# Answers to Chapter 06 Study Questions

1. exothermic, reactants
2. exothermic, reactants
3. endothermic, products
4. exothermic, reactants
5. exothermic, reactants
6. endothermic, products

1. $$\mathrm{\Delta H_f^\circ(C_3H_8) = -103.8\: kJ/mol}$$
2. $$\ce{3 C(s) + 4 H2(g) \rightarrow C3H8(g)}$$          $$\mathrm{\Delta H = -103.8\: kJ}$$
3. exothermic
4. $$\mathrm{1090\: kJ \times\dfrac{4\:mol\:H_2}{103.8\:kJ}=42.0\:moles\:H_2}$$
5. $$\mathrm{30.1\:g\:C_3H_8 \times\dfrac{1\:mol\:C_3H_8}{44.0\:g\:C_3H_8}\times\dfrac{103.8\:kJ}{1\:mol\:C_3H_8} =71.0\: kJ}$$
6. $$\ce{C3H8(g) + 5 O2(g) \rightarrow 3 CO2(g) + 4 H2O(l)}$$
7. $$\mathrm{\Delta H(reaction) = 3 \Delta H_f^\circ(CO_2) + 4 \Delta H_f^\circ(H_2O) -\Delta H_f^\circ(C_3H_8)}$$
$$\mathrm{=3 (-393.5\: kJ) + 4 (-285.8\: kJ) - (-103.8\: kJ)}$$
$$\mathrm{= -1180 + (-1143) + 103.8 = -2220\: kJ}$$

1. from the $$\mathrm{\Delta H_f^\circ}$$ Table:
1. $$\mathrm{\dfrac{1}{2} N_2(g) + \dfrac{1}{2} O_2(g) \rightarrow NO(g)}$$     $$\mathrm{\Delta H = + 90.4\: kJ}$$

2. $$\mathrm{\dfrac{1}{2} N_2(g) + O_2(g) \rightarrow NO_2(g)}$$     $$\mathrm{\Delta H = +33.9\: kJ}$$;

$$\mathrm{-2 \times (i) =}$$         $$\mathrm{2 NO(g) \rightarrow N_2(g) + O_2(g)}$$     $$\mathrm{\Delta H = -2(90.4) = -180.8\: kJ}$$
$$\mathrm{2 \times (ii) =}$$         $$\ce{N2(g) + 2 O2(g) \rightarrow 2 NO2(g)}$$    $$\mathrm{\Delta H = 2(33.9) = 67.8\: kJ}$$;
$$\mathrm{overall\: reaction = 2 NO(g) + O_2(g) \rightarrow 2 NO_2(g)}$$    $$\mathrm{\Delta H = -180.8 + 67.8 = -113.0\: kJ}$$
Exothermic

1. $$\mathrm{Pb(s) + 0.5 O_2(g) \rightarrow PbO(s)}$$         $$\mathrm{\Delta H = -217.9\: kJ}$$
2. $$\mathrm{3 Pb(s) + 2 O_2(g) \rightarrow Pb_3O_4(s)}$$      $$\mathrm{\Delta H = -734.7\: kJ}$$; therefore:

$$\mathrm{2 \times (ii) = 6 Pb(s) + 4 O_2(g) \rightarrow 2 Pb_3O_4(s)}$$    $$\mathrm{\Delta H = 2(-734.7) = -1469\: kJ}$$
$$\mathrm{-6 \times (i) = 6 PbO(s) \rightarrow 6 Pb(s) + 3 O_2(g)}$$    $$\mathrm{\Delta H = -6(-217.9) = +1307\: kJ}$$
$$\mathrm{overall\: reaction = 6 PbO(s) + O_2(g) \rightarrow 2 Pb_3O_4(s)}$$    $$\mathrm{\Delta H = -1469 + 1307 = -162\: kJ}$$
Exothermic

1. $$\mathrm{Q\: (J) = specific\: heat\: \left(\dfrac{J}{g\: ^\circ C}\right) \times mass\: (g)\times \Delta T \:(^\circ C)}$$;  $$\mathrm{\Delta T = 19.23 - 24.78 = -5.55\: ^\circ C}$$
$$\mathrm{Q = 4.18 \dfrac{J}{g\: ^\circ C} \times 60.0\: g \times -5.55\: ^\circ C = 1390\: J}$$
$$\mathrm{1\:mole\:NH_4Cl \times \dfrac{1390\:J}{5.03\:g\:NH_4Cl}\times\dfrac{53.5\:g\:NH_4Cl}{1\:mol\:NH_4Cl}=14,800\:J=14.8\: kJ/mole}$$
Endothermic

1. $$\mathrm{Q = 6485\: \dfrac{J}{^\circ C} \times 10.7^\circ C = 69,400\: J = 69.4\: kJ}$$
$$\mathrm{1\: mole\: C_2H_4 \times \dfrac{28.0\:g\:C_2H_4}{1\:mol\:C_2H_4}\times\dfrac{69.4\:kJ}{1.40\:g\:C_2H_4}= 1390\: kJ = 1.39 \times 10^3\: kJ}$$

1. $$\mathrm{2 NH_3(g) + 3 N_2O(g) \rightarrow 4 N_2(g) + 3 H_2O(l)} \hspace{38px} \mathrm{\Delta H = -1010\: kJ}$$

\begin{align} & \mathrm{3 N_2H_4(l) + 3 H_2O(l) \rightarrow 3 N_2O(g) + 9 H_2(g)} && \mathrm{\Delta H = 3(+317)\: kJ}\\ & \mathrm{N_2H_4(l) + H_2O(l) \rightarrow 2 NH_3(g) + \dfrac{1}{2} O_2(g)} && \mathrm{\Delta H = +143\: kJ} \end{align}
$$\underline{\mathrm{9 H_2(g) + 4\dfrac{1}{2} O_2(g)\rightarrow 9 H_2O(l)}\hspace{109px} \mathrm{\Delta H = 9(-286)\: kJ}}$$

$$\mathrm{4 N_2H_4(l) + 4 O_2(g) \rightarrow 4 N_2(g) + 8 H_2O(l)} \hspace{46px} \mathrm{\Delta H = -2490\: kJ}$$

for the reaction, $$\mathrm{N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2 H_2O(l)}$$,  $$\mathrm{\Delta H = \dfrac{(-2490)}{4} = -623\: kJ}$$