1. Enthalpy and Chemical Reactions
Combustion Reactions
1 mole
(Exothermic)
CH4(g) O2(g)  CO2(g) H2O(g)
kJ (heat) kJ
molCH4
Hcombustion =
(...per mole of substance burned)
Formation Reactions
1 mole
(Exothermic)
C(s) H2(g) ½ O2(g)  CH3OH(g)
kJ (heat) kJ
molCH3OH
Hformation =
(...per mole of substance formed)

2. Hcombustion Problem ( )   = -500. kJ
How much heat is produced when 10.0 grams of methane (CH4) burn in the presence of excess oxygen gas?
CH4(g) O2(g)  CO2(g) H2O(g)
kJ (heat)
(Exothermic)
10.0 grams
Excess
10.0 gCH4 1
gCH4
1 moleCH4
( ) kJ
molCH4  
= kJ
Negative sign tells us heat is released (exothermic) and...
...that the system’s energy has decreased.
What has happened to the energy of the surroundings? ...the universe?

3. Enthalpy and Chemical Reactions
Heats of fusion: Melting a solid into a liquid
(Endothermic)
1 mole
6.02 kJ +
H2O (s)  H2O(l) kJ
molH2O
Hfusion =
(...per mole of substance melted)
H2O (l)  H2O(g)
Heat of vaporization: Vaporizing a liquid into a gas.
(Endothermic)
1 mole
2260 kJ + kJ
molH2O
Hvaporization =
(...per mole of substance vaporized)
Why should it require so much more energy to vaporize water in comparison to melt it?
Because it requires much more energy to separate molecules to great distances found in gases.