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Chemistry LibreTexts

Answers to Chapter 06 Study Questions

  • Page ID
    11313

    1. 1.png

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