JPJC 2026 Reaction Kinetics Extention Answer

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 t he 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 initial 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+) Graph B (0.2 mol dm−3 H+)   0 15 30 x 0.5x  [I2] left at time t colourless brown colourless colourless

2 Absorbance  [I2] in the reaction mixture. • Graph A (or Graph B) is a decreasing straight line with constant gradient.  change in [I2] has no effect on the rate (rate remains constant)  rate is independent of [I2]  order of reaction w.r.t. [I2] = 0. • Comparing initial rates At time t = 0 s, Graph Absorbance  initial [I2] [H+] / mol dm−3 initial rate = gradient A x 0.1 0.5x 30 = 0.0167 x B x 0.2 0.5x 15 = 0.0333 x When [H+] doubles (or x 2), initial rate doubles (or x 2).  order of reaction w.r.t. [H+] is 1.

3 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. [4] (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 (t½)1 = 50s (t½)2= 50s 50 82 100 164 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. (0, 0.0160)   (82,0) (164,0) 

4 (a) • From the graph of [RBr] vs time with [OH-] = 0.4 mol dm-3 (t½)1 = 50 s, (t½)2 = 50 s Since half−lives are constant, the order of reaction w.r.t. [RBr] is 1. Note: Reject “since half-lives are equal/consistent….” • Comparing initial rates Draw tangent at t = 0 s and determinegradient of the tangent [RBr] / mol dm-3 [OH-] / mol dm-3 Initial rate / mol dm-3 s-1 0.0160 0.2 0.0160-0 0-164 = 9.76  10−5 0.0160 0.4 0.0160-0 0-82 = 1.95  10−4 When [OH-] doubles (or x 2), initial rate doubles (or x 2).  order of reaction w.r.t. [OH-] is 1. (b) rate = k [RBr][OH-] Note: “ rate equation = k [RBr][OH-] “ will be marked wrong! Using initial rate, 9.76  10-5 = k (0.0160)(0.2) (OR, 1.95  10-4 = k (0.0160)(0.4) ) mol dm-3 s-1 = k (mol dm-3)2  k = 0.0305 mol-1 dm3 s-1

5 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 [H2O2] 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 (t½)1 = 330 s 330 (t½)2 = 335 s 665 450 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]. (0, 0.02)  (450, 0) 

6 (a) By showing all working and clearly drawing any construction lines on the graph , use the graph to determine the: (i) order of reaction with respect to [H2O2], (t½)1 = 330 s, (t½)2 = 335 s  average t½ = 330+335 = 332.5 s2 Since half−lives are  constant, order of reaction w.r.t [H2O2] is 1. Note: Reject “since half-lives are equal/consistent,….” (ii) initial rate, in mol dm−3 s−1. Initial rate =gradient of tangent at t = 0 s =  -5 -3 -10.02-0 =4.44 10 mol dm s0-450 [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. initial [H2O2] / mol dm−3 Experiment initial [H+] / mol dm−3 initial [I−] / mol dm−3 initial rate / mol dm−3 s−1 0.0200 2 0.400 0.200 8.4  10−5 0.0200 3 0.300 0.200 6.3  10−5 0.0200 4 0.200 0.100 2.1  10−5 (b) Determine the orders with respect to [H+] and [I−]. Explain your reasoning. Let rate = k [H2O2] [H+]a [I–]b Comparing expts 2 & 3, rate  [H+]a since [H2O2] and [I-] are constant.    =         aa 2 3 (rate) 8.4 0.4 4 4= = a=1(rate) 6.3 0.3 3 3 OR, when [H+] x 4/3, rate x 4/3.  order of reaction w.r.t. to [H+] is 1. Comparing expts 2 & 4, rate  [H+][I-]b since [H2O2] is constant. ( )             1b b2 4 (rate) 8.4 0.4 0.2 = = 4= 2 2 b=1(rate) 2.1 0.2 0.1  order of reaction w.r.t. [I–] is 1. [2] t1/2 within one square are considered constant.

7 (c) Hence write the rate equatio