0.04 x 0.20 = 1.6 x 10^-2, which matches the given rate.
Practice questions
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Q1State the effect on the rate constant, k, of increasing the temperature of a reaction, all other factors unchanged.Show answer
Answer: k increases, because a larger proportion of molecular collisions now have energy greater than or equal to the activation energy.
Q2Write the expression for the equilibrium constant Kc for the reaction: N2(g) + 3H2(g) <=> 2NH3(g).Show answer
Answer: Kc = [NH3]^2 / ([N2] x [H2]^3).
Q3A heterogeneous equilibrium is: CaCO3(s) <=> CaO(s) + CO2(g). Explain why only CO2 appears in the expression for Kc.Show answer
Answer: CaCO3 and CaO are solids, and the concentration of a pure solid is constant (it does not change as the reaction proceeds), so solids are not included in the equilibrium constant expression; only species whose concentration can vary, here the gas CO2, appear.
Q4A gas mixture at equilibrium and a total pressure of 200 kPa contains 0.60 mol of gas A and 0.40 mol of gas B, with no other gases present. Calculate the partial pressure of gas A.Show answer
Answer: Mole fraction of A = 0.60 / (0.60 + 0.40) = 0.60. Partial pressure of A = mole fraction x total pressure = 0.60 x 200 = 120 kPa.
Q5A reaction has the experimentally determined rate equation rate = k[A][B]^0. State what this shows about the effect of changing the concentration of B on the rate, and what this implies about the reaction mechanism.Show answer
Answer: Since B is zero order, changing its concentration has no effect on the rate; this implies B does not take part in the rate-determining (slowest) step of the mechanism, even though it appears in the overall balanced equation for the reaction.
Q6Using the Arrhenius equation in the form ln k = ln A - Ea/(R x T), state what is plotted on each axis to obtain a straight-line graph, and what the gradient of this line represents.Show answer
Answer: ln k is plotted on the y-axis against 1/T on the x-axis; the gradient of the resulting straight line is equal to -Ea/R, so gradient x (-R) gives the activation energy, Ea.
Q7A reaction at a fixed temperature has Kc = 4.0 x 10^-3. State, with a reason, whether the position of equilibrium lies closer to the reactants or the products.Show answer
Answer: Closer to the reactants; a small value of Kc (much less than 1) means that at equilibrium the concentration of reactants is much greater than the concentration of products (raised to their respective powers).
Q8A catalyst is added to a reaction that has already reached equilibrium. State the effect, if any, on the value of Kc and on the time taken to reach a new equilibrium if the system is disturbed.Show answer
Answer: The value of Kc is unaffected, since a catalyst does not change the position of equilibrium or the value of K; the catalyst does decrease the time taken to reach a new equilibrium after a disturbance, because it increases the rate of both the forward and reverse reactions equally.
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
The rate constant for a reaction was measured at two temperatures: at T = 300 K, k = 2.10 x 10^-4 s^-1; at T = 320 K, k = 8.40 x 10^-4 s^-1. Using the Arrhenius equation in the form ln(k2/k1) = -(Ea/R) x (1/T2 - 1/T1), calculate the activation energy, Ea, of the reaction in kJ/mol. (R = 8.31 J/(mol K))
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At a certain temperature, 2.00 mol of PCl5(g) was allowed to reach equilibrium with PCl3(g) and Cl2(g) in a sealed 5.00 dm3 container, according to the equation PCl5(g) <=> PCl3(g) + Cl2(g). At equilibrium, the container was found to contain 0.60 mol of Cl2(g). (a) Calculate the equilibrium concentrations of PCl5, PCl3 and Cl2. (b) Calculate Kc for this reaction, including its units.
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A reaction between compounds P and Q has the experimentally determined rate equation rate = k[P]^2. A student proposes the following two-step mechanism: Step 1: P + P -> P2 (slow); Step 2: P2 + Q -> products (fast). Explain whether this mechanism is consistent with the rate equation.
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