RI 2024 Kinetics Tutorial - Self Check Ans

10 Suggested Answers for Self-Check 1 (a) 5C 2O42(aq) + 2MnO4(aq) + 16H+(aq)  2Mn2+(aq) + 10CO2(g) + 8H2O(l) Method 1: Measuring change in colour intensity with time MnO4 is purple. Since colour intensity  [MnO 4], the change in the colour intensity of MnO4 over time may be used to follow 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 are taken at regular time intervals. 2. Samples of KMnO 4(aq) of various concentrations (e.g., 0.001, 0.005, 0.100 mol dm 3) are prepared and their colour intensities meas ured using the colorimeter. A calibration curve is obtained by plotting colour intens ity against concentration. The concentration of KMnO4 corresponding to each colour intensity obtained in Step 1 is found using the calibration curve. The concentration of KMnO4 is then plotted against time. 3. Instantaneous rate is found by drawing a tangent to the curve and finding its gradient g1, where rate = g1. (Note: If the graph of colour intensity is plotted against time, then instantaneous rate is proportional to the gradient, g1, of the tangent drawn to the curve at that particular time, where rate  g1.) Method 2: Sampling, quenching and titration 1. The reactants are mixed, and the stopwatch is started simultaneously. 2. Aliquots (same volume) of the reactant mixture are withdrawn and quenched at regular time intervals (e.g., two-minute intervals) using excess K I(aq). I reacts with MnO 4 instantaneously and quantitatively to form I2. 3. The amount of I2 formed in each aliquot is found by titrating against a standard solution of Na2S2O3. 4. Volume of Na 2S2O3(aq)  amount of I2 formed  amount of unreacted MnO4  [MnO4] remaining (since each aliquot has the same volume). Hence a graph of volume Na2S2O3(aq) against time is plotted, which is similar to a graph of [MnO4] against time. 5. Instantaneous rate is proportional to the gradient, g 1, of the tangent drawn to the curve at that particular time, where rate  g1. (Other possible method: Volume of CO2 evolved can be measured over time.) (b) 2H2O2(aq)  2H2O(l) + O2(g) Method 1: Measuring volume of oxygen gas evolved Since a gas is evolved, the change in volume of oxygen evolved over time may be monitored. 1. Set up the apparatus as shown and measure the volume of O2 gas collected in the syringe at regular time intervals. 2. A graph of volume of O 2(g) against time is plotted. 3. Instantaneous rate is proportional to the gradient, g 1, of the tangent drawn to the curve at that particular time, where rate  g1. Method 2: Sampling, quenching and titration 1. Aliquots of the H 2O2 solution are pipetted at suitable time intervals into conical flasks. 2. At a suitable time, ice-cold water is added to the conical flask to slow down or quench the reaction.

11 3. Each quenched aliquot is then titrated against a standard solution of acidified KMnO 4. 4. Volume of KMnO 4  amount of unreacted H2O2  [H2O2] remaining (since each aliquot has the same volume). Hence, a graph of the volume of KMnO4 used against time can be plotted, which is similar to a graph of [H2O2] against time. 5. Instantaneous rate is proportional to the gradient, g 1, of the tangent drawn to the curve at that particular time, where rate  g1. 2 See last page. 3 Ans: C Change in pH cannot be used as the reaction is catalysed by H +, which implies that [H +] (and hence pH) remains unchanged in the reaction  A is incorrect Measuring rate of reaction several times but with a different concentration of ethyl ethanoate each time allows the order of reaction with respect to ethyl ethanoate (but not H+) to be found  B is incorrect Measuring rate of reaction several times but with a different concentration of H 2SO4 each time allows the order of reaction with respect to H+ to be found  C is correct Since the reaction is catalysed by H +, [H+] remains unchanged in the reaction. Hence, the removal of samples at various time intervals and titration against standard aq NaOH would not allow the order of reaction with respect to H+ to be found  D is incorrect 4 Ans: B Time/min Amount of XY2 Amount of X Amount of Y 0 n - - 45 0.5n 0.5n n 90 0.25n 0.5n + 0.25n = 0.75n n + 0.5n = 1.5n 135 0.125n 0.75n + 0.125n = 0.875n 1.5n+0.25n = 1.75n 180 0.0625n 0.875n + 0.0625n = 0.9375n 1.75n+0.125n = 1.875n At 90 min, mole ratio of XY2: Y is 0.25n : 1.5n which is 1 : 6. 5 rate = k[A]2[B] and original rate = R = k[A]02[B]0 (a) (b) (c) (d) change double partial pressures of A & B double partial pressure of A double volume of reacting vessel double total pressure with inert gas new rate 8R 4R ⅛R R reason doubling partial pressure  doubling concentration doubling partial pressure  doubling concentration doubling volume  halving concentration partial pressure of reactants remain constant = no change working new rate = k(2[A]0)2(2[B]0) = 8R new rate = k(2[A]0) 2([B]0) = 4R new rate = k(½[A]0)2(½[B]0) = ⅛R new rate = k([A]0)2([B]0) = R

12 6 rate = – d[I2]/dt = –gradient of graph For each experiment, –d[I2]/dt = constant. Order of reaction with respect to I2 = 0 To determine order of reaction wrt J, we compare rates of reaction when [ J] = a and [J] = 2a When [J] =a, rate = y/10 When [J] = 2a, rate = 2y/5 So, when [J]  2, rate  ቆ 2y 5 y 10 = 4ቇ  rate  [J]2 Hence order of reaction with respect to J = 2. 7 Ans: D When only [H2O2]  2, initial rate  2  Rxn is first order wrt H2O2. When [H2O2]  1.5 and [I−]  2, initial rate  3  Rxn is first order wrt I−. When [H2O2]  3, and both [I−] and [H+]  2, initial rate  6  Rxn is zero order wrt H+. Thus, rate = k[H2O2][I−], where units of k are mol–1 dm3 min–1  A is incorrect H+ is a reactant that is converted to H2O (i.e., H+ does not remain unchanged at the end of reaction) so it cannot be a catalyst. Furthermore, we have insufficient information as to the effect of H+ on the reaction (e.g. experiment with no H +). Thus, from the results obtained, we cannot conclude that H+ is a catalyst.  B is incorrect. The magnitude of the rate constant increases when a catalyst is used or when temperature is increased  C is incorrect When both [H2O2] and [I−]  3, the initial rate  9. Changing [H+] does not affect the initial rate  D is correct 8 Ans: C Note : For a mechanism to be valid, 1. the mechanism must be consistent with the rate law, and 2. the elementary steps must add up to the overall balanced equation. A Incorrect. rate = k[NO 2] Overall eqn: CO + NO2  CO2 + NO B Incorrect. rate = k[NO2]2 Overall eqn: CO + 2NO2  CO2 + 2NO + O C Correct. rate = k[NO2]2 Overall eqn: CO + NO2  CO2 + NO D Incorrect. rate = k[NO2][CO] Overall eqn: CO + NO2  CO2 + NO

13 2 Overall order of reaction Rate equation Common units of k Half-life Rate-[reactant] Graph [reactant]-Time Graph Other Graphs zero rate = k mol dm −3 s−1 t1/2 is not constant one rate = k [A] s −1 t1/2 = ln 2 k t1/2 is constant two rate = k [A] 2 mol−1 dm3 s−1 t1/2 is not constant 0 [A] rate k 0 Co Co [A]/mol dm time/s t12 t 3t -3 1 2 1 4 Co 1 8 Co t1 = t2 = t3 i.e. t is constant1 2 time/s -3 t1 = t2 = t3 i.e. t is constant1 2 0 Co [product]/mol dm t12 t 3t Co1 2 Co3 4 Co7 8 0 [A] rate r 4r a 2a 0 t1/2 1/2 t 1 2 t is not constant Co1 4 1 2 -3 time/s [reactant]/mol dm Co Co rate [A]20 gradient = k g r a d i en t = k 0 [ p r o d uc t ] t i m e t1/2 is not constant C0 ¾ C0 ½ C0