RI 2024 Kinetics Tutorial (Ans)

1 Tutorial 6: Reaction Kinetics – Suggested Answers 9 (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  [sucrose], 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 reaction 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 . 1 0 ) 2   kk 1.5 s

2 10 Compare experiments 1 and 2: When [NO 2] x 2, rate x 2  rate  [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 = kሺ0.040ሻሺ0.40ሻm kሺ0.020ሻሺ0.20ሻm = 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 2ൗ = ln 2 k' = ln 2 k[SO2] Since 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. 11 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. 12 (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.  rate  [H+], i.e. reaction is first order with respect to H+. Comparing experiments 1 and 3, When volume of CH 3COCH3 x 0.5 (corresponding to [CH3COCH3] x 0.5, rate  0.5.  rate  [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 ma thematical 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 mechanism 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 fits the observed kinetic data. In mechanism 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).  Mechanism 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) As 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 r eaction 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/rat e-determining step does not involve the breaking of the halogen-halogen bond.) 13 (a) amount of S 2O32 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  ሾI2] t Expt No Time for the appearance of deep blue colour/ s Initial rate  ሾI2] 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  [H2O2]  reaction is first order with respect to H2O2. Comparing experiments 2 and 3, when volume of KI  1.5 (corresponding to [KI] 1.5), initial rate  1.5.  rate  [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 with 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  reaction 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. 6 5 2 1 10005 10521 10105 51015    x x k k Rate Rate b b . . ))()(( ))()((

5 14 (a) Measuring change in colour intensity with time The reactants (i.e. ester and hydroxide i on) are colourless while the product, A, is yellow. Since colour intensity  [A], the change in the colour intensity of A over time may be used to measure the rate of reaction. 1. The reactants are mixed and the colour intensity of the reaction mixture may be measured using a colorimeter and the readings