JPJC 2026 Chemical Kinetics Revision Answer

1 2026 JC1 H2 Chemistry (9476) Topical Revision for Physical Chemistry (Topic 6: Reaction Kinetics) FOUNDATION KIT Question 1 2 3 4 5 6 7 8 9 10 Answer C B A C B C B B D D Question 11 12 13 14 15 Answer A D B D C 1 [SAJC Prelim 2016/P1/Q8] Option A is wrong as it is showing a product -time graph. NH4Cl is a reactant undergoing decomposition. So the correct reactant-time graph should be downward sloping. Option B is wrong as the constant half-life is not stated. Hence, it does not confirm whether it is a first order OR second order reaction OR even orders higher than 2nd order. Option C is correct. This is a rate vs concentration graph. Hence, rate is directly proportional to concentration of NH4Cl, since rate α [NH4Cl] Option D is wrong as rate should increase when concentration of NH4Cl is high. Ans: C 2 [HCI Prelim 2020/ P1/ Q12] Assume a simple first order reaction, A → product rate = k[A] Options A and C would refer to the graph But this graph means zero order for A, therefore Options A and C are incorrect. Option B would refer to the graph

2 The reactant concentration, [A], is described as decreasing exponentially with time. (The half-life would be constant too.) Option D is possible for an auto-catalysed reaction, but no such information is provided in the question. Ans: B 3 [SAJC Prelim 2012/ P1/ Q10] Using half-life method, when [As2O3] = 0.10 mol dm-3, half-life = 32 min. Constant half-life → 1st order wrt VO2+. Using initial rates method, when [As2O3] doubles from 0.05 to 0.10, the gradient roughly doubles too → 1st order wrt As2O3. Rate = k[As2O3][VO2+] Ans: A 4 [DHS Prelim 2013/ P1/ Q8] Option A: The units for k are mol–1 dm3 s–1. Option B: There is no intermediate, with only reactants and products in the equation. Intermediate do not appear in the overall chemical equation! Option C: In excess KOH, rate equation becomes rate = k’[CH3CH2Br], where k’ = k[OH–] For this pseudo first order reaction, t1/2= ' 2ln k = ][ 2ln −OHk Thus, t1/2 is inversely proportional to [OH–]. Option D: Finding half–life in the graph of [CH3CH2Br] against time shows order with respect to [CH3CH2Br]. Ans: C

3 5 [PJC Prelim 2014/ P1/ Q5] Notice that the total volume of the reaction mixture is kept constant at 40.0cm3 for all 4 experiments. As such, Vreactant α [reactant]. Also, since a fixed amount of sodium thiosulfate is reacted before the blue- black appears, therefore rate α 1/t Experiment number Volume used / cm3 Time / s rate 1/t / s-1 KI Na2S2O8 H2O 1 10.0 5.0 25.0 170 0.00588 2 15.0 5.0 20.0 113 0.00885 3 15.0 10.0 15.0 56.5 0.0177 4 20.0 20.0 0.0 z 1/z Comparing expt 1 and 2, [Na2S2O8] kept constant, when [KI] x 1.5, rate x 1.5. Hence, 1st order w.r.t [KI]. Comparing expt 2 and 3, [KI] kept constant, when [Na2S2O8] x 2, rate x 2. Hence, 1st order w.r.t [Na2S2O8] Therefore, rate = k[Na2S2O8][KI] Comparing expt 1 and 4, [KI] x 2, [Na2S2O8] x 4, rate should x 8. rate for expt 4 = 1/z = 0.00588 x 8 = 0.04704 Hence, z = 21.3 Ans: B 6 [JJC Prelim 2012/ P1/ Q9] Given t1/2 = 2.0 hour After first half-life, mass of X = ½(320) = 160 mg After the second half-life, mass of X = ½(160) = 80 mg After the third half-life, mass of X = ½(80) = 40 mg Hence, drug X will still be effective after 3 half-lives (3 x 2.0 = 6.0 hour). Ans: C 7 [ASRJC Prelim 2021/ P1/ Q13] Since the reaction is first order with respect to G, the rate equation can be written as rate = k[G]. Hence plotting the rate against concentration of G will give an upward slopping straight line passing through 0 as shown in the question. Hence 1 is correct. If half-life of G is plot against rate constant, using t1/2 = ln2 k it will correspond to a y = 1 x graph. Since the reaction is first order with respect to G, the rate of reaction decreases as the reaction proceeds due to the decreasing concentration of G. Plotting the concentration of H (product) against time, will give

4 an upward slopping curve with a decreasing gradient. The decreasing gradient indicates a slower rate as the reaction proceeds. Hence, 3 is incorrect. Ans: B 8 [TPJC Prelim 2012/P1/Q9] Option A: Incorrect. The rate equation is based on the slow step. Hence, rate = k[C][B]. However, C is an intermediate in the reaction since the overall reaction is 2A + 2B → A2B2. Hence, rate = k[A]2[B]. Option B: Correct. Since rate = k[A]2[B], hence k = rate / ([A]2[B]). Units of k = mol dm-3 s-1 / (mol dm-3)3 = mol-2 dm6 s-1 Option C: Incorrect. The order of reaction is w.r.t reactants, not intermediates or products. D is an intermediate as it does not appear in the overall equation! Option D: Incorrect. Overall equation is 2A + 2B → A2B2 Ans: B 9 [ACJC Prelim 2012/P1/Q8] For autocatalytic reaction, rate is slow at the start. Once the product (catalyst) is formed, rate will increase until it reaches a maximum. Then it will slow down since the reactants are slowly depleted. Ans: D 10 [HCI Prelim 2012/ P1/ Q31] Option 1: Incorrect. For a reaction that is zeroth order w.r.t a reagent, increasing the [reactant] will not lead to an increase of rate. Eg. rate = k[A]0 Option 2: Correct. rate = k[A] n and from the Arrhenius equation, k = Ae -Ea/RT, it can be seen that the rate constant k is affected by both temperature and catalyst. Using a catalyst would increase the value of k, and hence rate will increase. Option 3: Correct. As mentioned in option 2. Ans: D

5 11 [ACJC Prelim 2015/P1/Q32] Option 1: Correct. rate = k[NO]2[H2] Option 2: Correct. rate = k[N2O2][H2] = k[NO][NO][H2] = k[NO]2[H2] Option 3: Incorrect. rate = k[NO]2[H2]2 Ans: A 12 [EJC Prelim 2022/ P1/ Q7] 1 ✓: From the slow step, rate = k[(CH3)2C+(OH)][H2O] = k′[(CH3)2C(=O)][H3O+], k′ = kKc At low pH, [H3O+] is high and is essentially a constant. rate = k″[(CH3)2C(=O)] 2 : rate constant, aE RTk Ae −= depends only on the Ea and T. 3 ✓: I2 is only involved in a fast step after the slow (rate determining) step. Hence rate of reaction is independent of [I2]. Ans: D 13 [TJC Prelim 2022/ P1/ Q13] Graph of [W] against time is a downward-sloping straight line with constant gradient.  Rate of reaction is independent of [W].  Reaction is zero order wrt [W]. [W] / mol dm−3 [V] / mol dm−3 Rate of reaction / mol dm−3 min−1 0.10 1.0 0.10 20 = 0.0050 0.10 2.0 0.10 10 = 0.010 When [V]  2, while [W] is unchanged, rate of reaction  2  Reaction is first order wrt [V]. Order of reaction wrt [X] cannot be determined from the graph.  Only Statements 1 & 2 are correct. Ans: B

6 14 [NYJC Prelim 2020/ P1/ Q9] When the temperature increased, the Maxwell Boltzmann graph flattens out with the peak lower than T1. With the addition of catalyst, Ea(cat) is lower as more particles with energy  Ea. Ans: D 15 [ASRJC Prelim 2020/ P1/ Q8] Since rate of production of O 2 is faster in experiment 2 and lead(IV) oxide is recovered at the end of the experiment, lead(IV) oxide is a catalyst that speeds up the rate of the reaction. A catalyst increases the reaction rate by providing a different reaction pathway which has a lower activation energy (Ea). Since k = A 𝑒 −𝐸𝑎 𝑅𝑇⁄ , a lower Ea will decrease the value of rate constant k. Hence, experiment 2 has a lower Ea and higher k. Ans: C 16 [PJC 2011 Test Q1] (a)  volume of iodineinitial rate time taken for brown iodine to disappear (b) (i) Since the total volume of reaction mixture is constant at 30 cm3, the volume of each reagent used is directly proportional to its concentration in the mixture. Compare Experiment 1 and 2 , when [HCl] is doubled, keeping [CH3COCH3] and [KI] constant, the initial rate is doubled (from 2 68 to 2 34 or the time taken is halved), hence it is first order with respect to HCl. (ii) Compare Experim