CH 13 HW:
Hess’s Law:
1.
N2(g) + 2O2(g)  2NO2(g)
N2O4(g)  N2(g) + 2O2(g)
2NO2(g)  N2O4(g)
H= -9.16 KJ
H= -57.24 KJ
Enthalpy of formation = +66.4 kJ
Bond enthalpy = +285 kJ
Hess = +66.4 kJ
2.
N2(g) + 3H2(g)  2NH3(g)
2H2O(l) + NH3(g)  NO2(g) + 7/2H2(g)
1/2N2(g) + 2H2O(l)  NO2(g) + 2H2(g)
H= 650.97 KJ
H= 604.86 KJ
-92.2 kJ
-81.17 kJ
-92.22 kJ
3.
2Fe3O4(s) + CO2(g)  3Fe2O3(s) + CO(g)
2Fe(s) + 3CO2(g)  Fe2O3(s) + 3CO(g)
Fe3O4 + CO(g)  3FeO(s) + CO2(g)
Fe(s) + CO2(g)  FeO(s) + CO(g)
H= -541.21 KJ
H= 863.01 KJ
H= 10.99 KJ
+47.2 kJ
????
+36.45 kJ
4.
C3H8(g)  3 C(graphite) + 4H2(g)
C3H8(g) + 5O2(g)  3CO2 + 4 H2O(l)
CO2  C(graphite) + O2
H2(g) + 1/2O2(g)  H2O(l)
H= -575.49 KJ
H= 393.51 KJ
H= -285.83 KJ
+103.8 kJ, +2264.38 kJ, +1748 kJ
5.
C2H6(g)  C2H2(g) + 2H2(g)
C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(g)
H2O(g)  H2(g) + 1/2O2(g)
H= 241.82 KJ
2CO2(g) + 3H2O(g)  C2H6(g) + 7/2O2(g)
311.4 kJ
297 kJ
310.74 kJ
6.
P4(s) + 6Cl2(g)  4PCl3(g)
P4(s) + 10Cl2(g)  4PCl5(g)
PCl3(g) + Cl2(g)  PCl5(g)
-1148 kJ
-1255.2 kJ
-1148 kJ
H= -1255.54 KJ
H= 1427.8 KJ
P4 = 6 single
H= -1500 KJ
white P
H= -88 KJ
7.
2NO2(g)  N2(g) + 2O2(g)
N2(g) + 3H2(g)  2NH3(g)
H= -92.22 KJ
2NO2(g) + 7H2(g)  2NH3(g) + 4H2O(l)
H2O(l)  H2(g) + 1/2O2(g)
-66.4 kJ
-284.5 kJ
+66.4 kJ
H= -1169.14 KJ
H= 285.83 KJ
8. Which compound has the higher lattice energy: K2O or Na2O? KCl or SrCl2?
Na2O  smaller net radii; SrCl2  larger charges and more ions
9. Using the Kapustinskii equation, calculate the lattice energy for KCl and SrCl2
KCl = -672.38 kJ/mol, SrCl2 = -2134.62 kJ/mol
Standard Bond Energies
Single
Bonds
ΔH�*
Single
Bonds
ΔH�*
Multiple
Bonds
ΔH�*
H–H
104.2
B–F
150
C=C
146
C–C
83
B–O
125
N=N
109
N–N
38.4
C–N
73
O=O
119
O–O
35
N–CO
86
C=N
147
F–F
36.6
C–O
85.5
C=O (CO2)
192
Si–Si
52
O–CO
110
C=O (aldehyde)
177
P–P
50
C–S
65
C=O (ketone)
178
S–S
54
C–F
116
C=O (ester)
179
Cl–Cl
58
C–Cl
81
C=O (amide)
179
Br–Br
46
C–Br
68
C=O (halide)
177
I–I
36.
C–I
51
C=S (CS2)
138
H–C
99
C–B
90
N=O (HONO)
143
H–N
93
C–Si
76
P=O (POCl3)
110
H–O
111
C–P
70
P=S (PSCl3)
70
H–F
135
N–O
55
S=O (SO2)
128
H–Cl
103
S–O
87
S=O (DMSO)
93
H–Br
87.5
Si–F
135
P=P
84
H–I
71
Si–Cl
90
P≡P
117
H–B
90
Si–O
110
C≡O
258
H–S
81
P–Cl
79
C≡C
200
H–Si
75
P–Br
65
N≡N
226
H–P
77
P–O
90
C≡N
213
* Average Bond Dissociation Enthalpies in kcal mole-1
Lab: Hess’s Law
Part 1:
/30
1. Using a graduated cylinder, place 100.0 mL of water in a Styrofoam cup.
2. Obtain a cardboard lid containing 2 holes, a thermometer, and a stirring rod.
3. Record the initial temperature of the water.
4. Mass out 2.00g of sodium hydroxide and record.
5. Transfer the sodium hydroxide to the cup, place the cover on the top and stir.
Record the maximum temperature reached (approximately 5 minutes).
6. Dispose of solution 1.
Reaction: NaOH(s)  Na+ + OHPart 2:
1. Make 100 mL of .5 M hydrochloric acid solution and add it to your Styrofoam
cup.
2. Let this sit for 2-3 minutes and record the temperature.
3. Mass 2.00g of sodium hydroxide and record the mass.
4. Add the sodium hydroxide to the hydrochloric acid, cover, stir, and record the
maximum temperature reached (approximately 5 minutes)
5. Dispose of solution 2.
Reaction: NaOH(s) + H+ + Cl-  H2O(l) + Na+ + ClPart 3:
1. Make 50 mL of 1M hydrochloric acid and add to the Styrofoam cup.
2. Make 50 mL of 1M sodium hydroxide.
3. Let both solutions sit for 2-3 minutes and then determine and record the
temperature of both of the solutions.
4. Add the sodium hydroxide to the Styrofoam cup containing the hydrochloric acid.
Stir this mixture and record the highest temperature reached (2-3 minutes).
Reaction: Na+ + OH- + H+ + Cl-  H2O(l) + Na+ + Cl-
Calculations:
Part 1:
1. Mass of 100 mL of water
2. Temperature change of the water
3. Heat produced (kJ)
4. Moles of NaOH
5. Heat produced/mol NaOH (kJ/mol)
6. ΔH1 =
kJ/mol
Part 2:
1. Mass 100 mL of hydrochloric acid.
2. Temperature change of solution.
3. Heat produced (kJ).
4. Moles NaOH
5. Heat produced/mol NaOH (kJ/mol)
6. ΔH2 =
kJ/mol
Part 3:
1. Mass of 100 mL of solution
2. Average initial temperature of solutions.
3. Temperature change of solution.
4. Heat produced (kJ).
5. Moles sodium hydroxide used.
6. Heat produced/mol NaOH (kJ/mol)
7. ΔH3 =
kJ/mol
Questions:
1. Write the net ionic equation for each of the 3 reactions on page 1 along with the
appropriate enthalpy changes.
NaOH(s)  Na+ + OHNaOH(s) + H+  H2O(l) + Na+
OH- + H+  H2O(l)
2. Using Hess’s Law, compare the enthalpy change of reaction #2 to that of
reactions #1 + #3.
3. Calculate the percent error using reaction #2 as the actual and the net of #1 and #3
as the experimental.
4. The dissolving of ammonium nitrate in water is endothermic. How could this be
utilized for the treatment of athletic injuries?
5. If you repeated Part 1 of the lab but used twice as much sodium hydroxide, how
would this affect the amount of heat produced? How would it affect the ΔH
value? Would double assuming the solution is not saturated. Same because in
KJ/mol
6. If double the amount of NaOH were used in Part 2, without changing any other
concentrations or volumes, would this double the amount of heat produced in this
reaction? Explain. No. HCl is limiting
.1 L HCl x .5 mol HCl x 1 mol NaOH x 40 g NaOH = 2 g NaOH
1 L HCl
1 mol HCl
1 mol NaOH