Chemistry Answers Chapter 6

Chapter 6 Exercises 1 Reaction gives off a gas: change in volume could be measured. Reaction involves purple MnO4− ions being reduced to colourless Mn2+ ions: colorimetry could be used. Reaction involves a change in the concentration of ions (23 on the reactants side and 2 on the products side): conductivity could be used. 11 All these techniques enable continuous measurements to be made from which graphs could be plotted of the measured variable against time. The ashes must contain a catalyst that speeds up the reaction between sugar and oxygen. (Deduced from the fact that all other factors that affect reaction rate can be ruled out.) 12 (a) 2CO(g) + 2NO(g) → 2CO2(g) + N2(g) 2 C 3 (a) (i) Measure the decrease in the mass of flask + contents. 7 The reaction requiring the simultaneous collision of two particles is faster. The simultaneous collision of three particles is statistically less likely. 8 B (iii) Measure the increase in volume of gas collected. 0.180 0.160 −0.040 [H2O2] 0.140 0.080 (e) Catalytic converters reduce pollution from cars but do not remove it completely. As in (d), they are not effective when the engine first starts from cold, when an estimated 80% of pollution occurs. Other pollutants in car exhausts are not removed by the catalyst, e.g. ozone, sulfur oxides and particulates. Also the catalytic converter increases the output of CO2, a serious pollutant because of its greenhouse gas properties. 48s −0.022 0.060 60s 0.040 0 40 80 120 160 200 time/s At 60 s, rate = 8.3 × 10−4 mol dm−3 s−1 At 120 s, rate = 3.7 × 10−4 mol dm−3 s−1 5 D 6 A B (d) Catalytic activity involves the catalyst interacting with the gases, and the reaction occurring on its surface. As temperature increases, the increased kinetic energy of the gases increases the frequency with which they bind to the catalyst. 0.200 0.100 10 (c) The increased surface area of the catalyst in contact with exhaust gases will increase catalyst efficiency. (b) The rate of the reaction decreases with time because the concentration of the acid decreases. 0.120 B (b) CO is a toxic gas: it combines with haemoglobin in the blood and prevents it from carrying oxygen. NO is a primary air pollutant: it is oxidized in the air to form acidic oxides, leading to acid rain. It also reacts with other pollutants in the atmosphere, forming smog. (ii) Measure the increase in pH of the reaction mixture. 4 9 13 Experiment 1: rate = k[H2][I2] Experiment 2: rate = k[H2O2] 29 9 10:01 Answers Experiment 3: rate = k[S2O8 2−][I−] formal charges which have fractional values. The distribution of these formal charges in the transition state may help to interpret its stability and how it will react in the next step of the reaction mechanism. Experiment 4: rate = k[N2O5] 14 1st order with respect to NO; 1st order with respect to O3; 2nd order overall. 15 Rate = k[CH3Cl]2 Rate = k [CH3Cl][OH−] Practice questions Rate = k [OH−]2 16 (a) mol−1 dm3 s−1 (b) s−1 (c) mol dm s (d) mol−2 dm6 s−1 −3 −1 (e) mol−1 dm3 s−1 17 18 From the units of k, it must be 1st order. Rate = k[N2O5] k = 4.5 × 10−4/(2.0 × 10−3)2 = 1.1 × 102 mol−1 dm3 min−1 A 2 B 3 C 4 C 5 A 6 A 7 B 8 C 9 C 10 D 11 (a) catalyst; regenerated at end of reaction / OWTTE [2] (b) (i) N2O2 [1] (ii) ([H2] appears in rate expression so) step 2 rate-determining/rds/slow step [1] 19 C 20 NO: 2nd order; O2: 1st order Rate = k[NO]2[O2] 21 1 Experiment 2: rate = 1.5 × 10−2 mol dm−3 s−1 Experiment 3: rate = 1.5 × 10−2 mol dm−3 s−1 22 Rate = k[NO2][CO] 23 Yes, it fits the kinetic data and the overall stoichiometry. 24 C 25 (a) 2AB2 → A2 + 2B2 (b) Rate = k [AB2] C 29 134 kJ mol−1 27 D (c) (k2 >> k1 so) step 1 rate-determining / rds / slow step; two molecules of NO2 involved in step 1 consistent with rate expression / rate of overall reaction must equal rate of step 1 which is rate = k1[NO2]2 / OWTTE [2] (d) Ea = −R × m; measurement of gradient from two points on line 2 Accept a gradient in range –2.14 × 104 K to –2.27 × 104 K. (c) mol−1 dm3 s−1 26 Allow ‘since step 1 involves 2NO and step 2 involves H2 and as all 3 molecules are involved in rate expression, then two steps must have approximately same rate’ / OWTTE. 28 B correct answer for Ea; correct units kJ mol−1 / J mol−1 corresponding to answer [4] Allow kJ or J. Challenge yourself 1 2 Collecting a gas over warm water will cause its temperature and therefore its volume to increase. If the partially made/broken bonds are treated as containing only one electron, we can calculate A typical answer: Ea = 1.85 × 102 kJ mol−1. Allow answers for Ea in range 1.75 × 102 kJ mol−1 to 1.91 × 102 kJ mol−1. Award [4] for correct final answer with some working shown. Award [2 max] for correct final answer without any working shown. 30 10:01 12 (a) the concentration (of nitrogen(II) oxide) [1] (d) (i) (b) mol−2 dm6 s−1 / dm6 mol−2 s−1 Number of particles Award [0] if reference made to equilibrium. [1] Accept (mol−1 dm3)2 s−1. 13 (a) k increases with increase in T / k decreases with decrease in T [1] Ea (no catalyst) Energy Do not allow answers giving just the Arrhenius equation or involving lnk relationships. axes correctly labelled x = energy/ velocity/speed, y = number/% of molecules/particles/probability (b) gradient = −Ea/R; graph showing correct curve for Maxwell-Boltzmann distribution −30000 (K) = −Ea/R Allow value in range −28800 to −31300 (K). If two curves are drawn, first and second marks can still be scored, but not third. Ea = (30000 × 8.31=) 2.49 × 105 J mol−1/ 249 kJ mol−1 Curve(s) must begin at origin and not go up at high energy. Allow value in range 240–260 kJ mol−1. Allow [3] for correct final answer. two activation energies shown with Ecat shown lower [3] (c) 0.9 × 0.200 = 0.180 (mol dm−3); Award the mark for the final point if shown on an enthalpy level diagram. rate = (0.244 × (0.180)2 =) 7.91 × 10−3 mol dm−3 s−1 (ii) catalyst provides an alternative pathway of lower energy / OWTTE [1] Award [2] for correct final answer. Award [1 max] for either 9.76 × 10 mol dm−3 s−1 or 9.76 × 10−5 mol dm−3 s−1. −3 (a) to maintain a constant volume / OWTTE Accept catalyst lowers activation energy (of reaction). [1] (b) (i) [H+] order 1, [CH3COCH3] order 1, [I2] order 0; (rate = ) k[H+][CH3COCH3] 15 (a) XY + Z → X + YZ (b) [2] Award [2] for correct rate expression. Allow expressions including [I2]0. (ii) neither were correct / Alex was right about propanone and wrong about iodine / Hannah was right about propanone and hydrogen ions but wrong about iodine / OWTTE [1] (c) [CH3COCH3] = 0.100 mol dm−3 and [H+] = 0.100 mol dm−3 4.96 × 10−6 k= = 4.96 × 10−4 mol−1 dm−3 (0.100 × 0.100) s−1 [1] transition state transition 2 state 4 potential energy 14 Ea (with catalyst) 3 intermediate 5 Products ∆H Ea 1 Reactants extent of reaction Award [3] for all 5 correct, [2] for 4 correct, [1] for 3 correct. X–Y + Z, W 2 X–Y–W 3 W–Y 4 W–Y–Z Ignore calculation of [I2]. 5 W, Y–Z + X No ECF here for incorrect units. Award [3] for all 5 correct, [2] for 4 correct, [1] for 3 correct. (c) 1 31 10:01 Answers (d) 1st step is rate-determining step (highest energy of transition state); rate equation = k[W][XY] [2] (e) reaction is catalysed by W, which is not chemically changed at the end of the reaction [1] (f) see graph in (b) above [2] Chapter 7 Exercises 1 A 4 [NO2]2 (a) Kc = [NO]2[O2] 2 C 3 B [CH3OH][Cl−] (c) Kc = [CH3Cl][OH−] 5 (a) N2O4(g) 7 (d) Shift to the right 15 (a) Amount of CO will decrease (c) Amount of CO will increase CO(g) + 3H2(g) (a) 3F2(g) + Cl2(g) 2ClF3(g) [ClF3]2 Kc = [F2]3[Cl2] (b) 2NO(g) N2(g) + O2(g) [N2][O2] Kc = [NO]2 (c) CH4(g) + H2O(g) CO(g) + 3H2(g) [CO][H2]3 Kc = [CH4][H2O] (a) Mostly reactants (d) No change in CO 16 C 18 The Haber process is exothermic in the forward direction. Therefore, increasing temperature will decrease the value of Kc. This represents a decrease in the reaction yield. 19 (a) 17 (b) Mostly reactants 2HI(g) Kc = [HOCl]2 (a) [H O][Cl O] > K; not at equilibrium; reaction 2 2 proceeds to the left 9 (a) 7.73 × 104 (b) 3.60 × 10−3 (c) 6.00 × 10−2 10 B 13 (a) Shift to the left 11 D 12 C (b) Shift to the right (c) No shift in equilibrium 14 (a) Shift to the left H2(g) 0.0 +0.11 0.11 + I2(g) 0.0 +0.11 0.11 [H2][I2] (0.11)2 = 2.0 × 10−2 = [HI]2 (0.78)2 (b) At the higher temperature, the value of Kc is higher, so the reaction must be endothermic. proceeds to the left (b) At equilibrium [HOCl]2 > K; not at equilibrium; reaction (c) [H2O][Cl2O] B Initial: 1.0 Change: −0.22 Equilibrium: 0.78 (c) Mostly products 8 (e) Shift to the right (b) Amount of CO will decrease 2NO2(g) (b) CH4(g) + H2O(g) 6 (c) This is equivalent to an increase in pressure, so shifts to the left [NO2]4 [H2O]6 (b) Kc = [NH3]4[O2]7 20 Initial: N2(g) 1.6 + O2(g) 1.6 Change: −x Equilibrium: 1.6 − x −x 1.6 − x 2NO(g) 0.0 2x 2x As Kc is very small, 1.6 – x » 1.6 [NO]2 (2x)2 Kc = = (1.6)2 = 1.7 × 10−3 [N2][O2] x = 0.03298, so 2x = 0.066 [NO]eqm = 0.066 mol dm−3 (b) Shift to the right 32 10:01