Chemistry Answers Chapter 7

Answers 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 21 CO(g) + H2O(g) = H2(g) + CO2(g) (a) Initial: Change: Equilibrium: 4.0 −3.2 6.4 −3.2 0.0 +3.2 0.0 +3.2 0.8 3.2 3.2 3.2 [H2][CO2] (3.2)2 Kc = = 4.0 = (0.8)(3.2) [CO][H2 O] 3 The concentration of a pure solid or pure liquid is a constant, effectively its density, which is independent of its amount. These constant values therefore do not form part of the equilibrium expression, but are included in the value of K. 4 The value for Kc at 298 K for the reaction N2(g) + O2(g) 2NO(g) is extremely low, so the equilibrium mixture lies to the left with almost no production of NO. But at higher temperatures, such as in vehicle exhaust fumes, the reaction shifts to the right and a higher concentration of NO is produced. This gas is easily oxidized in the air, producing the brown gas NO2 which is responsible for the brownish haze: 2NO(g) + O2(g) → 2NO2(g). 5 The atom economies of the Haber process and the Contact process reactions described are both 100% as there is only one product. In other words, there is no waste. But this does not mean that all reactants are converted to product, so the stoichiometric yield is less than 100%. It is the goal of these industries to maximize yield and efficiency by choosing the optimum conditions, taking equilibrium and kinetic considerations into account. (b) Put the values into the equilibrium expression to determine Q: Q= (3.0)2 = 0.56 (4.0)2 This is not equal to the value of Kc so the reaction is not at equilibrium. As the value of this mixture is lower than Kc, the reaction will move to the right before equilibrium is established. 22 C 23 (a) 0 (b) Negative (c) Positive 24 (a) 79.8 kJ (b) Increasing temperature has increased the value of K so it must be an endothermic reaction. Challenge yourself Practice questions 1 2 Earth receives energy from the Sun and disperses energy, largely as heat. But exchange of matter is minimal – the only exceptions to Earth being a closed system are matter received from space such as asteroids and space dust, and matter lost to space such as spacecraft. The different values of Kc indicate different stabilities of the hydrogen halides. The bonding in HCl is the strongest and in HI the weakest. This is largely because of the size of the atoms. As I has a larger atomic radius than Cl, in HI the bonding pair is further from the nucleus than the bonding pair in HCl, and so experiences a weaker pull. The HI bond breaks more easily and so the dissociation reaction is favoured. 1 C 2 D 3 D 4 A 5 D 6 C 7 D 8 A 9 D 10 (a) (i) (K =) [SO3]2/[O2][SO2]2 [1] (ii) yield (of SO3) increases / equilibrium moves to right / more SO3 formed; 3 gaseous molecules → 2 gaseous molecules / decrease in volume of gaseous molecules / fewer gaseous molecules on right hand side [2] Do not allow ECF. (iii) yield (of SO3) decreases; 33 10:01 Answers forward reaction is exothermic / reverse/ backwards reaction is endothermic / equilibrium shifts to absorb (some of) the heat [2] (c) value of Kc increases; [SO2Cl2] increases; decrease in temperature favours (forward) reaction which is exothermic [3] Do not accept exothermic reaction or Le Châtelier’s principle. Do not allow ECF. (d) no effect on the value of Kc / depends only on temperature; Do not allow ECF. (iv) rates of both forward and reverse reactions increase equally; no effect on position of equilibrium; no effect on value of Kc [3] [SO2Cl2] decreases; increase in volume favours the reverse reaction which has more gaseous moles [3] Do not allow ECF. (b) 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) NO(g) Initial/ 0.100 mol dm−3 Change/ −0.038 mol dm−3 Equilibrium/ 0.062 mol dm−3 (e) no effect; H2(g) N2(g) H2O(g) 0.051 0.000 0.100 −0.038 +0.019 +0.038 0.013 0.019 0.138 catalyst increases the rate of forward and reverse reactions (equally) / catalyst decreases activation energies (equally) 12 increasing temperature favours endothermic/ reverse reaction; as yield decreases with increasing temperature [2 max] [N2] at equilibrium = 0.019 (mol dm ); −3 [H2O] at equilibrium = 0.138 (mol dm−3); (b) yield increases / equilibrium moves to the right / more ammonia; Kc = [N2][H2O]2/[NO]2[H2]2 = (0.019)(0.138)2/ [4] (0.062)2(0.013)2 = 5.6 × 102 increase in pressure favours the reaction which has fewer moles of gaseous products [2] [NH3]2 [1] (c) (Kc =) [N2][H2]3 Award [4] for final correct answer. Accept any value also in range 557–560. Do not penalize significant figures. (a) (Kc =) [SO2Cl2] [Cl2][SO2] [1] (d) [N2]: (at equilibrium = 1.00 − 0.031 =) 0.969 (mol dm−3); Ignore state symbols. [H2]: (at equilibrium = 3.00 − 3(0.031) =) 2.91 (mol dm−3); (0.062)2 [3] = 1.6(1) × 10−4 Kc = (0.969)(2.91)3 Square brackets [ ] required for the equilibrium expression. ( (b) 7.84 × 10−3 mol of SO2 and 7.84 × 10−3 mol of Cl2; Award [1] for 10.34. Award [3] for the correct final answer. [3] ) Ignore units. 7.84 × 10−3 mol dm−3 of SO2, 7.84 × 10−3 mol dm−3 of Cl2 and 7.65 × 10−4 mol dm−3 of SO2Cl2; 12.5 (a) exothermic Accept either of the following for the second mark. [H2] at equilibrium = 0.013 (mol dm−3); 11 [2] Award [1] for Kc = 1.4 × 10−4 (e) no effect 13 [1] (a) reactants and products in same phase/state; rate of forward reaction = rate of reverse reaction; 34 10:01 concentrations of reactants and products remain constant / macroscopic properties remain constant [2 max] (f) no effect (on the value of the equilibrium constant) as it speeds up forward and reverse reaction/ concentrations of reactants and products do not change / position of equilibrium does not change / no change in yield [2] Do not accept concentrations are equal. (b) (Kc =) [HI]2 [H2][I2] (c) no change to position of equilibrium [1] [1] (d) the reaction is exothermic / heat is given out/ ∆H is negative [1] (e) amount of H2 remaining at equilibrium 1.80 =1.60 − = 0.70 mol 2 amount of I2 remaining at equilibrium 1.80 = 0.10 mol =1.0 − 2 (1.80 / 4.0)2 1.802 Kc = / (0.70 / 4.00) × (0.10 / 4.00) 0.70 × 0.10 (1.80)2 [4] Kc = 0.70 × 0.10 = 46.3 Award [4] for correct final answer. 14 (a) ∆G° = 0 (b) ΔG = −70 kJ mol−1 The reaction has a very high value for K, so will go essentially to completion – from the equilibrium yield, this reaction is likely to give a high production of methanol. However, kinetic data are not available, so the rate cannot be deduced. 15 (a) [H2(g)] = 4.0 mol dm−3 [I2(g)] = 1.0 mol dm−3 [HI] = 4.0 mol dm−3 (b) (i) HI originally placed = 2.0 mol dm−3 (ii) I2 at equilibrium = 0.218 mol dm−3; HI at equilibrium = 1.56 mol dm−3 10:01