Honors Chemistry

Answers to Chapter 6 Study Questions

 

1. reactants

20

DH = -20 kJ products

0

 

2. a) exothermic, reactants b) exothermic, reactants c) endothermic, products

d) exothermic, reactants e) exothermic, reactants f) endothermic, products

 

3. a) DHf(C3H8) = -103.8 kJ/mol

b) 3 C(s) + 4 H2(g) C3H8(g) DH = -103.8 kJ

c) exothermic

d) = 42.0 moles H2

e) = 71.0 kJ

f) C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)

g) DH(reaction) = 3 DHf(CO2) + 4 DHf(H2O) - DHf(C3H8)

= 3 (-393.5 kJ) + 4 (-285.8 kJ) - (-103.8 kJ)

= -1180 + (-1143) + 103.8 = -2220 kJ

 

4. a) from the DHf Table:

(i) N2(g) + O2(g) NO(g) DH = + 90.4 kJ

(ii) N2(g) + O2(g) NO2(g) DH = +33.9 kJ;

-2 x (i) = 2 NO(g) N2(g) + O2(g) DH = -2(90.4) = -180.8 kJ

2 x (ii) = N2(g) + 2 O2(g) 2 NO2(g) DH = 2(33.9) = 67.8 kJ;

overall reaction = 2 NO(g) + O2(g) 2 NO2(g) DH = -180.8 + 67.8 = -113.0 kJ

Exothermic

 

b) (i) Pb(s) + 0.5 O2(g) PbO(s) DH = -217.9 kJ

(ii) 3 Pb(s) + 2 O2(g) Pb3O4(s) DH = -734.7 kJ; therefore:

2 x (ii) = 6 Pb(s) + 4 O2(g) 2 Pb3O4(s) DH = 2(-734.7) = -1469 kJ

-6 x (i) = 6 PbO(s) 6 Pb(s) + 3 O2(g) DH = -6(-217.9) = +1307 kJ

overall reaction = 6 PbO(s) + O2(g) 2 Pb3O4(s) DH = -1469 + 1307 = -162 kJ

Exothermic

 

 

5. Q (J) = specific heat (J/g C) x mass (g) x DT (C); DT = 19.23 - 24.78 = -5.55 C

Q = 4.18 J/g C x 60.0 g x -5.55 C = 1390 J

= 14,800 J = 14.8 kJ/mole

 

Endothermic

 

6. Q = 6485 J/C x 10.7C = 69,400 J = 69.4 kJ

= 1390 kJ = 1.39 x 103 kJ

 

7. 2 NH3(g) + 3 N2O(g) 4 N2(g) + 3 H2O(l) DH = -1010 kJ

3 N2H4(l) + 3 H2O(l) 3 N2O(g) + 9 H2(g) DH = 3(+317) kJ

N2H4(l) + H2O(l) 2 NH3(g) + O2(g) DH = +143 kJ

9 H2(g) + 4 O2(g) 9 H2O(l) DH = 9(-286) kJ

4 N2H4(l) + 4 O2(g) 4 N2(g) + 8 H2O(l) DH = -2490 kJ

for the reaction, N2H4(l) + O2(g) N2(g) + 2 H2O(l), DH = (-2490)/4 = -623 kJ