11.
Practice Problems
1.
2.
a.
10.0 g Fe2O3 x 1 mol x -1640 kJ = -51.3 kJ
160 g
2 mol
b.
-1.00 x 104 kJ x 4 mol x 55.8 g = 1360 g
-1640 kJ 1mol
a.
10.0 g CaSO4 x 1 mol x 224 kJ = 16.5 kJ
136 g 1 mol
b.
500 kJ x 1 mol/224 kJ x 100 g/1 mol = 223 g
3.
a.
Qwater = mcT = (100 g)(4.18 J/g•K)(-1.00 K) = -418 J
b.
1.51 g NH4Cl x 1 mol/53.5 g = 0.0282 mol
c.
0.418 kJ/0.0282 mol = 14.8 kJ/mol
4.
a.
Qwater = (250 g)(4.18 J/g•K)(33.1 – 23.8)K = 9720 J
b.
12.8 g MgSO4 x 1 mol/120. g = 0.107 mol
c.
-9.72 kJ/0.107 mol = -90.8 kJ
5.
H = (H–H) + (F–F) – 2(F–H)
H = 436 + 155 – 2(567) = -543 kJ
6.
7.
H = 2(C–H) + 1(CC) + 2(H–H) – 6(C–H) – 1(C–C)
H = 2(413) + 839 + 2(436) – 6(413) – 348 = -289 kJ
a.
Q = (C + mc)T
Q = [921 + (1000 g)(4.18 J/g•K)](9.3 K) = 4.8 x 104 J
b.
1.00 g C2H4 x 1 mol/28.0 g = 0.0357 mol
c.
-48 kJ/0.0357 mol = - 1300 kJ
d.
C2H4(g) + 3 O2(g)  2 CO2(g) + 2 H2O(g)
e.
H = 4(C–H) + 1(C=C) + 3(O=O) – 4(C=O) – 4(O–H)
H = 4(413) + 614 + 3(495) – 4(799) – 4(463) = -1297 kJ/mol
8.
Melting ice at 0oC
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Distilling alcohol-water mixture
>0
x
 0
<0
x
x
x
9.
H = H1 + 4H2 = -126 kJ + 4(-179 kJ) = -842 kJ
10.
H = ½H1 + -5H2 = ½(-2512 kJ) + -5(104 kJ) = -1776 kJ
H = H1 + -H2 + -½H3
H = -198.9 kJ + 142.3 kJ + -½(495.0 kJ) = -304.1 kJ
12. a.
2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g)
b.
H = 2 HfoCO2 + 4 HfoH2O – 2 HfoCH3OH – 3 HfoO2
H = 2(-393.5) + 4(-241.8) – 2(-238.7) – 3(0.0) = -1276.8 kJ
c.
1.00 g CH3OH x 1 mol x 1276.8 kJ = 20.0 kJ
32.0 g
2 mol
d.
Q = (C + mc)T
20000 J = [C + (1200 g)(4.18 J/g•K)](3.4 K)  C = 866 J/K
13. a.
H = HfoCaSO3•2 H2O – HfoCa – HfoSO3 – 2 HfoH2O
-795 = HfoCaSO3•2 H2O – 0 – (-395.7) – 2(-285.8)
HfoCaSO3•2 H2O = -1762 kJ
b.
So = SoCaSO3•2 H2O – SoCa – SoSO3 – 2SoH2O
-0.2535 = SoCaSO3•2 H2O – 0.0414 – 0.2567 – 2(0.0699)
SoCaSO3•2 H2O = 0.1844 kJ/K
14. a.
H = HfoCu2+ + 2 HfoNO2 + 2 HfoH2O – HfoCu – 4 HfoH+ – 2 HfoNO3
H = 64.8 + 2(33.2) + 2(-285.8) – 0 – 4(0) – 2(-205.0) = -30.4 kJ
b.
So = SoCu2+ + 2 SoNO2 + 2 SoH2O – SoCu – 4 SoH+ – 2 SoNO3So = -.0996 + 2(.2400) + 2(.0699) – .0332 – 2(.1464) = 0.1942 kJ/K
c.
Go = H –TS
Go = -30.4 kJ – (298 K)(0.1942 kJ/K) = -88.3 kJ
15. a.
H = HfoNH4+ + HfoNO3- – HfoNH4NO3
H = (-132.5) + (-205.0) – (-365.6) = 28.1 kJ  endo
b.
S = SoNH4+ + SoNO3- – SoNH4NO3
S = 0.1134 + 0.1464 – 0.1511 = 0.1087 kJ/K  increase
c.
Go = H –TS = 28.1 kJ – (298 K)( 0.1087kJ/K)
Go = -4.3 kJ  spontaneous
16. a.
Ho = 2 HfoSO3 – 2 HfoSO2 – HfoO2
Ho = 2(-395.7) – 2(-296.8) – 0 = -197.8 kJ
b.
So = 2 SoSO3 – 2 SoSO2 – SoO2
So = 2(0.2567) – 2(0.2481) – 0.2050 = -0.1878 kJ/K
c.
G = Ho –TSo
G = -197.8 – (400)(-0.1878) = -122.7 kJ
d.
Tthreshold = Ho/So = (-197.8)/(-0.1878) = 1053 K
 All temperatures below 1053 K
17. a.
Ho = 2 HfoCO2 + 3 HfoH2O – HfoC2H5OH – 3 HfoO2
Ho = 2(-393.5) + 3(-285.8) – (-277.6) – 3(0) = -1366.8 kJ
b.
So = 2 SoCO2 + 3 SoH2O – SoC2H5OH – 3 SoO2
So = 2(.2136) + 3(.0699) – (.1607) – 3(.2050) = -.1388 kJ/K
c.
G = Ho –TSo = -1366.8 – (293)(-0.1388)
G = -1326.1 kJ
d.
T = Ho/So = -1366.8/-0.1388 = 9847 K
 All temperatures below 9847 K
18.
H
S
G
+/–
–
+/–
–
Justification
Temperature increases
Can't predict about dissolving
dissolving occured
2.
19.
T = Hvap/Svap = 37.95/0.1078 = 352 K (79oC)
d.
T = Ho/So = -2183.5 kJ/-0.1173 kJ/K = 19,000 K
 < 19,000 K
a.
Go = Ho –TSo = -264 kJ – (298 K)(-0.278 kJ/K)
Go = -181 kJ
b.
T = Ho/So = -264 kJ/-0.278 kJ/K = 950 K and below
Practice Multiple Choice
c.
0.256 mol NF3 x -264 kJ/2 mol NF3 = -33.8 kJ
1.
B
2.
D
3.
B
H = (I-I) + 3(Cl-Cl) – 6(I-Cl)
H = (150) + 3(240) – 6(210) = -390 kJ
H = Hcomb – 2HH2O(l)
H = -1300 kJ – 2(45 kJ) = -1390 kJ
3.
H = HfoCO2 +2 HfoH2O – HfoCH4
-900 kJ = -400 kJ + 2(-300 kJ) – HfoCH4HfoCH4 = -100 kJ
4.
D
5.
D
the first and second reactions are reversed and the
third reaction is doubled H = -x + -y + 2z
Bonds are broken  +H, liquids more disordered
than solids  +S, threshold temperature G = 0
6.
D
Large positive S for reactions that produce many
more moles of gas compared to reactants.
7.
C
The total energy constant because insulation. Entropy
increases because melted ice has more disorder.
8.
B
9.
A
10.
C
Nonspontaneous at high temperature then H and S
are negative. Spontaneous at 298 K G is negative.
H = HfoC6H6 – 3HfoC2H2
H = (80 kJ) – 3(230 kJ) = -610 kJ
Temperature drop means +H. The gas product
means +S. The reaction occurs –G.
11.
D
Threshold temperature, G = 0.
G = H – TS = 0  H = TS.
12.
B
Ho and So are negative  range of spontaneous
temperatures is all temperatures below the threshold.
Practice Free Response
1.
a.
Ho = 4HfoCO2 + 4HfoH2O – HfoC3H7COOH – 5HfoO2
-2183.5 = 4(-393.5) + 4(-285.8) – HfoC3H7COOH – 5(0)
HfoC3H7COOH = -533.7 kJ/mol
b.
So = 4SoH2O + 4SoCO2 – SoC3H7COOH -5SoO2
So = 4(0.0699) + 4(0.2136) – (0.2263) – 5(0.2050)
So = -0.1173 kJ/K
c.
Go = Ho –TSo
Go = -2183.5 kJ – (298 K)(-0.1173 kJ/K)
Go = -2148.5 kJ
4.
d.
Ho = (NN) + 3(F–F) – 6(N–F)
-264 kJ = (946 kJ) + 3(F–F) – 6(272 kJ)
 (F–F) = 141 kJ/mol
a.
G is negative because AgNO3 is soluble in water, 
the solution process is spontaneous and G < 0.
b.
H is positive because heat is absorbed during
dissolving.
c.
S is positive because G = H – TS and G < 0 and
H > 0  S must be > 0.
a.
-3119.4 kJ
2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(l)
+1559.7 kJ
2 CO2(g) + 3 H2O(l) C2H6(g) + 7/2 O2(g) 
b.
C2H6(g) + 7/2 O2(g)  2 CO2(g) + 3 H2O(l)
-1559.7 kJ
3 H2O(l)  3 H2O(g)
3(+44.0 kJ)
C2H6(g) + 7/2 O2(g)  2 CO2(g) + 3 H2O(g)
-1427.7 kJ
c.
H = 2 HfoCO2 + 3 HfoH2O(l) – HfoC2H6 + 7/2 HfoO2
-1559.7 = 2(-393.5) + 3(-285.8) - HfoC2H6 + 7/2(0)
 HfoC2H6 = -84.7 kJ
d.
1.00 g C2H6 x 1 mol/30.0 g x -1559.7 kJ/mol = -52.0 kJ
e.
q = (C + mc)T
52,000 J = [C + (250 g)(4.18 J/goC)](13.13oC)
C = 2920 J/oC