RI 2025 Kinetics Tutorial Suggested Answer

1 Tutorial 6: Reaction Kinetics – Suggested Answers 8 (a) (I) The order of reaction with respect to a given reactant is the power to which the concentration of that reactant is raised in the rate equation. (II) The rate constant of a rection is the constant of proportionality in the rate equation of the reaction. (III)The half-life of a reaction is the time taken for the concentration of a reactant to decrease to half its initial value. (b) Comparing experiments I and II, when initial [sucrose] × (0.15/0.10 =) 1.5, initial rate × (0.036/0.024 =) 1.5 ⇒ rate ∝ [suc rose], i.e. reaction is first order with respect to sucrose. Comparing experiments I and III, when initial [HCl] × 2, initial rate × 2 ⇒ rate ∝ [HCl], i .e. reaction is first order with respect to HCl. ∴ rate equation: rate = k[sucrose][HCl] Using data from experiment I, k = 0.024 0.10 0.10× = 2.40 mol−1 dm3 s−1 (data from experiment II or III can be used as well) (c) rate = k[sucrose][HCl] 0.048 = 2.40(y)(0.25) ∴y = 0.08 (d) rate = k[sucrose][HCl] Since HC l is a catalyst in the reaction, its concentration is effectively constant during the reaction. The rate equation can be simplified to: rate = k’[sucrose], where k’ = k[HCl]. The reaction becomes a pseudo first-order r eaction with constant half-life: 1/ 2 In 2 In 2 In 2t 3.0 s' [HC ] (0.10)= = = =kk k l For experiment II, [HCl] is the same as experiment I. Hence, t1/2 of sucrose in experiment II = t1/2 of sucrose in experiment I = 3.0 s For experiment III, [HCl] is double that of experiment I. Hence, t1/2 of sucrose in experiment III 1/ 2 In 2 In 2 3.0t ' (2 x 0.10) 2= = = = =kk 1.5 s

2 9 Compare experiments 1 and 2: When [NO2] x 2, rate x 2 ⇒ r ate ∝ [NO2], i.e. reaction is first order with respect to NO2. Let rate = k[NO2][SO2]m where m is the order of reaction with respect to SO2. Compare experiments 2 and 3: rate3 rate2 = k(0.040)(0.40)m k(0.020)(0.20)m = 8 2 Solving, m = 1 So rate = k[NO2][SO2] Since [SO2] >> [NO2] in all 3 experiments, this is a pseudo first-order reaction and rate = k’[NO2], where k’ = k[SO2]. t1 2� = ln 2 k' = ln 2 k[SO2] S ince the [SO 2] in expt 1 and 2 are the same, the half-life should be the same at 48 s. Since [SO2] in expt 3 is twice that in expt 1, the half-life would be halved to 24 s. 10 Ans: B As the decomposition is a first order reaction, 1/ 2 In 2t = k . t½ is independent of [H2O2], i.e. doubling [H2O2] from 0.1 mol dm−3 to 0.2 mol dm−3 has no effect on t½. Since ln 2 and k are constants, t½ is constant and likewise, time taken for H2O2 to decompose by 10 % will also be constant at 5 min. 11 (a) Since rate of reaction ∝ Δ[I2] Δt ∝ volume of I2(aq) used time taken for solution to turn colourless, by calculating volume of I2(aq) used time taken for solution to turn colourless , we will be able to compare the rate of reaction for following experiments. Expt. No. vol. of CH3COCH3 / cm3 vol. of I2 / cm3 vol. of H2SO4(aq) / cm3 vol. of water / cm3 time taken for solution to decolourise / min vol of I2 time taken/ cm3 min–1 1 8.0 4.0 8.0 0.0 1.0 4.0 2 8.0 4.0 4.0 4.0 2.0 2.0 3 4.0 4.0 8.0 4.0 2.0 2.0 4 8.0 2.0 8.0 2.0 0.5 4.0 Comparing experiments 1 and 2, when volume of H2SO4 x 0.5 (corresponding to [H+] x 0.5), rate x 0.5. ⇒ r ate ∝ [H+], i.e. reaction is first order with respect to H+. Comparing experiments 1 and 3, When volume of CH3COCH3 x 0.5 (corresponding to [CH3COCH3] x 0.5, rate × 0.5. ⇒ r ate ∝ [CH3COCH3], i.e. reaction is first order with respect to CH3COCH3. Comparing experiments 1 and 4, When volume of I2 x 0.5 (corresponding to [I2] x 0.5, rate remains unchanged. ⇒ rate ∝ [I2]0, i.e. reaction is zero order with respect to I2. Alternative presentation Compare experiments 2 and 3: When [NO2] x 2 and [SO2] x 2, rate x 4 But since reaction is first order with respect to NO2, it means that rate x 2 due to [NO2] x 2 AND rate x 2 due to [SO2] x 2 ⇒ rate ∝ [SO2], i.e. reaction is first order with respect to SO2.

3 (b) (I) The rate equation for a reaction is a mathematical expression that relates the rate of reaction to the concentration of each reactant raised to the appropriate power. (II) The rate-determining step is the slowest step in the reaction mechanism of a multi-step reaction and it determines the overall reaction rate. (c) rate = k[CH3COCH3][H+]; Overall order = 2 (d) In m echanism A, the slow step (also the rate-determining step) is a bimolecular elementary reaction involving one molecule of CH 3COCH3 and one H+ ion. The rate equation is rate = k ’[CH3COCH3][H+], which is similar to the rate equation in (c). ⇒ Mechanism A fi ts the observed kinetic data. In m echanism B , the slow step is a bimolecular elementary reaction involving one molecule of CH 3COCH3 and one molecule of I2. The rate equation is rate = k’’[CH3COCH3][I2], which is different from the rate equation in (c). ⇒ M echanism B is inconsistent with the observed kinetic data. [Note: The reaction is zero order with respect to I2, so I2 is only involved in the fast step (which is not rate-determining).] (e) A s the reaction proceeds by a similar mechanism, bromine (like iodine) is not involved in the slow/rate-determining step. The reaction is therefore zero order with respect to bromine. Hence, the rate of reaction is similar to that of the reaction between propanone and iodine under the same experimental conditions. ( Note: Even though the Br−Br bond is stronger than the I−I bond, the rate of reaction will be similar, because slow/rate-determining step does not involve the breaking of the halogen-halogen bond.) 12 (a) amount of S2O32− used = 0.050 x 0.0020 = 0.000100 mol 2S2O32− + I2 → S4O62− + 2I− amount of I2 = ½ x 0.000100 = 5.00 x 10−5 mol (b) [I2] = = = 5.00 x 10−4 mol dm−3 Initial rate of reaction ≈ [I2] t Expt No Time for the appearance of deep blue colour/ s Initial rate ≈ [I2] t / mol dm−3 s−1 1 33 1.52 x 10−5 2 100 5.00 x 10−6 3 67 7.46 x 10−6 4 50 1.00 x 10−5 Comparing experiments 2 and 4, when volume of H2O2 × 2 (corresponding to [H2O2] × 2), initial rate × 2. ⇒ rate α [H2O2] ⇒ r eaction is first order with respect to H 2O2. Comparing experiments 2 and 3, when volume of KI × 1.5 (corresponding to [KI] ×1.5), initial rate × 1.5. ⇒ rate α [KI] ⇒ reaction is first order with respect to KI. mixture of volume iodine of amount 0.100 5.00x10 5-

4 Let rate equation be rate = k [H2O2][KI][HCl]b where b is the order of reaction w ith respect to HCl. Comparing experiments 1 and 2 and using the fact that [X] ∝ VX since total volume is a constant, ⇒ 3(0.5)b = 3.04 ⇒ b = 0 ⇒ r eaction is zero order with respect to HCl. (c) rate = k[H2O2][KI] ; units of rate constant is mol−1 dm3 s−1. (d) y = 33 s . Explanation (not required by question): The total volume is twice that of all the other experiments. The volume of each reagent used is twice that of Experiment 1 ⇒ initial concentrations of all reactants are the same as that of Experiment 1 ∴ time taken is the same as Experiment 1. 13 (i) Use the graphs to calculate the initial rates of reaction during the two runs. For run A, initial rate = − 0.0010 - 0 0 - 32 = 3.125 x 10–5 = 3.13 x 10–5 mol dm–3 min–1 For run B, initial rate = − 0.0010 - 0 0 - 16 = 6.25 x 10–5 mol dm–3 min–1 (Note: On the graph, label coordinates used to obtain the initial rates.) (ii) Comparing run A and run B, when [OH−] × 2, initial rate × 2 ⇒ rate α [OH−] ⇒ reaction is first order with respect to OH–