Chemistry 30
We know that chemical reactions actually occur in steps, and that by adding up the steps and enthalpy changes of a particular chemical reaction, we can determine its enthalpy change.
Scientists, however, do not calculate ΔH values for all of the reactions in the world. This would be a huge, unending task that could never be finished.
Instead of doing this, scientists record and use enthalpy changes for only one type of reaction – a reaction in which a compound is formed from its elements in their standard states.
The following is a description of what it means for a substance to be in its standard state:
The standard state of a substance is the normal physical state of the substance at 1 atm
pressure and 298K (25°C).
For example, in their standard states:
Iron is a solid (Fe)
Mercury is a liquid (Hg)
Oxygen is a diatomic gas (O
2
)
Examine the following reactions:
S
(s)
+ 3/2 O
2(g)
 SO
3(g)
ΔH° f
= -396 kJ
½ N
2(g)
+ O
2(g)
 NO
2(g)
ΔH° f
= + 33.2 kJ
The ΔH value that accompanies the reaction is called the standard enthalpy (heat) of
formation of the compound.
Definition:
Standard enthalpy of formation: the change in enthalpy that accompanies the formation of one mole of the compound in its standard state from its elements in their standard states.

Chemistry 30
Practice:
Write heats of formation reactions for each of the following compounds. Be sure to include the energy term with the equation, either as part of the equation or separately as Δ H.
CO
2 (g)
, CuCl
2 (g)
, CuCl
(g)
, N
2
H
4 (l)
, NH
4
Cl
(s)
.
C
(s)
+ O
2 (g)
→ CO
2 (g)
+ 393.5 kJ
Cu
(s)
+ Cl
2 (g)
→ CuCl
2 (s)
+ 220.1 kJ
or
or
Cu
(s)
+ ½ Cl
2 (g)
→ CuCl
2 (s)
+ 137.2kJ or
N
2 (g)
½N
+ 2H
2 (g)
2 (g)
+ 2H
+ 50.6 kJ → N
2 (g)
+ ½Cl
2 (g)
2
H
→ NH
4 (l)
4
Cl
(s)
or
+ 314.4 kJ
C
Cu
Cu
N
(s)
(s)
(s)
2 (g)
½N
+ O
2 (g)
2 (g)
+ Cl
→ CO
2 (g)
+ ½ Cl
+ 2H
→ CuCl
2 (g)
2 (g)
+ 2H
2 (g)
2 (g)
→ CuCl
→ N
2
H
+ ½Cl
2 (s)
4 (l)
2 (s)
2 (g)
→ NH
4
Δ H = -393.5 kJ
Δ H = -220.1 kJ
Δ H = -137.2 kJ
Δ H = +50.6 kJ
Cl
(s)
Δ H = -314.4 kJ
Practice #2:
The standard heat of formation, ΔH o f
, for sulfur dioxide (SO
2
) is -297 kJ/mol. How many kJ of energy are given off when 25.0 g of SO
2 (g)
is produced from its elements?
Calculate moles SO
2 molar mass of SO
2
= 32.1 + 2(16.0) = 64.1 g/mol moles SO
2
=
25.0 g
64.1 g/mol
= 0.390 mol

Chemistry 30
We know from the question that 297 kJ of energy is released for 1 mole of SO
2
— the definition of heat of formation. Determine how much energy will be released for 0.390 mol of SO
2
: kJ released = (0.39 mol)(-297 kJ/mol) = - 116 kJ