Chemistry Department
CHM 1220/1225 Winter 2006
Sample Exam New Material for Final Exam Answers
Note: All values for ΔHf0, S0, and ΔGf0 are from Appendix C
1. 270.8 kJ
0
2. a. ΔS will be positive because one gas molecule is breaking into two gas molecules. ΔS = +175.5 J/K (4 sf)
0
b. ΔS will be positive because three gas molecule are produced. ΔS = +507.0 J/K
0
c. ΔS will be positive because three gas molecule are produced. ΔS = +33.51 J/K
0
d. ΔS is difficult to predict. ΔS = –218.5 J/K
0
3. a. ΔH = 57.1 kJ (1 dp)
0
b. ΔH = +1411.2 kJ (1 dp)
0
c. ΔH = 11.6 kJ (1 dp)
0
d. ΔH =–976.3 kJ (1 dp)
0
4. a. ΔG = 4.52 kJ (2 dp)
0
b. ΔG = +1260.0 kJ (2 dp)
0
c. ΔG = 1.7 kJ (1 dp)
0
d. ΔG =–1012.1 kJ (1 dp)
5. a. ΔG = –166.3 kJ (1 dp)
0
0
d. ΔG = ΔH – TΔS = –1254.5 kJ (1 dp)
91
6. a. K = 6.00 x 10
c. K = 0.50
7. a. T = 325 K
b. T = 2783 K
c. T = 346 K
8. a. ΔH>0 and ΔS>0. ΔG depends on temperature (+ at low temperature; – at high temperature)
b. ΔH>0 and ΔS>0. ΔG depends on temperature (– at low temperature; + at high temperature)
9. a. ΔU = 56 J
b. ΔU = –71.0 J
c. ΔU = 0.84 J
d. ΔU = –1322 J
e. ΔU = –3140 J
f. ΔU = –8.65 kJ
10. pH = 2.17; pOH = 11.83
–5
11. Kb = 2.2 x 10
–5
12. a. Kb = 2.0 x 10
–11
b. Ka = 2.3 x 10
–3
13. [C5H5N] = 1.0 x 10 M; pH = 3.00
14. pH = 8.08
–2
15. [HC3H5O3] = 7.4 x 10 M
More detailed solutions on next pages.
CHM 1220/1225
2
Page 2 of 3
a.
N2O4(g) ––>
2 NO2(g)
0
Sf , J/(mol•K)
304.3 J/(mol•K) • 1 mol = 304.3 J/K
239.9 J/(mol•K) • 2 mol = 479.8 J/K
0
9.079 kJ/mol • 1mol = 9.079 kJ
33.1 kJ/mol • 2 mol = 66.2 kJ
ΔHf (kJ/mol)
0
97.72 kJ/mol • 1 mol = 97.72
51.12 kJ/mol • 2 mol = 102.24 kJ
ΔGf (kJ/mol)
0
ΔS = 479.8 J/K – 304.3 J/K = +175.5 J/K (4 sf)
0
ΔH = 66.2 kJ – 9.079 kJ = 57.121 kJ = 57.1 kJ (1 dp)
0
ΔG = 102.24 kJ – 97.72 kJ = 4.52 kJ (2 dp)
0
0
ΔG = ΔH – TΔS = 57.121 kJ – 1273 K • 0.1755 kJ/K = 57.121 kJ – 223.4115 kJ = –166.3 kJ (1 dp)
K=e
166.3x10 3 J
J
8.3145
•298K
mol•K
67.1181
29
=e
= 1.41 x 10
T = ΔH /ΔS = 57.1 kJ/0.1755 kJ/K = 325 K
€
0
0
b. 2 AlCl3(s) ––> 2 Al(s) + 3 Cl2(g)
2 AlCl3(s) ––>
0
Sf , J/(mol•K)
109.3 J/(mol•K) • 2 mol = 218.6 J/K
2 Al(s) +
3 Cl2(g)
28.28 J/(mol•K) • 2 mol 223.0 J/(mol•K) • 3 mol
= 56.56 J/K
= 669.0 J/K
0
–705.6 kJ/mol • 2 mol = –1411.2 kJ
0 kJ/mol • 2 mol = 0 kJ
0 kJ/mol • 3 mol = 0 kJ
ΔHf (kJ/mol)
0
–630.0 kJ/mol • 2 mol = –1260.0 kJ
0 kJ/mol • 2 mol = 0 kJ
0 kJ/mol • 3 mol = 0 kJ
ΔGf (kJ/mol)
0
ΔS = (56.56 + 669.0) J/K – 218.6 J/K = (725.56 – 218.6) J/K = +506.96 J/K = 507.0 J/K
0
ΔH = 0 kJ – (–1411.2 kJ) = +1411.2 kJ (1 dp)
0
ΔG = 0 kJ – (–1260.0 kJ) = +1260.0 kJ (2 dp)
0
0
T = ΔH /ΔS = 1411.2 kJ/0.5070 kJ/K = 2783 K
c.
Be(OH)2(s) ––> BeO(s) + H2O(l)
Be(OH)2(s) ––>
BeO(s) +
H2O(l)
0
Sf , J/(mol•K)
50.21 J/(mol•K) • 1 mol =
13.77 J/(mol•K) • 1 mol =
69.95 J/(mol•K) • 1 mol =
50.21 J/K
13.77 J/K
69.95 J/K
0
–905.8 kJ/mol • 1 mol =
–608.4 kJ/mol • 1 mol =
–285.8 kJ/mol • 1 mol =
ΔHf (kJ/mol)
–905.8 kJ
–608.4 kJ
–285.8 kJ
0
–817.9 kJ/mol • 1 mol =
–579.1 kJ/mol • 1 mol =
–237.1 kJ/mol • 1 mol =
ΔGf (kJ/mol)
–817.9 kJ
–579.1 kJ
–237.1 kJ
0
ΔS = (13.77 + 69.95) J/K – 50.21 J/K = (83.72 – 50.21) J/K = +33.51 J/K
0
ΔH = (–608.4 + –285.8) kJ – (–905.8 kJ) = –894.2 kJ – (–905.8 kJ) = 11.6 kJ (1 dp)
0
ΔG = (–579.1 + –237.1) kJ – (–817.9 kJ) = (–816.2 kJ) – (–817.9 kJ) = 1.7 kJ (1 dp)
K= e
–1.7x10 3 J
J
8.3145
•298K
mol•K
–0.6861
=e
= 0.50
T = ΔH /ΔS = 11.6 kJ/0.03351 kJ/K = 346 K
€
0
0
d. N2O4(g) + 4 H2(g) ––> N2(g) + 4 H2O(g)
N2O4(g) +
4 H2(g) ––>
N2(g) +
4 H2O(g)
0
Sf , J/(mol•K)
303.3 J/(mol•K) • 1
130.6 J/(mol•K) • 4
191.6 J/(mol•K) • 1
188.7 J/(mol•K) • 4
mol = 303.3 J/K
mol = 521.8 J/K
mol = 191.6 J/K
mol = 754.8 J/K
0
9.079
kJ/mol
•
1
mol
0
kJ/mol
•
4
mol
=
0
0
kJ/mol
•
1
mol
=
0
–241.8 kJ/mol • 4
ΔHf (kJ/mol)
= 9.079 kJ
kJ
kJ
mol = –967.2 kJ
0
97.72 kJ/mol • 1 mol 0 kJ/mol • 4 mol = 0
0 kJ/mol • 1 mol = 0
–228.6 kJ/mol • 4
ΔGf (kJ/mol)
= 97.72 kJ
kJ
kJ
mol = –914.4 kJ
0
ΔS = (191.6 + 754.8) J/K – (303.3 + 521.8) J/K = (946.4 – 1164.9) J/K = –218.5 J/K
0
ΔH = (0 + –967.2) kJ – (9.079 + 0)kJ) = –967.2 kJ – (9.079 kJ) = –976.279 kJ = –976.3 kJ (1 dp)
0
ΔG = (0 + –914.4) kJ – (97.72 + 0) kJ = (–914.4 kJ) – (97.72 kJ) = –1012.1 kJ (1 dp)
0
0
0
ΔG = ΔH – TΔS = –976.3 kJ – 1273 K • 0.2185 kJ/K = (–976.3 – 278.1505) kJ = –1254.5 kJ (1 dp)
9. a. q = 85 J; w = –29 J; ΔU = q + w = 56 J
b. q = –57.5 J; w = –13.5 J; ΔU = q + w = –71.0 J
c. ΔU = q + w = 0.84 J
CHM 1220/1225
Page 3 of 3
d. q = –900 J; w = –422 J; ΔU = q + w = –1322 J
e. q = –3140 J, w = 0 J; ΔU = q + w = –3140 J
f. q = –8.65 kJ; w = 0 J; ΔU = q + w = –8.65 kJ
10. [NH3] = 2.5 M
NH3(aq) +
H2O(l) ⇔
Start
2.5
Change
–x
Equilibrium
2.5 – x
Kb =
2
€
+
NH4 ( aq) +
0
+x
x
[NH+4 ][OH– ]
x2
=
= 1.8x10 –5 ; assuming (2.5 – x) = 2.5
[NH3 ]
(2.5 − x)
–5
-5
x = 2.5(1.8 x 10 ) = 4.5 x 10
–3
x = 6.7 x 10
+
-3
[H3O ] = x = 6.7 x 10 M
pH = 2.17; pOH = 11.83
11.
Bz(aq) +
Start
0.068
Change
–x
Equilibrium
0.068 – x
–
–3
pOH = 2.91; [OH ] = 1.2 x 10 M
+
H2O(l) ⇔
HBz ( aq) +
0
+x
x
[HBz + ][OH– ]
(1.2x10–3 ) 2
Kb =
=
= 2.2x10 –5
[Bz]
(0.068 − 0.0012)
–
€
–
OH (aq)
0
+x
x
–14
–10
–
OH (aq)
0
+x
x
–5
12. a. Kb for CN = Kw/Ka for HCN = 1.0 x 10 /4.9 x 10 = 2.0 x 10
+
–14
–4
–11
b. Ka for CH3NH3 = Kw/Kb for CH3NH2 = 1.0 x 10 /4.4 x 10 = 2.3 x 10
13.
+
+
C5H5NH (aq) +
H2O(l) ⇔
H3O ( aq)
C5H5N(aq)
Start
0.15
0
0
Change
–x
+x
+x
Equilibrium
0.15 – x
x
x
Assume that 0.15 – x = 0.15
x2
Kw
Ka =
=
= 7.14 x10–6
–9
0.15 1.4x10
2
14.
€
–6
x = 1.07 x 10
–3
–3
x = 1.0 x 10 M; pH = 3.00; [C5H5N] = 1.0 x 10 M
NH3(aq) +
0.10
–x
0.10 – x
Start
Change
Equilibrium
Kb =
H2O(l) ⇔
–5
–
–5
M; pOH = 5.92; pH = 8.08
15.
HLac(aq) +
H2O(l) ⇔
Start
C
Change
–x
Equilibrium
C–x
+
–2.51
–3
pH = 2.51; [H3O ] = 10
= 3.1 x 10 M
x2
Ka =
= 1.3x10 –4 ; x = 3.1 x 10–3 M
C− x
–2
7.4 x 10 = C – 1.3 x 10
–2
C = 7.4 x 10 M
€
–
OH (aq)
0
+x
x
[NH+4 ][OH– ] x(0.15 + x)
=
= 1.8x10 –5 ; assume (0.15 + x) = 0.15 and (0.10 – x) = 0.10
[NH3 ]
(0.10 − x)
x = 1.2 x 10 ; [OH ] = 1.2 x 10
€
+
NH4 ( aq) +
0.15
+x
0.15 + x
–4
€
+
H3O ( aq)
0
+x
x
(3.1x10 –3 ) 2
1.3x10–4
Lac(aq)
0
+x
x
= (C − 1.3x10 –4 )