Mr. Chapman
Chemistry 30
Norquay School
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.

Mr. Chapman
Chemistry 30
Definition:
Norquay School
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.
Exercise:
Predict the product of the following formation reactions, then use the table R-11 on page 975 to determine the ΔH° f
of the compound. Also, tell if the formation is exothermic / endothermic.
1.
Pb
(s)
+ Cl
2(g)
 PbCl
2(s)
2.
Mn(s) + S(s)  MnS(s)
3.
H
2(g)
+ ½ O
2(g)
 H
2
O
(g)
ΔH°
ΔH°
ΔH° f f f
= -359.4 kJ
= -214.2 kJ
= -241.8 kJ
Why Do We Care About Standard Enthalpies of Formation?
As mentioned before, the enthalpy change of some chemical reactions is impossible to determine by regular means (such as measuring temperature change). Using Hess’ Law, we can calculate the enthalpy changes in many chemical reactions by adding up the standard
formation equations.
On a loose leaf, copy down the following example from the overhead using standard formation
equations to determine an enthalpy change.
** Note: the standard heat of formation of an element in its standard state is zero. Because of this, we ignore elements in their standard states when calculating an enthalpy change.

Mr. Chapman
Chemistry 30
Norquay School
The stepwise procedure that we just used to calculate the enthalpy change showed how standard heats of formation can be used to produce the desired equation and its enthalpy change. The procedure can be summed up using the following formula:
Summation Equation:
H overall
= H of products
- H of reactants
After we complete practice problem #35 on page 541 in class, try to find the overall enthalpy change of the following reactions using the summation equation and the table of standard heats of formation given on the next page.
C
2
H
3
Cl (g) + HCl (g) H = ? C
2
H
4
(g) + Cl
2
Answer :- 107.5 kJ
4 NH
3
(g) + 5 O
2
(g)
Answer : - 907.6 kJ
4NO (g) + 6 H
2
O (g) H = ?

Mr. Chapman
Chemistry 30
Selected Heats of Formation
** Heats of formation of elements is 0 kJ/mol
Substance
CO
2
(g)
CO (g)
CH
4
(g)
C
2
H
4
(g)
C
2
H
6
(g)
C
3
H
8
(g)
C
4
H
10
(g)
C8H18 (l)
CH
3
OH ( l )
C
2
H
5
OH ( l )
C
2
H
3
Cl (g)
H
2
SO
4
(l)
Heat of Formation
H f (kJ/mol)
Substance
- 393.5 HCl ( g )
- 110.5 H
2
O ( l )
- 74.4 H
2
O ( g )
+52.5 H
2
O
2
( g )
- 83.8 SO
2
( g )
- 104.7 SO
3
( g )
- 125.6 NO ( g )
- 250.1 NH
4
Cl (s)
- 239.1 NO
2
( g )
- 235.2 NH
3
(g)
+37.3 H
2
S ( g )
- 814.0 HNO
3
(l)
Heat of
Formation
H f (kJ/mol)
- 92.3
- 285.8
- 242.0
- 187.8
- 296.8
- 395.7
+90.2
- 314.4
+33.2
- 45.9
- 20.6
- 174.1
Norquay School