KEY - Problem Set: Heat, Hess’s Law and Heats of Formation
1.
The production of iron and carbon dioxide from iron (II) oxide and carbon monoxide is an
exothermic reaction :
FeO (s) + CO(g)  Fe (s) + CO2 (g) + 26.3 kJ
How much kilojoules of heat are produced when 3.40 grams of FeO reacts with an excess
amount of CO?
1 mol Fe 2O3
26.3 kJ

 3.40 g  1.25 kJ
1 mol Fe 2O3
71.8 g
2. The burning of magnesium in oxygen is a very exothermic reaction:
2Mg + O2  2MgO + 1204 kJ
How many kilojoules are given off when 6.55 g of Mg reacts with an excess amount of
oxygen?
1204 kJ 1 mol Mg

 6.55 g  162 kJ
2 mol Mg 24.3 g
3. A considerable amount of heat is required for the decomposition of aluminum oxide:
2Al2O3  4Al + 3O2
H = +3352 kJ
how many grams of Al are produced when 5783 kJ of heat is absorbed by the reaction?
3352 kJ 5783 kJ

4 mol Al x mol Al
x  6.90095 mol Al
6.90095 mol Al 27.0 g

 186.3 g Al
1 mol
4. The combustion of ethane, C2H4, is an exothermic reaction:
C2H4 + O2  2CO2 2H2O
H = -1390 kJ
Calculate the heat liberated when 4.79 g of ethene burns.
1390 kJ 1 mol C 2 H 4

 4.79 g  238 kJ
1 mol C 2 H 4
28.0 g

5. Calculate H for the formation of lead (IV) chloride by the reaction of lead (II) chloride
with chlorine:
PbCl2 + Cl2  PbCl4
H = ?
Use the following thermochemical equations:
Pb + 2Cl2  PbCl4
H = -329.2 kJ
Pb + Cl2  PbCl2
H = -359.4 kJ flip
PbCl2  Pb + Cl2
H = +359.4 kJ
PbCl2 + Cl2  PbCl4
H = +30.2 kJ
6. Calculate the heat of reaction for : NO + ½ O2  NO2
From the following reactions…
½ N2 + ½ O2  NO
H = +90.4 kJ/mol flip
½ N2 + O2  NO2
H = +33.6 kJ/mol
NO  ½ N2 + ½ O2
H = -90.4 kJ/mol
NO + ½ O2  NO2
H =-56.8 kJ/mol
7. Calculate the heat change for the formation of copper (I) oxide from the elements:
Cu + ½ O2  CuO
Use the following two thermochemical equations to make the calculations:
CuO + Cu  Cu2O
H = -11.3 kJ
Cu2O + ½ O2  2CuO
H = -114.6 kJ
add as is
Cu + ½ O2  CuO
H =-125.9 kJ
8. Find the enthalpy change (H) for the formation of phosphorus pentachloride from its
elements.
2P + 5Cl2  2PCl5
Use the following equations:
PCl5  PCl3 + Cl2
H = +87.9 kJ x 2 and flip
2P + 3 Cl2  2PCl3
H = -574 KJ
2PCl3 + 2Cl2  2PCl5
H = -175.8 kJ
2P + 5Cl2  2PCl5
H =-750. kJ
9. Calculate the heat change, H, for the formation of nitrogen monoxide from its elements:
N2 + O2  2NO
Use these equations:
4NH3 + 3O2  2N2 + 6H2O
H = -1530 kJ flip
4NH3 + 5O2  4NO + 6H2O
H = -1170 kJ
2N2 + 6H2O  4NH3 + 3O2
H = +1530 kJ
2N2 + 2O2  4NO
H = +360 kJ divide by 2
N2 + O2  2NO
H =+180 kJ
10. Calculate the heat change, H, for the formation of nitrogen monoxide from its elements:
H2S(g) + 4F2(g)  2HF + SF6(g)
Use these equations:
½ H2(g) + ½ F2 (g)  HF(g)
H = -273 kJ multiply by 2
S(s) + 3F2 (g)  SF6(g)
H = -1220 kJ
H2 (g) + S(s)  H2S(g)
H =
-21 kJ flip
H2S(g)  H2 (g) + S(s)
H = +21 kJ
H2(g) + F2 (g)  2HF(g)
H = -546 kJ
H2S(g) + 4F2(g)  2HF + SF6(g)
H =-1745 kJ