# Answers to More Chapter 06 Study Questions

1. From the $$\mathrm{\Delta H_f^\circ}$$ Table:
1. $$\mathrm{0.5 H_2(g) + 0.5 Br_2(l) \rightarrow HBr(g)}$$    $$\mathrm{\Delta H = - 36.2\: kJ}$$
2. $$\mathrm{0.5 H_2(g) + 0.5 Cl_2(g) \rightarrow HCl(g)}$$    $$\mathrm{\Delta H = - 92.3\: kJ}$$

$$\mathrm{2 \times (i)}$$      $$\mathrm{H_2(g)+ Br_2(l) \rightarrow 2 HBr(g)}$$    $$\mathrm{\Delta H = 2(-36.2) = -72.4\: kJ}$$

$$\mathrm{-2 \times (ii)}$$      $$\mathrm{2 HCl(g) \rightarrow H_2(g)+ Cl_2(g)}$$    $$\mathrm{\Delta H = -2(-92.3) = 184.6\: kJ}$$

overall reaction:   $$\mathrm{2 HCl(g) + Br_2(l) \rightarrow 2 HBr(g) + Cl_2(g)}$$   $$\mathrm{\Delta H = -72.4 + 184.6 = 112.2\: kJ}$$

endothermic

1. $$\mathrm{0.5 N_2(g) + 1.5 H_2(g) \rightarrow NH_3(g)}$$       $$\mathrm{\Delta H = -46.2\: kJ}$$
2. $$\mathrm{H_2(g) + 0.5 O_2(g) \rightarrow H_2O(l)}$$         $$\mathrm{\Delta H = -285.8\: kJ}$$

$$\mathrm{-4 \times(i)}$$      $$\mathrm{4 NH_3(g)\rightarrow 2 N_2(g) + 6 H_2(g)}$$    $$\mathrm{\Delta H = -4(-46.2) = 184.8\: kJ}$$

$$\mathrm{6 \times (ii)}$$        $$\mathrm{6 H_2(g) + 3 O_2(g)\rightarrow 6 H_2O(l)}$$      $$\mathrm{\Delta H = 6(-285.8) = -1715\: kJ}$$

overall reaction:  $$\mathrm{4 NH_3(g) + 3 O_2(g) \rightarrow 2 N_2(g) + 6 H_2O(l)}$$  $$\mathrm{\Delta H = 184.8 - 1715 = -1530\: kJ}$$

exothermic

1. $$\mathrm{N_2(g) + 2 H_2(g) \rightarrow N_2H_4(l)}$$          $$\mathrm{\Delta H = + 50.6\: kJ}$$
2. $$\mathrm{0.5 N_2(g) + O_2(g) \rightarrow NO_2(g)}$$         $$\mathrm{\Delta H = + 33.9\: kJ}$$
3. $$\mathrm{H_2(g) + 0.5 O_2(g) \rightarrow H_2O(l)}$$         $$\mathrm{\Delta H = - 285.8\: kJ}$$

$$\mathrm{-1 \times (i)}$$      $$\mathrm{N_2H_4(l)\rightarrow N_2(g) + 2 H_2(g)}$$          $$\mathrm{\Delta H = -50.6\: kJ}$$

$$\mathrm{2 \times (ii)}$$        $$\mathrm{N_2(g) + 2 O_2(g) \rightarrow 2 NO_2(g)}$$         $$\mathrm{\Delta H = 2(33.9) = 67.8\: kJ}$$

$$\mathrm{2 \times (iii)}$$       $$\mathrm{2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l)}$$         $$\mathrm{\Delta H = 2(-285.8) = -571.6\: kJ}$$

overall reaction:  $$\mathrm{N_2H_4(g) + 3 O_2(g) \rightarrow 2 NO_2(g) + 2 H_2O(l)}$$  $$\mathrm{\Delta H = -554.4\: kJ}$$

exothermic

1. $$\mathrm{488\: kJ\: \times\dfrac{2\:mol\:P}{574\:kJ}=1.70\: moles\: P}$$

2. $$\mathrm{122\: g\: PCl_3 \times \dfrac{1\:mol\:PCl_3}{137.3\:g\:PCl_3}\times\dfrac{574\:kJ}{2\:mol\:PCl_3}=255\: kJ}$$

3. $$\mathrm{27.0\: kJ \times\dfrac{3\:mol\:Cl_2}{574\:kJ}\times\dfrac{70.9\:g\:Cl_2}{1\:mol\:Cl_2}= 10.0\: g\: Cl_2}$$

1. $$\mathrm{Q\: (J) = specific\: heat \left(\dfrac{J}{g\: ^\circ C}\right) \times mass\, (g) \times \Delta T\, (^\circ C)}$$;  $$\mathrm{\Delta T = 23.36 - 25.00 = -1.64^\circ C}$$.

$$\mathrm{Q = 4.18 \dfrac{J}{g\, ^\circ C} \times 50.0\: g \times (-1.64^\circ C) = -343\: joules}$$

$$\mathrm{1\: mole\: KClO_3 \times\dfrac{122.6\:g\:KClO_3}{1\:mol\:KClO_3}\times\dfrac{343\:joules}{1.00\:g\:KClO_3} = 42,100\: joules = 42.1\: kJ}$$

$$\mathrm{\Delta H = +42.1\: kJ}$$

1.                      $$\mathrm{2 C_6H_6(l) + 15 O_2(g) \rightarrow 12 CO_2(g) + 6 H_2O(l)}$$
1. $$\mathrm{6 C(s) + 3 H_2(g) \rightarrow C_6H_6(l)}$$          $$\mathrm{\Delta H = + 48.5\: kJ}$$
2. $$\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$$                $$\mathrm{\Delta H = - 393.5\: kJ}$$
3. $$\mathrm{H_2(g) + 0.5 O_2(g) \rightarrow H_2O(l)}$$         $$\mathrm{\Delta H = - 285.8\: kJ}$$

$$\mathrm{-2 \times (i)}$$      $$\mathrm{2 C_6H_6(l) \rightarrow 12 C(s) + 6 H_2(g)}$$  $$\mathrm{\Delta H = -2(+48.5) = -97.0\: kJ}$$

$$\mathrm{12 \times (ii)}$$    $$\mathrm{12 C(s) + 12 O_2(g)\rightarrow 12 CO_2(g)}$$  $$\mathrm{\Delta H = 12(-393.5) = -4722\: kJ}$$

$$\mathrm{6 \times (iii)}$$     $$\mathrm{6 H_2(g) + 3 O_2(g) \rightarrow 6 H_2O(l)}$$   $$\mathrm{\Delta H = 6(-285.8) = -1715\: kJ}$$

overall reaction: $$\mathrm{2 C_6H_6(l) + 15 O_2(g) \rightarrow 12 CO_2(g) + 6 H_2O(l)}$$   $$\mathrm{\Delta H = -6534\: kJ}$$

1. Given:  $$\mathrm{P(s) + 1.5 H_2(g) \rightarrow PH_3(g)}$$   $$\mathrm{\Delta H = +9.2\: kJ}$$

Find   $$\mathrm{\Delta H_f^\circ(P_4H_{10}) = x}$$

overall reaction: $$\mathrm{4 PH_3(g) + 8 O_2(g) \rightarrow P_4O_{10}(s) + 6 H_2O(g)}$$     $$\mathrm{\Delta H = -4500\: kJ}$$

1. $$\mathrm{P(s) + 1.5 H_2(g) \rightarrow PH_3(g)}$$       $$\mathrm{\Delta H = +9.2\: kJ}$$
2. $$\mathrm{4 P(s) + 5 O_2(g) \rightarrow P_4O_{10}(s)}$$      $$\mathrm{\Delta H = x}$$
3. $$\mathrm{H_2(g) + 0.5 O_2(g) \rightarrow H_2O(g)}$$    $$\mathrm{\Delta H = -241.8\: kJ}$$

$$\mathrm{-4 \times (i)}$$      $$\mathrm{4 PH_3(g) \rightarrow 4 P(s) + 6 H_2(g)}$$     $$\mathrm{\Delta H = -4(9.2) = -36.8\: kJ}$$

$$\mathrm{1 \times (ii)}$$        $$\mathrm{4 P(s) + 5 O_2(g) \rightarrow P_4O_{10}(s)}$$     $$\mathrm{\Delta H = x}$$

$$\mathrm{6 \times (iii)}$$       $$\mathrm{6 H_2(g) + 3 O_2(g) \rightarrow 6 H_2O(g)}$$  $$\mathrm{\Delta H = 6(- 241.8) = -1451\: kJ}$$

overall reaction: $$\mathrm{4 PH_3(g) + 8 O_2(g) \rightarrow P_4O_{10}(s) + 6 H_2O(g)}$$    $$\mathrm{\Delta H = -4500\: kJ}$$

$$\mathrm{-4500\: kJ = (-36.8 + x + -1451)\: kJ}$$;  $$\mathrm{x = -4500 + 36.8 + 1451\: kJ = -3012\: kJ}$$

$$\mathrm{\Delta H_f^\circ(P_4H_{10}) = -3012\: kJ}$$