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 co