JPJC 2026 Reaction Kinetics Extension

1 Name: _______________________ Class: _________ Date: __________ Reaction Kinetics: Extension Exercise 1 Consider the reaction between propanone and iodine in aqueous solution. CH3COCH3(aq) + I2(aq) ⎯⎯⎯⎯⎯→ CH3COCH2I(aq) + H+(aq) + I−(aq) In the experiment, a large excess of propanone is used and the concentration of iodine remaining for this reaction is measured by using a photoelectric colorimeter. Graph A is obtained when [H+] is 0.1 mol dm−3. Graph B is obtained by repeating the experiment using 0.2 mol dm −3 H+, all other conditions being identical. Use the graphs to determine the order of reaction with respect to [ I2] and [H+] respectively. [4] Thought Process rate = k [CH3COCH3]a [I2]b [H+]c • [CH3COCH3] is used in large excess  [CH3COCH3] remains effectively constant in each expt as only a small fraction of CH3COCH3 is reacted. • H+ is a catalyst which is regenerated and not consumed  [H+] remains constant in each expt. • Thus, rate equation can be simplified as: rate = k’ [I2]b where k’ = k [CH3COCH3]a [H+]c = constant  A plot of [I2] vs time graph gives the order of reaction w.r.t. [I2]. To determine order of reaction w.r.t. [I2] Refer to [I2] vs time graph.  If a decreasing straight line graph is obtained  zero order w.r.t. [I2]  If a decreasing curve with constant t½ is obtained  first order w.r.t. [I2] To determine order of reaction w.r.t. [H+] Compare i nitial rates (at t = 0 ) for both experiments where [H+] is different and concentration of all other reactants are kept constant. H+ (catalyst) absorbance time /s Graph A (0.1 mol dm−3 H+)   m Graph B (0.2 mol dm−3 H+)   0 15 30 x 0.5x

2 2 A bromoalkane, RBr, undergoes a reaction with aqueous OH− ions. RBr + OH− → ROH + Br− The kinetics of the reaction was studied twice using solutions of different initial [OH−] at the same temperature. The following results were obtained. (a) Use the graphs to determine the order of reaction with respect to [RBr] and [OH −] respectively. [4] (b) Construct a rate equation for the reaction and use it to calculate a value for the rate constant, stating its units. [3] 0 40 80 120 160 [RBr] / mol dm–3 0.0020 0.0040 0.0060 0.0080 0.0100 Time / s with [OH–] = 0.20 mol dm–3 with [OH–] = 0.40 mol dm–3 0.0120 0.0140 0.0160 Thought Process: rate = k [RBr]a [OH–]b Note the initial [RBr] and initial [OH–]. For each experiment, [OH–] is used in large excess and thus [OH–] remains effectively constant during the course of the reaction as only a small fraction of OH – is reacted. The rate equation can be simplified as: rate = k’ [RBr]a where k’ = k [OH–]b.  To determine order w.r.t. [RBr], refer to [RBr] vs time graph. If a decreasing straight line graph is obtained  zero order w.r.t. [RBr] If a decreasing curve with constant t½ is obtained  first order w.r.t. [RBr]  To determine order w.r.t. [OH–], Compare initial rates where [OH –] is different and [RBr] is kept same.  draw tangent at t=0 and determine gradient of tangent.

3 3 The Harcourt and Esson reaction is that between H2O2 and acidified KI. H2O2(aq) + 2H+(aq) + 2I−(aq) → 2H2O(l) + I2(aq) The rate of reaction can be followed by measuring the amount of iodine produced at various times, from which the concentration of H2O2 remaining can be calculated. Experiment initial [H+] / mol dm−3 initial [I−] / mol dm−3 initial [H2O2] / mol dm−3 1 0.200 0.200 0.0200 Experiment 1 was performed and the grid below shows the experimental data of [H 2O2] remaining at various times of the reaction. time /s 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 0 100 200 300 400 500 600 700 800 [H2O2] / mol dm-3 Thought Process: rate = k [H2O2]a [H+]b [I–]c Note that the initial [H+] and initial [I–] are 10 times higher than initial [H2O2]  they are used in large excess and only a small fraction of H+ and I– are reacted during the reaction. H ence their concentrations remained effectively constant during the reaction.  the rate equation can be simplified as: rate = k’ [H2O2]a where k’ = k [H+]b [I–]c = constant  A plot of [H2O2] vs time graph gives the order of reaction w.r.t. [H2O2].

4 (a) By showing all working and clearly drawing any con struction lines on the graph, use the graph to determine the: (i) order of reaction with respect to [H2O2], (ii) initial rate, in mol dm−3 s−1. [4] Further experiments were carried out by changing [H +] and [ I−] but keeping the initial [H2O2] the same as before. The following results were obtained. Experiment initial [H+] / mol dm−3 initial [I−] / mol dm−3 initial rate / mol dm−3 s−1 2 0.400 0.200 8.4  10−5 3 0.300 0.200 6.3  10−5 4 0.200 0.100 2.1  10−5 (b) Determine the orders with respect to [H+] and [I−]. Explain your reasoning. [2] (c) Hence write the rate equation for the reaction and calculate a value for the rate constant. Include units in your answer. [3] (d) With reference to your answer in (a), predict the half-life of H2O2 in experiment 2. [1] (e) The reaction between hydrogen peroxide and iodide ions in acidic solution is considered to proceed by the following steps. step 1 H2O2 + I− → IO− + H2O step 2 IO− + H+ → HOI step 3 HOI + H+ + I− → 2H2O + I2 Based on your rate equation obtained in part (c), state and explain which of the three steps is the rate-determining step. [2]

5 4 A student investigated the rate of reaction between sodium and ethanol. 2CH3CH2OH(l) + 2Na(s) → 2CH3CH2ONa(ethanol) + H2(g) A freshly cut piece of sodium was weighed and added to a large excess of ethanol. The experiment was conducted at room temperature and pressure. The total volume of gas liberated was recorded every minute. The results are tabulated below. At the end of the reaction, total volume of gas liberated is 60.0 cm3. Time / min 0 1 2 3 4 5 6 7 Total volume of gas / cm3 0.0 23.0 36.5 46.0 51.0 54.5 57.0 58.5 Thought Process: Since CH3CH2OH is used in large excess, only a small fraction of CH3CH2OH is reacted and hence [CH3CH2OH] remains effectively constant during the reaction. Hence, Na is the limiting reagent and the volume of H₂ produced is proportional to the amount of Na that has reacted. Therefore, the volume of H₂ yet to be produced, 60.0 − 𝑉𝑡, is proportional to the amount of Na remaining. If half-lives are constant, the reaction exhibits first-order kinetics. Let A: initial amount of limiting reactant (Na) present Vmax: maximum volume of H2 gas formed when all the limiting reactant (Na) has reacted. At t: 0 t1/2 2t1/2 3t1/2 Reactant (Na) : A ½ A ¼ A 1/8 A Product (H2): 0 ½ Vmax ¾ Vmax 7/8 Vmax

6 (a) (i) Plot the graph of total volume of gas against time from the experimental results on the grid below. [1] (ii) Use the graph to show that the overall kinetics are first order. Show all your working and draw clearly all construction lines on your graph. [2] (iii) In this experiment, the kinetics appear to be zero order with respect to [ethanol]. Suggest a reason for this. [1] (b) In the atmosphere, the oxidation of SO2 to SO3 can be catalysed by N