NJC 7 Kinetics H1 Tutorial Ans

1 Reaction Kinetics Discussion Questions 1 The initial rate of the reaction X + Z ⎯→ P + Q was measured as a function of the concentration of each reactant. The following results were obtained: Expt [X] / mol dm−3 [Z] / mol dm−3 Initial rate / mol dm−3 s−1 1 0.20 0.30 9.0  10−3 2 0.20 0.40 12.0  10−3 3 0.40 0.40 48.0  10−3 (a) Determine the order of reaction with respect to X, Y and Z. Comparing expt 1 and 2 where [X] and [Y] are the same. When [Z]expt2 [Z]expt1 = 0.40 0.30 = 4 3 , Rate2 Rate1 = 12 9 = 4 3 Hence 1st order wrt Z. Comparing expt 2 and 3 where [Y] and [Z] are the same. When [X]expt3 [X]expt2 = 2 , Rate3 Rate2 = 4  Rate  [X]2 Hence 2nd order wrt X (b) Write the rate equation. Rate = k[X]2[Z]

2 2 The kinetics of the reaction between iodide and peroxodisulfate can be investigated by varying the volume of the reactants used. The two reactants are mixed in the presence of a known amount of Na2S2O3 and a little starch. The time taken for an intense blue colour to be observed is then determined. Experiment Volume used/cm3 t/s 1.0 mol dm−3 KI 0.040 mol dm−3 Na2S2O8 H2O 1 10.0 5.0 25.0 170 2 15.0 5.0 20.0 113 3 15.0 10.0 15.0 57 4 20.0 20.0 0.0 x What is the value of x? A 21 B 28 C 85 D 63 Rate ∝ 1 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 Expt Volume used/cm3 t/s 1 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 1.0 mol dm−3 KI 0.040 mol dm−3 Na2S2O8 H2O 1 10.0 5.0 25.0 170 1 170 = 0.00588 2 15.0 5.0 20.0 113 1 113 = 0.00885 3 15.0 10.0 15.0 57 1 57 = 0.01754 4 20.0 20.0 0.0 x Comparing expt 1 and 2 where [Na2S2O8] is the same. When [KI] × 1.5, rate also × 1.5. Hence it is 1st order wrt KI. Comparing expt 2 and 3 where [KI] is the same. When [Na2S2O8] × 2, rate also × 2. Hence it is 1st order wrt Na2S2O8 Rate = k[KI][ Na2S2O8] Comparing Expt 1 and 4, When [KI] × 2, and [Na2S2O8] × 4, rate is 2×4 = 8 times of expt 1. Rate of expt 4 = 8 × 0.00588 = 0.0470 1 𝑥 = 0.0470, x = 21.3 s Answer is A

3 3 The following reaction is monitored by measuring the changes in total pressure during the reaction, with initial pressure of O2 at 300 kPa. 3O2(g) → 2O3 (g) Determine the order of reaction with respect to oxygen As shown in the graph, t1/2 (1) is 55s (from 300 kPa to 150 kPa) t1/2 (2) is 59s (from 250 kPa @ t =14 to 125 kPa @ t=73) As the reaction proceeds, PO2 decreases and hence [O2] decreases. It is observed that the rate of reaction decreases with a constant half-life of 57s. Hence it is first order w.r.t O2.

4 4 Explain, using collision theory, how the following changes can affect the initial reaction rate of a gaseous system. Illustrate your answer with the Boltzmann distribution curve for parts (ii) and (iii). (i) increasing the pressure There is greater no of particle per unit volume leading to greater collision frequency therefore frequency of effective collision increases leading to increase in rate. Note: Assumption made that increase in total pressure comes from smaller volume. (ii) decreasing the temperature As the temperature decreases, the average kinetic energy of the particles decreases. There are less particles with kinetic energy equal or greater than the activation energy as shown in the Boltzmann distribution. This result in lower frequency of effective collisions. Hence rate of reaction decreases. (iii) addition of a catalyst In the presence of catalyst, the reaction proceeds with an alternative reaction pathway of lower activation energy(Ea’). There are more particles with kinetic energy equal or greater than the lowered activation energy as shown in the Boltzmann distribution curve. This result in greater frequency of effective collisions. Hence rate of reaction increases.

5 5 The decomposition of hydrogen peroxide is a first order reaction. H2O2 ⎯→ H2O + ½ O2 ΔH = −98 kJ mol−1 The uncatalysed reaction has an activation energy of 79 kJ mol−1. (a) Use the data provided to construct a reaction pathway diagram for this uncatalysed reaction. (b) Explain what is meant by the term activation energy. Activation energy ( Ea) is the minimum amount of energy that reactants must possess before a reaction can occur. (c) The enzyme catalase act as catalysts, and speed up the reaction. What effect will the presence of the catalyst have on the rate constant for this reaction? Explain your answer. A catalyst provides an alternative reaction path of lower activation energy than that of the uncatalysed reaction, rate constant increases and rate increases Multiple Choice Questions 6 Lead is the final product formed by a series of changes in which the rate -determining stage is the radioactive decay of uranium -238. This radioactive decay is a first order reaction with a half-life of 4.5 × 109 years. What would be the age of the rock sa mple, originally lead-free, in which the molar proportion of uranium to lead is now 1:7? Time Amt of Uranium left Amt of Lead formed Mole ratio of U : Pb 0, Inital 1 0 1 : 0 One t1/2 0.5 0.5 1 :1 Two t1/2 0.25 0.75 1 : 3 Three t1/2 0.125 0.875 1 : 7 Mole ratio of U : Pb = 1:7 → 3 half lives Half-life = 4.5 × 109 years Age of rock = 3 × 4.5 × 109 years = 13.5 × 109 years

6 7 The decomposition 2N2O5 ⎯→ 4NO2 + O2 is first order with respect to N2O5. In an experiment, 0.10 mol of pure N 2O5 was put into an evacuated flask. It was found that there was 0.025 mol of N2O5 left 34 minutes later. Which statement is true? A It took 17 minutes for the amount of NO2 to rise from 0 mol to 0.10 mol B There was 0.0625 mol of N2O5 left after 17 minutes. C There was 0.0125 mol of N2O5 left after 68 minutes. D The amount of NO2 in the flask went up by four times in the first 34 minutes. Since it is a first order reaction, half lives are constant. Time 0 1st t1/2 2nd t1/2 3rd t1/2 4th t1/2 Amount of N2O5 /mol (reactant) 0.1 → 0.05 → 0.025 → 0.0125 → 0.00625 Amount of NO2 /mol (product) 0 → 2 ×(0.05) = 0.10 → 2 × 0.075 = 0.15 Since there are 0.025 mol of N2O5 left after 34 min, we know that two half-lives = 34 min Hence, t1/2 = 17 min After 17min, amount of N2O5 reacted = 0.05 mol After 17min, amount of NO2 formed = 2(0.05) = 0.10 mol Option A is true. After 17 min, there was 0.05 mol of N2O5 left. After 68 min (after four t1/2), there was 0.00625 mol (0.1 × ½ × ½ × ½ × ½) of N2O5 left. Option B & C are false. After 34 min, amount of NO2 formed = 2(0.10−0.025) = 0.15 mol Therefore the amount of NO2 went up from 0 to 0.15 mol. Option D is false.

7 8 When temperature decreases from T2 to T1, maximum of the curve is displaced to the Left. Area L increase, M decreases, N decreases Ans : C

8 9 A B C D 1, 2 & 3 are correct 1 & 2 only are correct 2 & 3 only are correct 1 only are correct When temperature increases from T 1 to T2, maximum of the curve is displaced to the right. Option 1 is correct. The number of molecules with energy above any given value increases, E.g. K.E.  Ea .(as can be seen from the greater shaded area in the diagram above). Option 2 is correct. However, the proportion of molecules with lower energy decreases. Option 3 is incorrect. Answer: (B)