3.1.11 RATE EQUATIONS

EQ: In another experiment, the effect of temperature on the rate of a reaction is investigated. Table 1 shows the results. (a) Complete Table 1. (2 MARKS) (b) The Arrhenius equation can be written in the form ln k = -Ea/RT + lnA Use the data in Table 1 to plot a graph of ln k against 1/T on the grid in Figure 2. Calculate the activation energy, Ea, in kJ mol−1 The gas constant, R = 8.31 J K−1 mol−1. (5 MARKS)

(a) - 0.00314 6 - −14.2 (b) - Vertical axis with sensible scales (plotted points must take up more than half the grid) NOT M1 if y-axis in wrong direction - all points plotted correctly (within ±0.5 small square) - Best fit straight line based on the student's data (ignoring anomalous point if relevant) - Gradient = −13 125 (Gradient calculated within range: 12876 - 13598) - -13125 = -Ea/R - Ea = 13 125 x 8.31 = 109 069 = 109 (kJ mol−1 )

EQ: A second series of experiments was carried out to investigate how the rate of the reaction varies with temperature. The results were used to obtain a value for the activation energy of the reaction, Ea Identical amounts of reagents were mixed at different temperatures. The time taken, t, for a fixed amount of bromine to be formed was measured at different temperatures. The results are shown in Table 3. (a) Complete Table 3. (2 MARKS) (b) The Arrhenius equation can be written as lnk = -(Ea/R)(1/T) + C1 In this experiment, the rate constant, k, is directly proportional to 1/t Therefore ln(1/t) = -(Ea/R)(1/T) + C2 where C1 and C2 are constants. Use values from Table 3 to plot a graph of ln(1/t) (y axis) against 1/T on the grid. Use your graph to calculate a value for the activation energy, in kJ mol-1 , for this reaction. The value of the gas constant, R = 8.31 J K-1 mol-1 (6 MARKS)

(a) - 1/T value 3.31(1) × 10^-3 or 0.00331(1) - ln(1/t) value -3.30 or -3.297 (b) - y axis labelled with values (no units) and plotted points use over half of the axis - points plotted correctly - best fit straight line (minimum 3 points plotted) - gradient = 6.64 × 103 (K) or -6640 (K) - Ea = gradient × 8.31 - = 55.2 kJ mol1

Q: Iodine reacts slowly with propanone in the presence of an acid catalyst according to the equation CH3COCH3 + l2 ⟶ CH3COCH2l + Hl The rate of this reaction can be followed by preparing mixtures in which only the initial concentration of propanone is varied. At suitable time intervals, a small sample of the mixture is removed and titrated with sodium thiosulfate solution. This allows determination of the concentration of iodine remaining at that time.The rate of this reaction can be followed by preparing mixtures in which only the initial concentration of propanone is varied. At suitable time intervals, a small sample of the mixture is removed and titrated with sodium thiosulfate solution. This allows determination of the concentration of iodine remaining at that time. Five mixtures, A, B, C, D and E, are prepared as shown in Table 1. (a) Calculate the initial concentration, in mol dm−3, of the propanone in mixture A. (2 MARKS) b) State and explain why different volumes of water are added to mixtures B, C, D and E. (2 MARKS) c) Calculate the volume of 0.0100 mol dm−3 sodium thiosulfate solution required to react with all of the iodine in a 10.0 cm3 sample of mixture E, before the iodine reacts with propanone. The equation for the reaction in the titration is 2Na2S2O3(aq) + I2(aq) ⟶ Na2S4O6(aq) + 2Nal(aq). (4 MARKS0 (d) The results for mixture E are shown in Table 2. V is the volume of 0.0100 mol dm−3 sodium thiosulfate solution needed, at different times, t, to react with the iodine in a 10.0 cm3 sample of E. Use these data and your answer to part (c) to plot a graph of V (y-axis) against t (x-axis) for mixture E. Draw a best-fit straight line through your points and calculate the gradient of this line. (e) The gradients for similar graphs produced by mixtures A, B, C and D are shownin Table 3. Each gradient is a measure of the rate of the reaction between iodine and propanone. Use information from Table 1 and Table 3 to deduce the order with respect to propanone. Explain your answer. (2 MARKS) (f) Each sample taken from the reaction mixtures is immediately added to an excess of sodium hydrogencarbonate solution before being titrated with sodium thiosulfate solution. Suggest the purpose of this addition. Explain your answer. (2 MARKS)

(a) - Amount of propanone = (25.0 × 1) / 1000 = 0.025 mol - Concentration in mixture A = 0.025 / (90.0/1000) = 0.278 mol dm−3 (b) - To make volumes constant for all mixtures. - So that volume of propanone is proportional to concentration. (c) - Amount of iodine in mixture E = (40.0 × 0.02) / 1000 = 8.0 × 10−4 mol - Amount of iodine in sample = 8.0 × 10−4 × (10 / 90) = 8.89 × 10−5 mol - Amount of thio required = 2 × 8.89 × 10−5 = 1.78 × 10−4 mol - Volume of thio = (1.78 × 10−4 / 0.01) × 1000 = 17.8 cm3 (d) - Scale Graph: must cover at least half the grid and axes must be plotted in correct orientation but ignore labeling. - Points: Must be correctly plotted to within ± half a small square. - Best-fit straight line: Must be the best-fit line possible if point(s) are plotted incorrectly. - Gradient = y / x = −0.060 (e) - Gradients / rates are proportional to volumes / concentrations of propanone or B to D show that gradient / rate halves when vol / concentration halves. - So first order with regard to propanone. (f) - To stop / quench the reaction at that time. - By removing the acid catalyst for the reaction (by neutralisation).

Q: The mechanisms of some reactions can be deduced from kinetic data. The table shows the results of five experiments involving the reaction between benzaldehyde (C6H5CHO) and potassium hydroxide solution. (a) Calculate the value of [C6H5CHO]2 for each experiment. Write your answers in the table above. Plot a graph of initial rate against [C6H5CHO]^2 on the grid below. Label the axes with units. (4 MARKS) (b) Deduce the order of the reaction with respect to C6H5CHO from the graph. Explain your answer. (2 MARKS)

(a) - Calculates value of [C6H5CHO]^2: - he graph labels with units and appropriate scales and using sensible proportion of graph (plotted points must cover at least half the printed grid). - for the plotting of 5 points. - the line of best fit. (b) - 2nd order - (since) [C6H5CHO]^2 plotted against rate is straight line / directly proportional.

Q: The initial rate of the reaction between compounds A and B was measured in a series of experiments at a fixed temperature. The following rate equation was deduced. rate = k[A][B]^2 (a) Complete the table of data below for the reaction between A and B. (3 MARKS) (b) Using the data for experiment 1, calculate a value for the rate constant, k, and state its units. (3 MARKS) (c) State how the value of the rate constant, k, would change, if at all, if the concentration of A were increased in a series of experiments. (1 MARK)

(a) - Experiment 2 2.60 × 10-3 - Experiment 3 0.60 × 10-2 - Experiment 4 11.4 × 10-2 (b) - k = (10.4 x 10^-3) / [4.8 x 20^-2][6.60 x 10^-2]^2 - 49.7 - mol-2 dm6 s-1 (c) No change

Q: The initial rate of the reaction between substances P and Q was measured in a series of experiments and the following rate equation was deduced. rate = k[P]^2 [Q] (a) Complete the table of data below for the reaction between P and Q. (2 MARKS) (b) Using the data from experiment 1, calculate a value for the rate constant, k, and deduce its units. (6 MARKS) (c) What change in the reaction conditions would cause the value of the rate constant to change? (1 MARK)

(a) - Experiment 2: 0.4(0) × 10-3 - Experiment 3: 0.15 - Experiment 4: 0.28 (b) k = 4.8 x 10^-3 / (0.20)^2 x (0.30) = 0.40 mol-2 . dm6 . s-1 (c) (change in) temperature

EQ: Substances P and Q react in solution at a constant temperature. The initial rate of reaction was studied in three experiments by measuring the change in concentration of P over the first five seconds of the reaction. The data obtained are shown in Table 1. (a) Complete Table 2 to show the initial rate of reaction of P in each experiment (1 MARK) (b) Determine the order of reaction with respect to P and the order of reaction with respect to Q. (2 MARKS) (c) A reaction between substances R and S was second order with respect to R and second order with respect to S. At a given temperature, the initial rate of reaction was 1.20 × 10-3 mol dm-3 s-1 when the initial concentration of R was 1.00 × 10-2 mol dm-3 and the initial concentration of S was 2.45 × 10-2 mol dm-3 Calculate a value for the rate constant, k, for the reaction at this temperature. Give the units for k. (3 MARKS)

(a) - Expt 2: 3.2 × 10^-4 - Expt 3: 3.2 × 10^-4 (b) - P order = 1 - Q order = 2 (c) - ( Rate = k[R]^2 [S]^2 ) - k = Rate/[R]^2 [S]^2 - k = 19992 = 2.00 × 10^4 - Units mol-3 dm9 s-1

EQ: Iodide ions are oxidised to iodine by hydrogen peroxide in acidic conditions. H2O2(aq) + 2H+ (aq) + 2I- (aq) → I2(aq) + 2H2O(l) The rate equation for this reaction can be written as rate = k [H2O2]^a [I- ]^b [H+ ]^c In an experiment to determine the order with respect to H+ (aq), a reaction mixture is made containing H+ (aq) with a concentration of 0.500 mol dm-3 A large excess of both H2O2 and I- is used in this reaction mixture so that the rate equation can be simplified to rate = k1[H+ ]^c (a) Explain why the use of a large excess of H2O2 and I- means that the rate of reaction at a fixed temperature depends only on the concentration of H+ (aq). (2 MARKS) (b) Samples of the reaction mixture are removed at timed intervals and titrated with alkali to determine the concentration of H+ (aq). State and explain what must be done to each sample before it is titrated with alkali. (2 MARKS)

(a) - H2O2 and/or I - concentration change is negligible / H2O2 and/or I - concentration (effectively) constant - so have a constant/no effect on the rate / so is zero order (w.r.t. H2O2 and I- ) / a and b are zero (b) - Stop the reaction / quench - By dilution / cooling / adding a reagent to react with H2O2/I-

EQ: Hydrogen peroxide is a powerful oxidising agent. Acidified hydrogen peroxide reacts with iodide ions to form iodine according to the following equation. H2O2(aq) + 2H+ (aq) + 2I− (aq) → I2(aq) + 2H2O(l) The initial rate of this reaction is investigated by measuring the time taken to produce sufficient iodine to give a blue colour with starch solution. A series of experiments was carried out, in which the concentration of iodide ions was varied, while keeping the concentrations of all of the other reagents the same. In each experiment the time taken (t) for the reaction mixture to turn blue was recorded. The initial rate of the reaction can be represented as ( 1/t), and the initial concentration of iodide ions can be represented by the volume of potassium iodide solution used. A graph of log10 (1/t) on the y-axis against log10 (volume of KI(aq)) is a straight line. The gradient of this straight line is equal to the order of the reaction with respect to iodide ions. The results obtained are given in the table below. The time taken for each mixture to turn blue was recorded on a stopclock graduated in seconds. (a) Use the results given in the table to plot a graph of log10 (1/t) on the y-axis against log10 (volume of KI(aq)). Draw a straight line of best fit on the graph, ignoring any anomalous points. (b) Determine the gradient of the line you have drawn. Give your answer to two decimal places. Show your working. (3 MARKS) (c) Deduce the order of reaction with respect to iodide ions. (1 MARK) (d) A student carried out the experiment using a flask on the laboratory bench. The student recorded the time taken for the reaction mixture to turn blue. State one way this method could be improved, other than by repeating the experiment or by improving the precision of time or volume measurements. Explain why the accuracy of the experiment would be improved. (2 MARKS)

(a) - Log (1 / time) on the y-axis + log (vol) on x-axis - Sensible scales - Plots points correctly ± one square - Line through the points is smooth - Line through the points is best fit - ignores last point (b) - Uses appropriate x and y readings - Correctly calculates gradient 0.95 ± 0.02 (c) First order or order is 1 (d) - Thermostat the mixture / constant temperature / use a water bath or Colorimeter / uv-visible spectrometer / light sensor to monitor colour change - Reaction / rate affected by temperature change or Eliminates human error in timing / more accurate time of colour change

Q: The initial rate of the reaction between gases D and E was measured in a series of experiments at a constant temperature. The results are shown in the table. (a) Deduce the order of reaction with respect to D and the order with respect to E. (2 MARKS) (b) Suggest why initial rates of reaction are used to determine these orders rather than rates of reaction at other times during the experiments. (1 MARK) (c) State how the initial rate is obtained from a graph of the concentration of the product against time. (2 MARKS)

(a) - Order wrt D = 1 OR first OR [D] OR [D]1 - Order wrt E = 2 OR second OR [E]2 (b) - (At time zero/start) the concentrations are known (c) - (Calculate) gradient (of tangent/curve/graph) - at t=0 or at start of graph/curve

EQ: This question is about rates of reaction. Iodine and propanone react together in an acid-catalysed reaction CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + HI(aq) A student completed a series of experiments to determine the order of reaction with respect to iodine. Method • Transfer 25 cm3 of 1.0 mol dm-3 propanone solution into a conical flask. • Add 10 cm3 of 1.0 mol dm-3 HCl(aq) • Add 25 cm3 of 5.0 × 10-3 mol dm-3 I2(aq) and start a timer. • At intervals of 1 minute, remove a 1.0 cm3 sample of the mixture and add each sample to a separate beaker containing an excess of NaHCO3(aq) • Titrate the contents of each beaker with a standard solution of sodium thiosulfate and record the volume of sodium thiosulfate used. (a) Suggest why the 1.0 cm3 portions of the reaction mixture are added to an excess of NaHCO3 solution. (2 MARKS) (b) Suggest why the order of this reaction with respect to propanone can be ignored in this experiment. (2 MARKS)

(a) - The sodium hydrogencarbonate solution neutralises the acid (catalyst) - So stops the reaction (b) - The concentration/amount of propanone is much larger than/200 times larger than the concentration/amount of iodine - Concentration of propanone is (almost) constant

EQ: Iodide ions are oxidised to iodine by hydrogen peroxide in acidic conditions. H2O2(aq) + 2H+ (aq) + 2I- (aq) → I2(aq) + 2H2O(l) The rate equation for this reaction can be written as rate = k [H2O2]^a [I- ]^b [H+ ]^c In an experiment to determine the order with respect to H+ (aq), a reaction mixture is made containing H+ (aq) with a concentration of 0.500 mol dm-3 A large excess of both H2O2 and I- is used in this reaction mixture so that the rate equation can be simplified to rate = k1[H+ ]^c A graph of the results is shown in Figure 1. (a) Explain how the graph shows that the order with respect to H+ (aq) is zero. (2 MARKS) (b) Use the graph in Figure 1 to calculate the value of k1 Give the units of k1. (3 MARKS)

(a) - constant gradient OR change/decrease in concentration is proportional to time - as [H+ ] changes/decreases (b) - evidence of attempt at calculation of gradient via ∆y/∆x - k1 = 0.0012 / 1.2 x 10-3 - units = mol dm-3 s-1

Q: Iodine and propanone react in acid solution according to the equation I2 + CH3COCH3 → CH3COCH2I + HI The rate equation for the reaction is found to be rate = k [CH3COCH3][H+ ] (a) At the start of the experiment, the rate of reaction was found to be 2.00 × 10-5 mol dm-3 s-1 when the concentrations of the reactants were as shown below. Use these data to calculate a value for the rate constant and deduce its units. (3 MARKS) (b) How can you tell that H+ acts as a catalyst in this reaction? (2 MARKS) (c) Calculate the initial rate of reaction if the experiment were to be repeated at the same temperature and with the same concentrations of iodine and propanone as in part (a) but at a pH of 1.25. (2 MARKS)

(a) - k = (2 x 10^-5) / [q.5][3 x 10^-2] - Units: mol-1 dm3 s-1 (b) - Appears in rate equation - does not appear in (stoichiometric / overall) equation (c) - pH = -log10 [H+ ] = 1.25 - [H+ ] = 0.056(2) - rate = (4.44 × 10-4) × (1.50) × (0.0562) - = 3.75 × 10-5 - (3.7 — 3.8)

Q: The reaction between propanone and iodine in the presence of hydrochloric acid was studied at a constant temperature. CH3COCH3 + I2 → CH3COCH2I + HI The following rate equation was deduced. rate = k [CH3COCH3][H+] In an experiment the initial concentrations of propanone, iodine and hydrochloric acid were as shown in the table. The initial rate of reaction in this experiment was 8.64 × 10-7 mol dm-3 s-1. (a) Use the data in the table and the rate equation to calculate a value for the rate constant at this temperature. Give units with your answer. (2 MARKS) (b) A series of experiments was carried out using concentrations of propanone approximately 100 times the concentrations of iodine and hydrochloric acid. Suggest the rate equation under these conditions. Explain your answer. (2 MARKS)

(a) - k = (8.64 x 10^-7) / [5.82 x 10^-2][4.76 x 10^-1] - mol-1 dm+3 s-1 (b) - Rate = k [H+] - (Large excess of propanone) so [CH3COCH3] is (effectively) constant

EQ: A series of experiments is carried out with compounds C and D. Using the data obtained, the rate equation for the reaction between the two compounds is deduced to be rate = k [C][D] In one experiment at 25 °C, the initial rate of reaction is 3.1 × 10−3 mol dm−3 s−1 when the initial concentration of C is 0.48 mol dm−3 and the initial concentration of D is 0.23 mol dm−3 (a) Calculate a value for the rate constant at this temperature and give its units. (3 MARKS) (b) An equation that relates the rate constant, k, to the activation energy, Ea, and the temperature, T, is lnk = -Ea/RT + lnA Use this equation and your answer from Question (a) to calculate a value, in kJ mol−1 , for the activation energy of this reaction at 25 °C. For this reaction lnA = 16.9 The gas constant R = 8.31 J K−1 mol−1. (4 MARKS)

(a) - k = (rate/[C][D]) - 2.8 × 10-2 min 2sfs - mol-1 dm3 s-1 (b) - ln k = ln 2.8 × 10-2 (= - 3.58) - Ea = RT(ln A - ln k) - OR Ea = RT(ln k - ln A) - Ea = 8.31 × 298 (16.9 + 3.58) ( = 50716 J mol-1 ) - Ea = 51 kJ mol-1

EQ: Bromate(V) ions and bromide ions react in acid conditions according to the equation BrO3 − (aq) + 5Br − (aq) + 6H+ (aq) → 3Br2(aq) + 3H2O(l) A series of experiments was carried out at a given temperature. The results were used to deduce the rate equation for the reaction. rate = k [BrO3 − ][Br − ][H+ ] ^2 Table 2 shows an incomplete set of results. (a) Use the data from Experiment 1 to calculate a value for the rate constant, k, at this temperature and give its units. Give your answer to an appropriate number of significant figures (3 MARKS) (b) Complete Table 2. (3 MARKS)

(a) - k = 2.4 x 10^-2 / (0.1)(0.2)(0.3)^2 - 13 (must be 2 sfs) - Units mol-3.dm9.s-1 (b) - Experiment 2 [BrO3 ] = 0.15 - Experiment 3 rate = 0.26 or 0.27 - Experiment 4 [H+ ] = 0.45 or 0.46

EQ: The following table shows the results of three experiments carried out at the same temperature to investigate the rate of the reaction between compounds P and Q. (a) Use the data in the table to deduce the order with respect to P and the order with respect to Q. (2 MARKS) (b) In a reaction between R and S, the order of reaction with respect to R is one, the order of reaction with respect to S is two and the rate constant at temperature T1 has a value of 4.2 × 10-4 mol-2 dm6 s-1 . (i) Write a rate equation for the reaction. Calculate a value for the initial rate of reaction when the initial concentration of R is 0.16 mol dm-3 and that of S is 0.84 mol dm-3 . (ii) In a second experiment performed at a different temperature, T2, the initial rate of reaction is 8.1 × 10-5 mol dm-3s -1 when the initial concentration of R is 0.76 mol dm-3 and that of S is 0.98 mol dm-3. Calculate the value of the rate constant at temperature T2. (iii) Deduce which of T1 and T2 is the higher temperature. (6 MARKS)

(a) - order with respect to P is 2 - order with respect to Q is 1 (b)(i) - rate = k[R][S]^2 - rate = (4.2 × 10-4) × 0.16 × 0.84^2 - = 4.7 × 10-5 (mol dm-3 s-1) (ii) - k = rate / [R][S]^2 - 1.1 x 10^-4 (iii) T1

Q: Compound A, HCOOCH2CH2CH3, is an ester. (a) Name this ester and write an equation for its reaction with aqueous sodium hydroxide. (2 MARKS) (b) The initial rate of reaction between ester A and aqueous sodium hydroxide was measured in a series of experiments at a constant temperature. The data obtained are shown below. Use the data in the table to deduce the order of reaction with respect to A and the order of reaction with respect to NaOH. Hence calculate the initial rate of reaction in Experiment 4. (3 MARKS) (c) ) In a further experiment at a different temperature, the initial rate of reaction was found to be 9.0 × 10-3 mol dm-3 s -1 when the initial concentration of A was 0.020 mol dm-3 and the initial concentration of NaOH was 2.00 mol dm-3 . Under these new conditions with the much higher concentration of sodium hydroxide, the reaction is first order with respect to A and appears to be zero order with respect to sodium hydroxide. (i) Write a rate equation for the reaction under these new conditions. (ii) Calculate a value for the rate constant under these new conditions and state its units. (iii) Suggest why the order of reaction with respect to sodium hydroxide appears to be zero under these new conditions. (6 MARKS)

(a) - propyl methanoate; - HCOOC3H7 + OH- → HCOO- + C3H7OH 1 - OR HCOOC3H7 + NaOH → HCOONa + C3H7OH; (b) - order wrt A = 1; - order wrt NaOH = 1; - Initial rate in Exp 4 = 2.4 × 10-3; (c) (i) r(ate) = k[A] OR r(ate) = k[A][NaOH]0 ; (ii) k = (9.0 x 10^-3)/0..02 = 0.45 s-1 (iii) - (large) excess of OH- or [OH- ] is large/high; - [OH- ] is (effectively) constant OR [A] is the limiting factor

EQ: This question involves the use of kinetic data to deduce the order of a reaction and calculate a value for a rate constant. The data in Table 2 were obtained in two experiments on the rate of the reaction between compounds C and D at a constant temperature. The rate equation for this reaction is rate = k[C]^2 [D] (a) Use the data from experiment 4 to calculate a value for the rate constant, k, at this temperature. Deduce the units of k. (3 MARKS) (b) Calculate a value for the initial rate in experiment 5. (1 MARK)

(a) - rate = k [C]^2 [D] therefore k = rate / [C]^2 [D] - k = (7.2 x 10^-4) / [(1.9 x 10^-2) x (3.5 x 10^2)] - = 57.0 - mol-2.dm6.s-1 (b) rate = 57.0 × (3.6 × 10-2 ) 2 × 5.4 × 10^-2 = 3.99 × 10^-3 (mol dm-3 s -1 )

EQ: Butadiene dimerises according to the equation 2C4H6 --> C8H12 The kinetics of the dimerisation are studied and the graph of the concentration of a sample of butadiene is plotted against time. The graph is shown in Figure 1. (a) Draw a tangent to the curve when the concentration of butadiene is 0.0120 mol dm-3 . (1 MARK) (b) The initial rate of reaction in this experiment has the value 4.57 × 10-6 mol dm-3 s-1 . Use this value, together with a rate obtained from your tangent, to justify that the order of the reaction is 2 with respect to butadiene. (5 MARKS)

(a) Gradient drawn on graph (b) Stage 1: Rate of reaction when concentration = 0.0120 mol dm-3 - From the tangent - Change in [butadiene] = -0.0160 - 0 and change in time = 7800 - 0 - Gradient = -(0.0160 - 0)/(7800 - 0) = -2.05 × 10-6 - Rate = 2.05 × 10-6 (mol dm-3 s-1 ) Stage 2: Comparison of rates and concentrations - Initial rate/rate at 0.0120 = (4.57 × 10-6 )/(2.05 × 10-6 ) = 2.23 - Inital concentration/concentration at point where tangent drawn = 0.018/0.012 = 1.5 Stage 3: Deduction of order - If order is 2, rate should increase by factor of (1.5)2 = 2.25 this is approximately equal to 2.23 therefore order is 2nd with respect to butadiene

EQ: The rate of the reaction between substance A and substance B was studied in a series of experiments carried out at the same temperature. In each experiment the initial rate was measured using different concentrations of A and B. These results were used to deduce the order of reaction with respect to A and the order of reaction with respect to B. (a) What is meant by the term order of reaction with respect to A? (1 MARK) (b) When the concentrations of A and B were both doubled, the initial rate increased by a factor of 4. Deduce the overall order of the reaction. (1 MARK) (c) In another experiment, the concentration of A was increased by a factor of three and the concentration of B was halved. This caused the initial rate to increase by a factor of nine. (i) Deduce the order of reaction with respect to A and the order with respect to B. (ii) Using your answers from part (c)(i), write a rate equation for the reaction and suggest suitable units for the rate constant. (4 MARKS)

(a) Power (or index or shown as x in [ ]x ) to which the concentration of a substance is raised (b) 2 (c) (i) - Order with respect to A: 2 - Order with respect to B: 0 (ii) - Rate equation: (rate =) k [A]^2 - Units for rate constant: mol-1 dm3 s-1

Example 2: 2NO + O2 → 2NO2 Step 1: NO + NO → N2O2 = slow Step 2: N2O2 + O2 → 2NO2 = fast a) Which is the rate-determining step? b) What is the rate equation?

(a) Step 1 because it is slow (b) Rate = K [NO]2 because there are 2 molecules involved in the rate-determining step

EQ: The rate equation for the reaction between compounds A and B is rate = k [A]^2 [B] Figure 2 shows how, in an experiment, the concentration of A changes with time, t, in this reaction. (a) Draw a tangent to the curve at t = 0 (1 MARK) (b) Use this tangent to deduce the initial rate of the reaction. (1 MARK) (c) The experiment was repeated at the same temperature and with the same initial concentration of B but with a different initial concentration of A. The new initial rate was 1.7 times greater than in the original experiment. Calculate the new initial concentration of A. (2 MARK)

(a) Straight line through (0.00, 0.50) which cuts time axis at between 5 and 12.5 secs OR conc 0.3 at time between 2s and 5s (b) Mark is for correct calculation of their gradient : e.g. 0.50/11 = 0.045 or 4.5 × 10-2 (mol dm-3 s-1 ) (c) - [A] increases by √1.7 = new[A] = 1.30 × 0.50 = 0.65 (mol dm-3 ) 2 sfs min

Q: Propanone and iodine react in acidic conditions according to the following equation. CH3COCH3 + I2 → ICH2COCH3 + HI A student studied the kinetics of this reaction using hydrochloric acid and a solution containing propanone and iodine. From the results the following rate equation was deduced. rate = k[CH3COCH3][H+ ] (a) When the initial concentrations of the reactants were as shown in the table below, the initial rate of reaction was found to be 1.24 × 10-4 mol dm-3 s-1 . Use these data to calculate a value for the rate constant, k, for the reaction and give its units. (3 MARKS) (b) Deduce how the initial rate of reaction changes when the concentration of iodine is doubled but the concentrations of propanone and of hydrochloric acid are unchanged. (1 MARK) (c) The following mechanism for the overall reaction has been proposed. Use the rate equation to suggest which of the four steps could be the rate-determining step. Explain your answer. (2 MARKS) (d) Use your understanding of reaction mechanisms to predict a mechanism for Step 2 by adding one or more curly arrows as necessary to the structure of the carbocation below. (1 MARK)

(a) k = (1.24 x 10^-4) / [4.4][0.82] - 3.44 × 10^-5 (min 3sfs) - mol-1 dm3 s-1 (b) no change or no effect or stays the same or 1.24 × 10^-4 (c) - 1 or 2 or 1 and 2 - rate equ doesn't involve I2 or only step which includes 2 species in rate equ (d) H being removed by being joined to C.

Q: The rate of hydrolysis of an ester X (HCOOCH2CH2CH3) was studied in alkaline conditions at a given temperature. The rate was found to be first order with respect to the ester and first order with respect to hydroxide ions. (a) Name ester X. (1 MARK) (b) Using X to represent the ester, write a rate equation for this hydrolysis reaction. (1 MARK) (c) When the initial concentration of X was 0.024 mol dm-3 and the initial concentration of hydroxide ions was 0.035 mol dm-3, the initial rate of the reaction was 8.5 × 10-5 mol dm-3 s-1 . Calculate a value for the rate constant at this temperature and give its units. (3 MARKS) (d) In a second experiment at the same temperature, water was added to the original reaction mixture so that the total volume was doubled. Calculate the initial rate of reaction in this second experiment. (1 MARK) (e) In a third experiment at the same temperature, the concentration of X was half that used in the experiment in part (a) (iii) and the concentration of hydroxide ions was three times the original value. Calculate the initial rate of reaction in this third experiment. (1 MARK) (f) State the effect, if any, on the value of the rate constant k when the temperature is lowered but all other conditions are kept constant. Explain your answer. (2 MARKS)

(a) propyl methanoate (b) rate = k[X][OH- ] (c) - k = 8.5 x 10^-6 / (0.024)(0.035) - = 0.10(12) 2sf minimum - mol-1 dm3 s-1 (d) 2.1(3) × 10^-5 (e) 1.3 ×10^-4 (1.28 ×10^-4) (f) - Lowered - fewer particles/collisions have energy > Ea OR fewer have sufficient (activation) energy (to react)

EQ: This question is about rates of reaction. Iodine and propanone react together in an acid-catalysed reaction CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + HI(aq) A student completed a series of experiments to determine the order of reaction with respect to iodine. Method • Transfer 25 cm3 of 1.0 mol dm-3 propanone solution into a conical flask. • Add 10 cm3 of 1.0 mol dm-3 HCl(aq) • Add 25 cm3 of 5.0 × 10-3 mol dm-3 I2(aq) and start a timer. • At intervals of 1 minute, remove a 1.0 cm3 sample of the mixture and add each sample to a separate beaker containing an excess of NaHCO3(aq) • Titrate the contents of each beaker with a standard solution of sodium thiosulfate and record the volume of sodium thiosulfate used. The volume of sodium thiosulfate solution used in each titration is proportional to the concentration of iodine in each beaker. The table below shows the results of the experiment. (a) Use the results in the table above to draw a graph of volume of sodium thiosulfate solution against time. Draw a line of best fit. (3 MARKS) (b) Explain how the graph shows that the reaction is zero-order with respect to iodine in the reaction between propanone and iodine. (2 MARKS)

(c) - Suitable axes (plotted points must take up at least half of the grid) - For all points correctly plotted to ± 1 /2 small square - For straight line of best fit which avoids the anomalous plot M1 (d) - The graph is a straight line / has a constant gradient - So the rate of reaction does not change as the concentration (of iodine) changes / the iodine is being used up at a constant rate.

EQ: This question is about rate equations. The initial rate of reaction between compounds P and Q is measured in a series of experiments at constant temperature. The rate equation deduced is Rate: k[P][Q]^2 Table 1 shows some rate data. (i) Complete table 1. (3 MARKS) (ii) Sketch a graph to show how k changes as [P] is increased. (1 MARK)

(i) - 5.25 x 10-4 - 7.20 x 10-2 - 6.60 x 10-2 (ii) Look at image.

EQ: This question is about reaction kinetics. Substances A and B react together in acidic conditions. The overall equation can be represented by 2A + B -› C + D The rate equation for the reaction is Rate = K[A][H*] (i) Suggest the role of H* ions in this reaction. Use the overall equation and the rate equation to explain your answer. (2 MARKS) (ii) Give one piece of evidence that the overall reaction between A and B must have more than one step. (1 MARK)

(i) - catalyst - not present in the overall equation but present in the rate equation (ii) the rate is independent of B Or rate equation only involves A or [H*] is in the rate equation but not in the overall equation

EQ: The compound (CH3)3CBr reacts with aqueous sodium hydroxide as shown in the following equation. (CH3)3CBr + OH- → (CH3)3COH + Br- This reaction was found to be first order with respect to (CH3)3CBr but zero order with respect to hydroxide ions. The following two-step process was suggested. Step 1: (CH3)3CBr → (CH3)3C+ + Br- Step 2: (CH3)3C+ + OH- → (CH3)3COH (i) Explain how the rate data helps to support the suggested mechanism. (1 MARK) (ii) Suggest how the rate of the reaction between (CH3)3CCl would compare with the rate of the reaction in step 1. Explain your answer. (2 MARKS)

(i) Hydroxide ions must react in a step after the rate determining step (ii) - Slower - C-Cl bond is stronger

How do you prove orders of reaction using conc vs time graphs?

- Complete 2 sets of gradients for 2 different concentrations - In the example to the right, order = 1st order.

EQ: A slow reaction has a rate constant k = 6.51 × 10-3 mol-1 dm3 at 300 K. Use the equation ln k = ln A - Ea/RT to calculate a value, in kJ mol-1 , for the activation energy of this reaction. The constant A = 2.57 × 1010 mol-1 dm3 . The gas constant R = 8.31 J K-1 mol-1 . (2 MARKS)

- Ea = RT(lnA - lnk)/1000 - Ea = 8.31 × 300 (23.97 - (-5.03))/1000 = 72.3 (kJ mol-1 )

REQUIRED PRACTICAL 8: Measuring rate by gas collection (continuous rate method) --> sources of error

- Gas escaping through gaps in apparatus - Syringe sticking - Gas escaping before bung replaced - Some gases are water soluble → not all gas reaches the syringe

EQ: The Arrhenius equation can be written as lnk = -Ea/RT + lnA The figure below shows a graph of ln k against 1/T for the reaction 2 HI(g) → H2(g) + I2(g) Use the figure above to calculate a value for the activation energy (Ea), in kJ mol−1 , for this reaction. The gas constant R = 8.31 J K−1 mol−1. (3 MARKS)

- Gradient = (−14.1 − −2.8) / (0.00180 − 0.00128) = −11.3 / 0.00052 = −21731 - Gradient = −Ea / R - −Ea = their answer x 8.31 ( = 180583 J mol-1 ) - Ea = M2 ÷ 1000 (= 181 kJ mol-1 )

What is the Arrhenius equation for rate? State the units for each.

- K = Rate constant - A = Arrhenius constant (Pre-exponential factor) - e = Exponent (inverse of ln) - Ea = Activation energy (J.mol-1) - R = Gas constant (8.314 J.mol-1.K-1 - T = Temperature (K) R and T give the average kinetic energy. The -Ea/RT bit... - The ratio between Ea and average kinetic energy - The lower Ea, and the higher the kinetic energy, the faster the reaction - Ie, smaller ratio = faster reaction, hence - sign - Essentially e-Ea/RT calculates the fraction of molecules that possess enough energy to react A: Accounts for the fraction of molecules that would react if Ea = 0 (unlikely) You can also take ln of both sides to remove the exponent: lnK = -Ea/RT + LnA

REQUIRED PRACTICAL 8: Measuring rate by loss in Mass (Continous Rate) --> sources of error

- Liquid escaping - Balance errors - Water soluble gases remain in the flask

What are the units for rate.

- Mol.dm-3 - Mol.dm-3.min-1

EQ: The rate equation for a reaction is rate = k[E] Explain qualitatively why doubling the temperature has a much greater effect on the rate of the reaction than doubling the concentration of E. (3 MARKS)

- Reaction occurs when molecules have E≥Ea - Doubling T causes many more molecules to have this E - Whereas doubling [E] only doubles the number with this E

Q: Gases A and B react as shown in the following equation. 2A(g)+ B(9) --> C(g) + D(g) The initial rate of the reaction was measured in a series of experiments at a constant temperature. The following rate equation was determined. rate = k[A]^2 The reaction is zero order with respect to B. State the significance of this zero order for the mechanism of the reaction. (1 MARK)

- Slow step or rds involves only A OR B does not appear in the slow step or the rds OR B only appears after the slow step or the rds

H2 + 2ICl → I2 + 2HCl Rate = K [H2] [ICl] Write the steps to this.

- Step 1: H2 + ICl → HCl + HI = slow - Step 2: HI + ICl → I2 + HCl = fast

Q: Compound A reacts with compound B as shown by the overall equation A + 3B → AB3 The rate equation for the reaction is rate = k[A][B]^2 A suggested mechanism for the reaction is Step 1: A + B → AB Step 2: AB + B → AB2 Step 3: AB2 + B → AB3 Deduce which one of the three steps is the rate-determining step. Explain your answer. (2 MARKS)

- Step 2 - (this step with previous) involves one mol/molecule/particle A and two Bs or 1:2 ratio or same amounts (of reactants) as in rate equation

REQUIRED PRACTICAL 8: Measuring rate by gas collection (continuous rate method) --> analysing graph

- The gradient represents the rate of the reaction - The reaction is fastest at the start where the gradient is steepest - The rate drops as the reactants are used up and their concentration drops - Graph will become horizontal/plateu and gradient becomes zero → reaction has stopped Higher the concentration/temperature/surface area → faster the rate [steeper the gradient]

Example 1: A + B → C + D Rate is given as K [ A ]. What are the steps of this reaction?

- This means that one molecule of A is involved in the RDS, NO B! - If they are in the rate equation, they are involved in the rate determining step (slow) - Step 1: A → C + X = slow - Step 2: X + B → D = fast

Q: The rate equation for the hydrogenation of ethene C2H4(g) + H2(g) C2H6(g) is Rate = k[C2H4][H2] At a fixed temperature, the reaction mixture is compressed to triple the original pressure. What is the factor by which the rate of reaction changes? A 6 B 9 C 12 D 27

B;

EQ: This question is about rates of reaction. Phosphinate ions (H2PO2 - ) react with hydroxide ions to produce hydrogen gas as shown. H2PO2 - + OH− →HPO3 2- + H2 Another student reacted different initial concentrations of phosphinate ions with an excess of hydroxide ions. The student measured the time (t) taken to collect 15 cm3 of hydrogen gas. Each experiment was carried out at the same temperature. Table 1 shows the results. State the relationship between the initial concentration of phosphinate and time (t). Deduce the order of the reaction with respect to phosphinate. (2 MARKS)

- [H2PO2 −]^2 is proportional to 1/t - Order = 2

EQ: The rate equation for a different reaction is rate = k[R] Explain why doubling the temperature has a greater effect on the rate of reaction than doubling the concentration of R. (2 MARKS)

- doubling the temperature greatly increases the number of particles with E > Ea - doubling the concentration has a small/negligible effect on the number of particles with E > Ea

EQ: lodine is a reactant, but does not appear in the rate equation. Explain why the reaction must occur in more than one step. (2 MARKS)

- iodine cannot be involved in the rate determining step - iodine must react in a step after the rate determining step / in a later step

EQ: Cisplatin, [Pt(NH3)2Cl2], is used as an anti-cancer drug. When the complex ion B reacts with DNA, the water molecule is replaced as a bond forms between platinum and a nitrogen atom in a guanine nucleotide. The remaining chloride ligand is also replaced as a bond forms between platinum and a nitrogen atom in another guanine nucleotide. An experiment is done to investigate the rate of reaction. During the experiment the concentration of cisplatin is measured at one-minute intervals. Explain how graphical methods can be used to process the measured results, to confirm that the reaction is first order. (3 MARKS)

- plot concentration (y-axis) against time (x-axis) and take tangents / (calculate the) gradients (to calculate rates) - Plot rate/gradients against conc - straight line through origin / directly proportional confirms first order

EQ: This question is about rates of reaction. Phosphinate ions (H2PO2 - ) react with hydroxide ions to produce hydrogen gas as shown. H2PO2 - + OH− →HPO3 2- + H2 A student completed an experiment to determine the initial rate of this reaction. The student used a solution containing phosphinate ions and measured the volume of hydrogen gas collected every 20 seconds at a constant temperature. Figure 1 shows a graph of the student's results. Use the graph in Figure 1 to determine the initial rate of reaction for this experiment. State its units. Show your working on the graph. (3 MARKS)

- tangent drawn to the curve at 0,0 - Evidence of value used in calculation leading to initial rate = 5.5 - cm3 s−1

How do you calculate the rate using a conc vs time graph.

Gradient of the curve at any given point = delta [conc ] / delta time

EQ: Propanone reacts with iodine in the presence of hydrochloric acid as shown CH3COCH3 + I2 → CH3COCH2I + HI The rate equation for this reaction can be written as Rate = k[H+]a[CH3COCH3]b where a and b are whole numbers. A student does a series of experiments to determine the order of reaction with respect to H* In these experiments the student uses a large excess of propanone and a large excess of iodine. Under these conditions the rate equation for the reaction is Rate = k1[H+]a Deduce why propanone is now not included in the rate equation. (2 MARKS)

- the concentration of propanone is effectively constant Or the change in concentration of propanone will be negligible - (so) the concentration of propanone has no effect on the rate of reaction Or (so) the order will be zero with respect to propanone

How do you find the order of a reaction using data?

1. Find 2 experiments/runs where only one [ ] changes 2. Determine relationship between [ ] and rate change.

Use the rate data to determine the rate expression for the reaction and hence the value of the rate constant.

1. Find orders! (a) [CH3COCH3] experiments 1+2 = rate double when [ ] doubles → 1st order (b) [I2] experiments 2+3 = rate stays the same when [ ] doubles → 0 order 2. Rate expression = K [CH3COCH3]1 a) K = rate / [CH3COCH3] 3. Pick an experiment and substitute, e.g. experiment 1 a) K = (6.8 x 10-5) / 0.51 = 1.36 x 10-4 4. Units! K = (mol.dm-3.s-1) / (mol.dm-3) = s-1

EQ: Iodide ions react hydrogen peroxide to form iodine as shown. H2O2 + 2H+ + 2I- → I2 + 2H2O A student reacted iodide ions with hydrogen peroxide at different temperatures to determine the activation energy (Ea) for this reaction. The student measured the time taken for a set amount of iodine to be formed at different temperatures. The results from the experiment are shown in Table 2. An equation that relates activation energy (Ea) to the time taken (time) and temperature (T), where C is a constant, is shown. Suggest why the student did not repeat the experiment at temperatures higher than 336K (1 MARK).

At higher temperatures, the times would be very short

Q: The rate equation for the acid-catalysed reaction between iodine and propanone is: rate = k [H+ ] [C3H6O] The rate of reaction was measured for a mixture of iodine, propanone and sulfuric acid at pH = 0.70 In a second mixture the concentration of the sulfuric acid was different but the concentrations of iodine and propanone were unchanged. The new rate of reaction was a quarter of the original rate. What was the pH of the second mixture? A 1.00 B 1.30 C 1.40 D 2.80 (1 MARK)

B

REQUIRED PRACTICAL 8: Measuring rate by Iodine Clock (Initial rate) --> method

1. Rinse a clean 25.0 cm° pipette with sulfuric acid. Then, use this pipette to transfer 25.0 cm' sulfuric acid, of known concentration (e.g. 0.25 mol dm*), to a clean 250 cm' beaker. This beaker is your reaction vessel. 2. Using a clean pipette or measuring cylinder, add 20.0 cm of distilled water to the beaker containing the sulfuric acid. 3. Using a dropping pipette, add a few drops (or about 1 cm) of starch solution to the same beaker. 4. Measure out 5.0 cm of potassium iodide solution of a known concentration (e.g. 0.1 mol dm- ), using either a 5.0 cm' pipette or a burette, rinsed with potassium iodide solution. Transfer this volume to the reaction vessel. 5. Next, using a pipette rinsed with sodium thiosulfate solution, or a clean measuring cylinder, add 5.0 cm* sodium thiosulfate to the reaction vessel. Swirl the contents of the beaker so all the solutions are evenly mixed. 6. Finally, rinse a 10.0 cm pipette with hydrogen peroxide solution. Then, use the pipette to transfer 10.0 cm3 hydrogen peroxide solution to the reaction vessel and simultaneously start a stop watch. Stir the contents of the beaker throughout the reaction, using a glass rod. 7. Stop the stop watch when the contents of the beaker turn from colourless to blue-black, this marks the endpoint. Record this time in a results table, along with the quantities of sulfuric acid, water, potassium iodide and sodium thiosulfate solutions you used in that experiment. 8. Repeat the experiment varying the volume of potassium iodide solution. Use varying amounts of distilled water in each experiment so the overall volume of the reaction mixture remains constant.

EQ: The rate constant, k, for a reaction varies with temperature as shown by the equation k = Ae^(-Ea/RT) For this reaction, at 25 °C, k = 3.46 × 10−8 s−1 The activation energy Ea = 96.2 kJ mol−1 The gas constant R = 8.31 J K−1 mol−1 Calculate a value for the Arrhenius constant, A, for this reaction. Give the units for A. (4 MARKS)

A = 2571138293. units = s-1

Define and state the rate equation.

A mathematical representation of how changing concentration affects the rate of reaction. It involves the rate constant (k) Eg, A + B → C + D r (rate) = K [ A ]order [ B ]order - Order can be 0, 1 or 2 - If 0 order, it is eliminated from the expression - K is the rate constant → it stays constant at a fixed temperature and increases if temperature is increased - Unit for r is usually mol.dm-3.s-1

EQ: The results of an investigation of the reaction between P and Q are shown in this table. The rate equation is: rate = k [P] [Q]^2 What is the initial concentration of Q in experiment 2? A 0.167 B 0.333 C 0.408 D 0.612 (1 MARK)

C - Because P is 1st order and P increases by x3, so new rate should be 1.2, but is 0.8. - so difference in Q must be 2/3 of original as difference between 1.2 and 0.8 is 2/3. - But Q is second order so difference is squared, so difference is actually root of 2/3 - so 5 x root of 2/3 = 0.408

Q: The equation and rate law for the reaction of substance P with substance Q are given below. 2P + Q → R + S rate = k[P]^2[H+] Under which one of the following conditions, all at the same temperature, would the rate of reaction be slowest? Look at image. (1 MARK)

C;

What effect does a substance with 1st order have on the initial rate of a reaction?

Change in [ ] has a directly proportional effect on initial rate

What effect does a substance with 2nd order have on the initial rate of a reaction?

Change in [ ] has an exponential effect on initial rate (concentration of A Rate is 2nd order with respect to [ B ] → so if [ B ] doubles, rate increases by a factor of a (2^2) → If [ B ] triples, rate increases by a factor of 9 (3^2)

What effect does a substance with 0 order have on the inotial rate of a reaction?

Change in [ ] has no effect on initial rate of reaction

EQ: This question involves the use of kinetic data to deduce the order of a reaction and calculate a value for a rate constant. The data in Table 1 were obtained in a series of experiments on the rate of the reaction between compounds A and B at a constant temperature. Show how these data can be used to deduce the rate expression for the reaction between A and B. (3 MARKS)

Consider experiments 1 and 2: [B constant] - [A] increases × 3: rate increases by 32 therefore 2nd order with respect to A Consider experiments 2 and 3: - [A] increases × 2: rate should increase × 22 but only increases × 2 - Therefore, halving [B] halves rate and so 1st order with respect to B - Rate equation: rate = k[A]2 [B]

Q: Rate = k [A]^2 [B] Correct units for the rate constant in the rate equation above are A mol dm−3 s−1 B mol−1 dm−3 s−1 C mol2 dm−6 s−1 D mol−2 dm6 s−1 (1 MARK)

D;

TIP!

Dont forget the units of k include mol, dm3 AND s-1 mol.^-dm^+.s^-1

Find the orders of reaction using data for this example (1).

Example 1: finding A (experiment 1 + 2) - [ A ] doubles rate x4 (or 22) - → order w.r.t (with respect to) [ A ] is 2nd order Example 1: finding B (experiment 2 + 3) - [ B ] doubles rate - no change - Order w.r.t. (with respect to) [ B ] is 0 order

Find the orders of reaction using data for this example (2).

Example 2: finding B (experiment 1 + 2) - [ B ] increases x4 and rate increases x4 → order w.r.t. (with respect to) [ B ] = 1st order Example 2: finding A (experiment 2 + 3) - [ A ] halves (/2) - Rate halves (/2) - Account for [ B ] first - Ie, [ B ] doubles so rate will double so rate should be 6.4 x 10-3 - However, rate in experiment is actually 4x slower - Ie 6.4 x 10-3 / 4 = 1.6 x 10-3 - [ A ] halves so order w.r.t [ A ] is 2nd order Example 2: finding C (experiment 1 + 4) - Account for [ B ] (1st order) - 0.8 x 10-3 x 3 = 2.4 x 10-3 - Accounting for [ A ] (2nd order) - 2.4 x 10-3 / 4 = 0.6 x 10-3 - [ C ] halved → no change, it is 0.6 x 10-3 Therefore order w.r.t. [ C ] = 0 order.

Q: Consider the graphs E, F, G and H below. Write in the box below the letter of the graph that shows how the rate constant k varies with temperature. (1 MARK)

G

EQ: A general equation for a reaction is shown. A(aq) + B(aq) + C(aq) → D(aq) + E(aq) In aqueous solution, A, B, C and D are all colourless but E is dark blue. A reagent (X) is available that reacts rapidly with E. This means that, if a small amount of X is included in the initial reaction mixture, it will react with any E produced until all of the X has been used up. Explain, giving brief experimental details, how you could use a series of experiments to determine the order of this reaction with respect to A. In each experiment you should obtain a measure of the initial rate of reaction. (6 MARKS)

Indicative Chemistry content Method 1 Stage 1 Preparation - Measure (suitable/known volumes of) some reagents (ignore quoted values for volume) - Measure (known amount of) X / use a colorimeter - into separate container(s) - (allow up to two reagents and X measured together into one container); reference to A, B or C added last. NOT if X added last. Stage 2 Procedure - Start clock/timer at the point of mixing (don't allow if only 2 reagents mixed) (allow even if X not added or added last) - Time recorded for appearance of blue colour/specific reading on colorimeter/disappearing cross - Use of same concentration of B and C / same total volume / same volume/amount of X - Same temperature/use water bath - Repeat with different concentrations of A (can be implied through different volumes of A and same total volume) Stage 3 Use of Results - 1/time taken is a measure of the rate - plot of 1/time against volumes/concentrations of A or plot log(1/time) vs log(volume or concentration of A) - description of interpreting order from shape of 1/time vs volume or concentration graph / gradient of log plot gives order / allow interpretation of time vs concentration graph / ratio between change in concentration and change in rate (e.g, 2x[A] = 2 × rate so 1st order) I Indicative Chemistry content - Alternative Method Using Colorimetry and repeated Continuous Monitoring Stage 1 Preparation - Measure (suitable/known volumes of) A, B and C (ignore quoted values for volume) - Use of colorimeter - into separate container(s) - (allow up to two reagents measured together into one container) - ignore use of X Stage 2 Procedure - Start clock/timer at the point of mixing - Take series of colorimeter readings at regular time intervals - Use of same concentration of B and C / same total volume / (same volume/amount of X) - Same temperature - Repeat with different concentrations of A (can be implied through different volumes of A and same total volume) Stage 3 Use of Results - Plot absorbance vs time and measure/calculate gradient at time=0 - plot of gradient against volumes/concentrations of A or plot log(1/time) vs log(volume or concentration of A) - description of interpreting order from shape of 1/time vs volume or concentration graph / gradient of log plot gives order

Q: The reaction between propanone and iodine in the presence of hydrochloric acid was studied at a constant temperature. CH3COCH3 + I2 → CH3COCH2I + HI The following rate equation was deduced. rate = k [CH3COCH3][H+] Suggest why the order with respect to iodine is zero. (1 MARK)

Iodine is not involved in (or before) the rate determining / slow(est) / limiting step (in the mechanism)

Draw the graph for the Arrhenius equation.

K = Ae^-(Ea/RT) 1. Take ln of both sides lnK = -Ea/RT + lnA 2. Extract T lnK = -Ea/R x 1/T + lnA Y = m x + c I.e. a straight line graph - LnK = y-axis - 1/T = x-axis - lnA = y-axis intercept - NB: A = elnA - -Ea/R = gradient of the line - R = 8.314 J.K-1..mol-1 Therefore Ea [units J.mol-1) = Gradient x R

State the relationship between K (in rate equation) and temperature.

K and temperature are proportional to each other.

Draw a concentration-time graph and a rate-concentration graph for a zero order reaction.

Look at image.

EQ: This question is about rates of reaction. Phosphinate ions (H2PO2 - ) react with hydroxide ions to produce hydrogen gas as shown. H2PO2 - + OH− →HPO3 2- + H2 A student completed an experiment to determine the initial rate of this reaction. The student used a solution containing phosphinate ions and measured the volume of hydrogen gas collected every 20 seconds at a constant temperature. different initial concentrations of phosphinate ions with an excess of hydroxide ions. The student measured the time (t) taken to collect 15 cm3 of hydrogen gas. Each experiment was carried out at the same temperature. Complete the diagram in Figure 2 to show how the hydrogen gas could be collected and measured in the experiments 1 and 2. (1 MARK)

Look at image.

Draw a concentration-time graph and a rate-concentration graph for a 2nd order reaction.

Look at image. - conc vs time graph is a steeper curve than in 1st order (levels out sooner)

Draw a concentration-time graph and a rate-concentration graph for a 1st order reaction.

Look at image. - conc vs time graph is a curve that has a consistent T 1/2

REQUIRED PRACTICAL 8: Measuring rate by loss in Mass (Continous Rate) --> method

Measuring the rate of reaction by recording the loss of mass of reactants/products over time Eg, CaCO3 (s) + 2HCL(aq) → CaCL2 (s) + H2O(l) + CO2 (g) Equipment: - Conical flask - Reactants in conical flask - Top pan balance - Cotton wool Method: 1. Place conical flask containing known conc./vol HCl with cotton wool on the "zeroed" valance with the CaCO3 (s) by its side (= starting mass) 2. Combine reactants, replace cotton wool + start timer 3. Record mass at regular time intervals until reaction completion (repeat) 4. Plot graph of loss of mass and time. Ratte units: g.s-1

REQUIRED PRACTICAL 8: Measuring rate by gas collection (continuous rate method) --> method

Measuring the rate of reaction by recording the volume of gas evolved over time during the course of a reaction Eg, Mg(s) + 2HCl(aq) → MgCl2 (aq) + H2 (g) Equipment: - Conical flask - Reactants into conical flask, one in excess - Bond - Gas syringe Method: - Add reactants to the flask, one in excess [50 cm3 1.0 mol.dm.3 HCl, weight 0.2 g Mg and add], plave bong firmly and start timer - Record the volume of gas evolved at set time intervals (every 15 seconds for 3 minutes) until the reaction completes - Repeat, changing independent variable, e.g.: conc, SA, temperature - Rate units = cm3.s-1 → plot graph of results to find rate - Initial rate = gradient from the origin (time = 0)

What is the overall order of a reaction?

Sum of all orders of reaction. E.g.: 1st order with respect to [ A ] and 2nd order with respect to [ B ] = 1 + 2 = 3 (overall)

What is the order of a reaction, with respect to a given substance in a reaction.

The number/exponent that describes the effect that its change in concentration has on the initial rate of reaction.

REQUIRED PRACTICAL 8: Measuring rate by Iodine Clock (Initial rate)

The reaction: H2O2 (aq) + 2I-(aq) + 2H+(aq) → 2H2O(l) + I2 (aq) - Starch indicator used to detect I2 (aq) → colourless to blue/black The reaction is very fast, so Na2S2O3 reacts with I2 to slow it down: 2S2O3 2-(aq) + I2 (aq) → S4O6 2-(aq) + 2I-(aq) - Small amount of S2O3 2- is added to the reaction mixture (containing starch and excess of H2O2) before I-(aq), then iodide ions are added. - Any I2 produced is instantly converted back to I- (delays reaction) - When S2O3 2- runs out 2I- → I2 and the solution turns blue/black - S2O3 2- acts like a delta, extending the time taken and increasing accuracy Changing [H2O2] - Time taken for colour change recorded using different [H2O2] - 1/T calculated for each [H2O2] = rate

EQ: Define the term overall order of reaction. (1 MARK)

The sum of powers/indices (to which the concentrations are raised in the rate equation)

Q: The Arrhenius equation can be written as: lnk = -Ea/RT + lnA The table below shows the value of the rate constant at different temperatures for a reaction. i) Complete the table by calculating the values of ln k and at each temperature. ii) Plot a graph of ln k against 1/T on the grid opposite. Use your graph to calculate a value for the activation energy, in kJ mol−1, for this reaction. To gain full marks you must show all your working. The gas constant R = 8.31 J K−1 mol−1. (8 MARKS)

lnk in order: (1 MARK) - -9.7 - -8.2 - -7.1 - -4.8 1/T in order: (1 MARK) - 1.43 x 10^-3 - 1.38 x 10^-3 - 1.34 x 10^-3 - 1.26 x 10^-4 - M3: plots points on graph - M4: Draws line of best fit - Calculates gradient: change in y / change in x - Answer = -2.88 x 10^4 - Calculates Ea - Recognizes that gradient = -2.88 x 10^4 is -Ea/R - Ea = 2.88 x 10^4 x 8.314 - /1000 = 239 kJ.mol-1

Many reactions involve a multistep mechanism. Each step happens at a different rate. The slowest of these steps is the ___________ step!

rate-determining (RDS)

EQ: The student determines the order of reaction with respect to propanone in a series of experiments. Table 1 shows the volumes of solution in each experiment. Suggest why the student adds distilled water to experiments 2, 3, 4 and 5. (1 MARK)

to keep the total volume constant Or to keep the concentrations of iodine and hydrochloric acid constant