A Data Booklet may be used.
Source: Learners' Alley, Reaction Kinetics slides
Source: Learners' Alley, Reaction Kinetics slides
| expt | [A] / mol dm−3 | [B] / mol dm−3 | initial rate / mol dm−3 s−1 |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10−3 |
| 2 | 0.20 | 0.10 | 2.0 × 10−3 |
| 3 | 0.10 | 0.30 | 1.8 × 10−2 |
Source: Learners' Alley, Reaction Kinetics slides
Source: Learners' Alley, Reaction Kinetics slides
Source: Learners' Alley, Reaction Kinetics slides
Source: Learners' Alley, Reaction Kinetics slides
Source: Learners' Alley, Reaction Kinetics slides
CommonTwo colourless substances, X and Y, react to give a coloured substance Z. The time, t, taken for various initial concentrations of X and Y to produce Z is recorded in the table.
| [X] / mol dm−3 | [Y] / mol dm−3 | t / s |
|---|---|---|
| 0.05 | 0.05 | 44 |
| 0.05 | 0.10 | 22 |
| 0.10 | 0.05 | 44 |
What is the rate equation?
Source: HCI 2007 Prelim P1 Q9, via EJC 2022 JC1 Reaction Kinetics Tutorial, A4
Propanone reacts with bromine in the presence of an acid catalyst. CH3COCH3 + Br2 → CH3COCH2Br + H+ + Br−
The progress of the reaction was studied by finding the time taken for Br2 to decolourise using different volumes of reactants.
| expt | volume of Br2 added / cm3 | volume of CH3COCH3 added / cm3 | volume of H+ added / cm3 | volume of H2O added / cm3 | relative rate |
|---|---|---|---|---|---|
| 1 | 10 | 20 | 10 | 10 | 2 |
| 2 | 10 | 10 | 10 | 20 | 1 |
| 3 | 10 | 20 | 5 | 15 | 1 |
| 4 | 10 | 40 | 20 | 30 | 2 |
What is the rate equation for the reaction?
Source: EJC 2018 Promo, via EJC 2022 Reaction Kinetics Topical Revision, Q7
CommonAt 700 °C, nitrogen monoxide and hydrogen react as follows. 2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
| expt | initial [NO] / mol dm−3 | initial [H2] / mol dm−3 | initial rate / mol dm−3 s−1 |
|---|---|---|---|
| 1 | 0.0020 | 0.012 | 0.0033 |
| 2 | 0.0040 | 0.012 | 0.013 |
| 3 | 0.0060 | 0.012 | 0.030 |
| 4 | 0.012 | 0.0020 | 0.020 |
| 5 | 0.012 | 0.0040 | 0.040 |
| 6 | 0.012 | 0.0060 | 0.060 |
Source: EJC 2022 JC1 Reaction Kinetics Tutorial, A5
The reaction between nitric oxide and oxygen is third order overall, and it has been suggested that the mechanism involves the following two steps. 2NO → N2O2 (fast) N2O2 + O2 → 2NO2 (slow)
Source: A Level 2005 P3 Q1, via EJC 2022 JC1 Reaction Kinetics Tutorial, A11
The rate equation for the reaction 3A(g) + B(g) → C(g) + D(g) is rate = k[A]2[B]. The rate is R for certain pressures of A and B, at constant temperature. Write down, in terms of R, the rate of reaction when
Source: RJC H2 Chemistry Tutorial 6 (Reaction Kinetics), Self-check Q4
ChallengeSucrose is hydrolysed to glucose and fructose: C12H22O11 + H2O → C6H12O6 + C6H12O6. The reaction can be catalysed by dilute acid. The following results were obtained using hydrochloric acid as the catalyst.
| expt | initial [HCl] / mol dm−3 | initial [sucrose] / mol dm−3 | initial rate / mol dm−3 s−1 |
|---|---|---|---|
| I | 0.10 | 0.10 | 0.024 |
| II | 0.10 | 0.15 | 0.036 |
| III | 0.20 | 0.10 | 0.048 |
| IV | 0.25 | y | 0.048 |
Source: RJC H2 Chemistry Tutorial 6 (Reaction Kinetics), Standard Q1
42 × 2.0 × 10−4 = 4.0 × 10−4 mol dm−3 s−1[1]
Overall order 3: k = mol dm−3 min−1(mol dm−3)3 = mol−2 dm6 min−1[1]
Let rate = k[A]a[B]b. Experiments 1 and 2: 2.0 × 10−32.0 × 10−3 = k(0.20)a(0.10)bk(0.10)a(0.10)b, so 1 = 2a and a = 0. The reaction is zero order with respect to A.[1]
Experiments 1 and 3: 1.8 × 10−22.0 × 10−3 = k(0.10)a(0.30)bk(0.10)a(0.10)b, so 9 = 3b and b = 2. The reaction is second order with respect to B.[1]
or Comparing experiments 1 and 2: when [A] is doubled and [B] is kept constant, the initial rate is unchanged, so the reaction is zero order with respect to A.
Comparing experiments 1 and 3: when [B] is tripled and [A] is kept constant, the initial rate increases 9 times, so the reaction is second order with respect to B.
Rate equation: rate = k[B]2[1]
As temperature increases, the average kinetic energy of the reactant particles increases, so a greater proportion of reactant particles have energy greater than or equal to the activation energy, Ea.[1]
Hence, the frequency of effective collisions increases, and the rate of reaction increases.[1]
The power of its concentration term in the experimentally determined rate equation.[1]
k = ln 2 ÷ 20 = 0.0347 s−1[1]
A catalyst in the same phase as the reactants.[1]
Arate = k[Y][1]
Since rate ∝ 1t, let rate = k[X]x[Y]y. Rows 1 and 3: 1/441/44 = k(0.10)x(0.05)yk(0.05)x(0.05)y, so 1 = 2x and x = 0
Rows 1 and 2: 1/221/44 = k(0.05)x(0.10)yk(0.05)x(0.05)y, so 2 = 2y and y = 1
or Comparing experiments 1 and 3: when [X] was doubled and [Y] kept constant, the time taken remained unchanged, hence the rate remained unchanged. The reaction is zero order with respect to X.
Comparing experiments 1 and 2: when [Y] was doubled and [X] kept constant, the time taken was halved, hence the rate doubled. The reaction is first order with respect to Y.
Arate = k[CH3COCH3][H+][1]
To keep the total volume of every experiment the same, halve the volume of all reactants for experiment 4: 5, 20, 10 and 15 cm3, a total of 50 cm3. The rate remains unchanged, as the proportion of each reactant in the solution does not change when the volumes are halved.
Let relative rate = k(Vpropanone)m(VH+)n(VBr2)p at the same total volume. Experiments 1 and 2: 12 = (1020)m, so m = 1. Experiments 1 and 3: 12 = (510)n, so n = 1
Experiment 4 with all volumes halved (5, 20, 10, 15 cm3) and experiment 1: 22 = (510)p, so p = 0
or By comparing experiments 1 and 2, when the volume of CH3COCH3 is halved, the rate is halved. The reaction is thus first order with respect to CH3COCH3.
By comparing experiments 1 and 3, when the volume of H+ is halved, the rate is halved. The reaction is thus first order with respect to H+.
By comparing experiment 1 and experiment 4 (after halving all volumes), when the volume of Br2 is halved, the rate remains the same. The reaction is thus zero order with respect to Br2.
Let rate = k[NO]x[H2]y. Experiments 1 and 2: 0.0130.0033 = k(0.0040)x(0.012)yk(0.0020)x(0.012)y, so 4 = 2x and x = 2. The reaction is second order with respect to NO.[1]
Experiments 4 and 5: 0.0400.020 = k(0.012)x(0.0040)yk(0.012)x(0.0020)y, so 2 = 2y and y = 1. The reaction is first order with respect to H2.[1]
or Comparing experiments 1 and 2: when [NO] was doubled and [H2] kept constant, the rate increased to 4 times the original rate, so the reaction is second order with respect to NO.
Comparing experiments 4 and 5: when [H2] was doubled and [NO] kept constant, the rate doubled, so the reaction is first order with respect to H2.
rate = k[NO]2[H2][1]
Using values from experiment 6: 0.060 = k(0.012)2(0.0060)
k = 6.94 × 104 mol−2 dm6 s−1[1]
The rate constant is a proportionality constant for a given reaction at a specified temperature, denoted by k in (i).[1]
The order of reaction is the power to which the concentration of that reactant is raised in the experimentally determined rate equation. The order of reaction with respect to NO is 2 and with respect to O2 is 1.[1]
The rate-determining step is the slowest step in a sequence of elementary reactions, and it determines the rate of the reaction. Here, it is N2O2 + O2 → 2NO2.[1]
A catalyst lowers the Ea by providing an alternative pathway for the reaction.[1]
A greater proportion of reactant particles have energy greater than or equal to the lowered Ea. Hence, the frequency of effective collisions increases and the reaction rate increases.[1]
A catalyst increases the rate of reaction. From the rate equation, rate = k[A], where A is the reactant, if the rate increases while all concentrations remain the same, then the value of k must also increase. Hence, the rate constant is increased by the presence of a catalyst.[1]
or From the Arrhenius equation, k = Ae−Ea/RT, the rate constant is affected by Ea.
Let rate = k[sucrose]x[HCl]y. Experiments I and II: 0.0360.024 = k(0.15)x(0.10)yk(0.10)x(0.10)y, so 1.5 = 1.5x and x = 1. The reaction is first order with respect to sucrose.[1]
Experiments I and III: 0.0480.024 = k(0.10)x(0.20)yk(0.10)x(0.10)y, so 2 = 2y and y = 1. The reaction is first order with respect to HCl. Hence rate = k[sucrose][HCl].[1]
or Comparing experiments I and II: when the initial [sucrose] was increased 1.5 times, the initial rate increased 1.5 times, so rate ∝ [sucrose]. The reaction is first order with respect to sucrose.
Comparing experiments I and III: when the initial [HCl] was doubled, the initial rate also doubled, so rate ∝ [HCl]. The reaction is first order with respect to HCl.
Using data from experiment I, k = 0.024(0.10)(0.10) = 2.40 mol−1 dm3 s−1[1]
Since HCl is a catalyst, its concentration is effectively constant during the reaction, so rate = k′[sucrose], where k′ = k[HCl]. The reaction is pseudo first order, with a constant half-life.[1]
In experiment II, [HCl] is the same as in experiment I, so t½ = 3.0 s. In experiment III, [HCl] is double that of experiment I, so t½ = 1.5 s.[1]