A Data Booklet may be used.
CommonA certain first order reaction took 42 minutes for the initial concentration of the reactant to decrease from 0.48 mol dm−3 to 0.06 mol dm−3. How long will it take for the concentration of the same reactant to decrease from 0.40 mol dm−3 to 0.10 mol dm−3?
Source: ACJC 2018 Promo, via EJC 2022 Reaction Kinetics Topical Revision, Q1
The dissociation of a compound RS is shown below. RS(g) → R(g) + S(g)
The rate equation for the reaction is rate = k[RS] and the rate constant is 3.6 × 10−3 s−1. If the initial concentration of RS is 2.0 × 10−2 mol dm−3, what will be the concentration of RS after 385 seconds?
Source: PJC 2007 Prelim P1 Q12, via EJC 2022 JC1 Reaction Kinetics Tutorial, A2
Which statements about order of reaction are correct?
Source: A Level 2006 P1 Q34, via EJC 2022 JC1 Reaction Kinetics Tutorial, A1
The following data were obtained from studies of the reaction between NO and O2 in a vessel at constant temperature.
| experiment | I | II | III | IV |
|---|---|---|---|---|
| total initial pressure / atm | 1.00 | 1.60 | 2.00 | x |
| initial pressure of NO / atm | 0.40 | 0.40 | 0.80 | 0.20 |
| initial rate of reaction / atm s−1 | 1.08 | 2.16 | 8.64 | 1.08 |
Which of the following statements is correct regarding the above system?
Source: HCI 2018 Promo, via EJC 2022 Reaction Kinetics Topical Revision, Q4
The reaction of acidified aqueous potassium iodide with aqueous hydrogen peroxide, 2I−(aq) + H2O2(aq) + 2H+(aq) → I2(aq) + 2H2O(l), is thought to involve the following steps.
| step 1 | H2O2 + I− → H2O + OI− | slow |
| step 2 | OI− + H+ → HOI | |
| step 3 | HOI + H+ + I− → I2 + H2O |
Which of the following conclusions can be drawn from this information?
Source: HCI 2018 Promo, via EJC 2022 Reaction Kinetics Topical Revision, Q6
CommonAt 800 K, nitrogen monoxide reacts with hydrogen according to the equation 2H2(g) + 2NO(g) → 2H2O(l) + N2(g). The table shows how the initial rate of this reaction depends on the initial partial pressures of the reactants.
| experiment | pH2 / atm | pNO / atm | initial rate / atm s−1 |
|---|---|---|---|
| 1 | 0.64 | 1.60 | 1.50 × 10−7 |
| 2 | 0.64 | 0.80 | 3.75 × 10−8 |
| 3 | 0.32 | 1.60 | 7.50 × 10−8 |
Source: 2023 VJC J2 Prelim P2 Q4(a)
Isoprene, CH2=C(CH3)CH=CH2, reacts with bromine. To determine the rate equation for this reaction, a student obtained the following data.
| expt | [isoprene] / mol dm−3 | [Br2] / mol dm−3 | initial rate / mol dm−3 s−1 |
|---|---|---|---|
| 1 | 0.04 | 0.064 | 0.88 × 10−3 |
| 2 | 0.08 | 0.064 | 1.76 × 10−3 |
| 3 | 0.08 | 0.128 | 3.52 × 10−3 |
Source: 2023 YIJC J2 Prelim P3 Q2(b)(i) and (ii)
The rate of reaction between propanone, CH3COCH3, and acidified iodine to give iodopropanone, CH3COCH2I, is found to be independent of [I2], but directly proportional to [H+] and directly proportional to [propanone].
| A | CH3COCH3 + H+ → X (slow) | X + I2 → products (fast) |
| B | CH3COCH3 + I2 → Y (slow) | Y + H+ → products (fast) |
Source: RJC H2 Chemistry Tutorial 6 (Reaction Kinetics), Standard Q5
Source: 2024 DHS J2 Prelim P3 Q2(d)(ii)
The reaction between iron(III) ions and iodide ions is shown. The initial concentration of Fe3+(aq) is 0.150 mol dm−3.
Fe3+(aq) + I−(aq) → Fe2+(aq) + ½I2(aq)
The rate of this reaction was followed by the change in the concentration of I−(aq), shown in the graph.
Source: 2024 VJC J2 Prelim P2 Q2(a)
In the presence of chlorine atoms, ozone decomposes exothermically to produce oxygen. The reaction mechanism has two steps.
step 1 O3 + Cl• → ClO• + O2 step 2 O3 + ClO• → Cl• + 2O2
Source: 2025 NYJC J2 Prelim P2 Q5(d)(ii) and (iii)
ChallengeExplain each of the following observations as fully as you can.
Source: RJC H2 Chemistry Tutorial 6 (Reaction Kinetics), Standard Q6(b) and (d)
A28 minutes[1]
[Reactant] falls from 0.48 to 0.24 to 0.12 to 0.06 mol dm−3 in three half-lives, 14 min each.
[Reactant] falls from 0.40 to 0.20 to 0.10 mol dm−3 in two half-lives of 14 min each, 28 min in total.
B5.0 × 10−3 mol dm−3[1]
t½ = ln 2k = 0.6933.6 × 10−3 = 192.5 s
Number of half-lives = 385192.5 = 2
ct = c0(½)n = (2.0 × 10−2) × (½)2 = 5.0 × 10−3 mol dm−3
B1 and 2 only[1]
CThe rate constant k has units of atm−2 s−1.[1]
Initial pressure of O2 = initial total pressure − initial pressure of NO = 0.60, 1.20, 1.20 and (x − 0.20) atm.
Let rate = kpNOxpO2y. Experiments I and II: 2.161.08 = k(0.40)x(1.20)yk(0.40)x(0.60)y, so 2 = 2y and y = 1
Experiments II and III: 8.642.16 = k(0.80)x(1.20)yk(0.40)x(1.20)y, so 4 = 2x and x = 2
or From experiments I and II, the pressure of NO remained unchanged while the pressure of O2 doubled. The rate doubled, so the reaction is first order with respect to O2.
From experiments II and III, the pressure of O2 remained unchanged while the pressure of NO doubled. The rate quadrupled, so the reaction is second order with respect to NO.
D2, 3 and 4 only[1]
Let rate = kpNOxpH2y. Experiments 1 and 2: 3.75 × 10−81.50 × 10−7 = k(0.80)x(0.64)yk(1.60)x(0.64)y, so 14 = (12)x and x = 2. The order with respect to NO is 2.[1]
Experiments 1 and 3: 7.50 × 10−81.50 × 10−7 = k(1.60)x(0.32)yk(1.60)x(0.64)y, so 12 = (12)y and y = 1. The order with respect to H2 is 1.[1]
or From experiments 1 and 2, as pNO halves and pH2 is kept constant, the rate decreases by 4 times. Hence, the order with respect to NO is 2.
From experiments 1 and 3, as pH2 halves and pNO is kept constant, the rate decreases by 2 times. Hence, the order with respect to H2 is 1.
rate = kpNO2pH2[1]
Experiment 1: k = 1.50 × 10−7(1.60)2(0.64)
= 9.16 × 10−8[1] atm−2 s−1[1]
Let rate = k[isoprene]x[Br2]y. Experiments 2 and 3: 3.52 × 10−31.76 × 10−3 = k(0.08)x(0.128)yk(0.08)x(0.064)y, so 2 = 2y and y = 1. The order of reaction with respect to Br2 is one.[1]
Experiments 1 and 2: 1.76 × 10−30.88 × 10−3 = k(0.08)x(0.064)yk(0.04)x(0.064)y, so 2 = 2x and x = 1. The order of reaction with respect to isoprene is one.[1]
or Using experiments 2 and 3: keeping [isoprene] constant, when [Br2] doubled, the initial rate doubled. The order of reaction with respect to Br2 is one.
Using experiments 1 and 2: keeping [Br2] constant, when [isoprene] doubled, the initial rate doubled. The order of reaction with respect to isoprene is one.
Mechanism A fits the observed kinetic data. The slow step, which is the rate-determining step, is a bimolecular elementary reaction between propanone and the acid. Its rate equation is of the form rate = k′[CH3COCH3][H+], which is similar to the experimentally determined rate equation.[1]
The reaction is zero order with respect to iodine, so iodine is only involved in the fast step (which is not rate-determining).[1]
The rate-determining step in mechanism B is a bimolecular elementary reaction between propanone and iodine, with rate = k″[CH3COCH3][I2]. Since the rate is found to be independent of [I2], mechanism B is inconsistent with the observed kinetic data.[1]
Let the original [NO2] be c mol dm−3. Then [NO2] after the 1st half-life = c2 mol dm−3.
1st half-life = 1kc, 2nd half-life = 2kc
Total time taken = 3kc = 30.54c = 46.3[1]
c = 0.120 mol dm−3[1]
Initial rate = −4.00 × 10−3 − 12.0 × 10−323.0 − 0 = 3.48 × 10−4 mol dm−3 s−1[1]
Rate at 20 s = −4.00 × 10−3 − 10.3 × 10−347.0 − 0 = 1.34 × 10−4 mol dm−3 s−1[1]
Let n be the order with respect to I−. At 20 s, [I−] = 7.60 × 10−3 mol dm−3.
initial raterate at 20 s = 3.48 × 10−41.34 × 10−4 = (12.0 × 10−37.60 × 10−3)n
n = 2.09, so the reaction is second order with respect to I−.[1]
3.48 × 10−4 = k(0.150)(12.0 × 10−3)2
k = 16.1[1] mol−2 dm6 s−1[1]
Correct shape: two steps, with the overall reaction exothermic.[1]
The axis, the balanced species and the enthalpy change are labelled correctly.[1]
NoteStep 1 is likely to be endothermic, as more energy is taken in to break the stronger O–O bond in ozone than is released in forming the weaker Cl–O bond. Step 2 must then be exothermic for the overall ΔH to be negative.
Platinum acts as a heterogeneous catalyst in the decomposition of NH3 into N2 and H2.
At very low pressures, the active sites on the catalyst surface are not saturated with ammonia molecules. Hence, the rate of decomposition of ammonia depends only on the partial pressure of ammonia, and the reaction is first order with respect to ammonia.[1]
At moderate or high pressures, most, if not all, of the active sites on the platinum surface are taken up by ammonia molecules.[1]
Any increase in the partial pressure of ammonia then has no effect on the rate of decomposition, and the reaction is zero order with respect to ammonia. The limiting factor is the availability of active sites.[1]
As temperature increases, the average speed of the reactant particles increases. This leads to a higher collision frequency, which contributes to an increase in the reaction rate.[1]
However, the large increase in reaction rate from a small rise in temperature is mainly because, at a higher temperature, the proportion of reactant particles with energy greater than or equal to the activation energy increases significantly, as the Maxwell–Boltzmann distribution curve shows.[1]
Consequently, the frequency of effective collisions increases accordingly, and hence the reaction rate increases. An increase in temperature also leads to a larger rate constant.[1]