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Answers to More Chapter 06 Study Questions

1. 

  1. From the ΔHf° Table: 
    1. 0.5 H2(g)  +  0.5 Br2(l)   →  HBr(g)    ΔH = - 36.2 kJ
    2. 0.5 H2(g)  +  0.5 Cl2(g)   →  HCl(g)    ΔH = - 92.3 kJ

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

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

overall reaction:   2 HCl(g)  +  Br2(l)   →   2 HBr(g)  +  Cl2(g)   ΔH = -72.4 + 184.6 = 112.2 kJ

endothermic

 

  1.    
    1. 0.5 N2(g)  +  1.5 H2(g)   →   NH3(g)       ΔH = -46.2 kJ
    2. H2(g)  +  0.5 O2(g)   →   H2O(l)         ΔH  =  -285.8 kJ

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

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

overall reaction:  4 NH3(g)  +  3 O2(g)   →   2 N2(g)  +  6 H2O(l)  ΔH = 184.8 -1715 = -1530 kJ

exothermic

 

  1.     
    1. N2(g)  +  2 H2(g)   →   N2H4(l)          ΔH = + 50.6 kJ
    2. 0.5 N2(g)  + O2(g)   →   NO2(g)         ΔH = + 33.9 kJ
    3. H2(g)  +  0.5 O2(g)   →   H2O(l)         ΔH = - 285.8 kJ

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

      \(\mathrm{2 \times (ii)}\)        \(\mathrm{N_2(g) + 2 O_2(g) \rightarrow 2 NO_2(g)}\)         ΔH = 2(33.9) =  67.8 kJ

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

overall reaction:  N2H4(g)  +  3 O2(g)   →   2 NO2(g)  +  2 H2O(l)  ΔH = -554.4 kJ

exothermic

 

2. 

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

 

3.  Q (J)  =  specific heat (J/g °C)   x   mass (g)   x   ΔT (°C);  ΔT  =  23.36 -25.00 = -1.64°C.

      Q  =  4.18 J/g °C  x  50.0 g  x  (-1.64°C)  =  -343 joules

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

      ΔH  =  +42.1 kJ

 

4.                     2 C6H6(l)  +  15 O2(g)   →   12 CO2(g)  +  6 H2O(l)

  1. 6 C(s)  +  3 H2(g)   →   C6H6(l)          ΔH = + 48.5 kJ
  2. C(s)  +  O2(g)   →   CO2(g)                ΔH = - 393.5 kJ
  3. H2(g)  +  0.5 O2(g)   →   H2O(l)         ΔH = - 285.8 kJ

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

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

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

overall reaction: 2 C6H6(l)  +  15 O2(g)   →   12 CO2(g)  +  6 H2O(l)   ΔH = -6534 kJ

 

5.  Given:  P(s)  +  1.5 H2(g)   →   PH3(g)   ΔH = +9.2 kJ

      Find   ΔHf°(P4H10) = x

overall reaction: 4 PH3(g)  +  8 O2(g)   →   P4O10(s)  +  6 H2O(g)     ΔH = -4500 kJ

  1. P(s)  +  1.5 H2(g)   →   PH3(g)       ΔH = +9.2 kJ
  2. 4 P(s)  +  5 O2(g)   →   P4O10(s)      ΔH = x
  3. H2(g)  +  0.5 O2(g)   →   H2O(g)    ΔH = -241.8 kJ

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

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

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

overall reaction: 4 PH3(g)  +  8 O2(g)   →   P4O10(s)  +  6 H2O(g)    ΔH = -4500 kJ

      -4500 kJ  =  (-36.8  +  x  +  -1451) kJ;  x  =  -4500 + 36.8 + 1451 kJ = -3012 kJ

      ΔHf°(P4H10)   =   -3012 kJ