Heat in Changes of State
What happens when you place an ice cube on
a table in a warm room?
Molar Heat of Fusion (ΔHfus): heat absorbed by
one mole of a substance in melting from a solid to a
liquid at a constant temperature
Molar heat of Solidification (ΔHsolid): heat lost when
one mole of a liquid solidifies at a constant temperature
ΔHfus = - ΔHsolid
Molar Heat of Vaporization (ΔHvap): amount of heat
necessary to vaporize one mole of a given liquid
H2O (l)  H2O (g)
ΔHvap = 40.7 kJ/mol
Molar Heat of Condensation (ΔHcond): amount of heat
released when 1 mol of vapor condenses
ΔHvap = - ΔHcond
High Enthalpy
condensation
vaporization
Vapor
ΔHvap
-ΔHcond
Solid
solidification
fusion
Liquid
-ΔHfus
Low Enthalpy
-ΔHsolid
The Heating Curve of
Water
ΔHvap
ΔHfus
http://netcamp.prn.bc.ca/nuggets/heatingcurve.swf
Molar Heat of Solution (ΔHsoln): heat change caused
by dissolution of one mole of substance
DHsoln = Hsoln - Hcomponents
Exothermic Reaction:
H2O(l)
CaCl2 (s)  Ca2+(aq) + 2Cl-(aq)
ΔHsoln = -445.1 kJ/mol
Endothermic Reaction:
H2O(l)
NH4NO3 (s)  NH4+ (aq) + NO3- (aq)
ΔHsoln = 25.7 kJ/mol
How much heat (in kJ) is released when 2.500 mol NaOH (s)
is dissovled in water?
ΔHsoln = -445.1 kJ/mol
List Knowns and Unknowns
Knowns:
ΔHsoln = -445.1 kJ/mol
Amount of NaOH(s): 2.500 mol
Unknown:
ΔH = ?kJ
Solve:
ΔH = 2.500 mol NaOH (s) x
-445.1 kJ = -1113 kJ
1 mol NaOH
Hess’s Law
Hess’s Law of heat summation: If you add two or more
thermochemical equations to give a final equation,
then you can also add the heats of reaction to give
the final heat of reaction.
Example: C (diamond)  C (graphite)
Use Hess’s Law to find the enthalpy changes for the conversion
of diamond to graphite by using the following combustion
rxns
C (s, graphite) + O2 (g)  CO2 (g) ΔH = -393.5 kJ
C (s, diamond) + O2 (g)  CO2 (g) ΔH = -395.4 kJ
1st Step: Write equation the first equation in reverse
CO2 (g)  C (s, graphite) + O2 (g) ΔH = 393.5 kJ
**when you write an equation in reverse, change sign**
2nd Step: Add the two equations together
CO2 (g)  C (s, graphite) + O2 (g) ΔH = 393.5 kJ
C (s, diamond) + O2 (g)  CO2 (g) ΔH = -395.4 kJ
C (s, diamond)  C (s, graphite)
ΔH = -1.9 kJ
Use Hess’s Law to find the enthalpy change for the
formation of carbon monoxide from its elements.
C (s, graphite) + O2 (g)  CO2 (g) ΔH = -393.5 kJ
CO (g) + ½ O2 (g)  CO2 (g)
ΔH = -283.0 kJ
1st Step: Write the first equation in reverse
CO2 (g)  CO (g) + ½ O2
(g) ΔH = 283.0 kJ
2nd step: Add the equations together
CO2 (g)  CO (g) + ½ O2 (g)
ΔH = 283.0 kJ
C (s, graphite) + O2 (g)  CO2 (g) ΔH = -393.5 kJ
C (s, graphite) + ½ O2 (g)  CO (g) ΔH = -110.5 kJ
Standard Heat of Formation (ΔHf0): the change in
enthalpy that accompanies the formation of one
mole of a compound from its elements with all
substances in their standard states at 25°C
ΔHf0 of a free element in its standard state is
arbitrarily set at 0.
ΔHf0 = 0 for diatomic molecules
Standard Heat of Reaction (ΔH0rxn ):heats of
reaction at standard conditions
ΔH0rxn = ΔHf0 (products) - ΔHf0 (reactants)