### Example 1

```Thermochemistry
Part 4: Phase Changes &
Enthalpies of Formation
Specific Heat

Specific heat: The amount of heat that
must be added to a stated mass of a
substance to raise it’s temperature, with
no change in state.
q  mc T 
Example: How much heat is released by 250.0 g of H2O
as it cools from 85.0oC to 40.0oC? (Remember, specific
heat of water = 4.18 J/goC)
q = mcT
q = (250.0 g)(4.18 J/goC)(40.0-85.0)
q = -47,000 J = -47.0 kJ
But what if there is a phase change?
LATENT HEAT OF FUSION, Hfus

Definition: the enthalpy change (energy
absorbed) when a compound is converted
from a solid to a liquid without a change in
temperature.

“Latent” means hidden; the heat
absorbed/released during a phase change
does not cause the temperature to change.

Note: Hfus for water is 334 J/g
LATENT HEAT OF FUSION, Hfus
A = solid
B = melting (solid + liquid)
C = liquid
D = boiling (liquid + gas)
E = gas
LATENT HEAT OF VAPORIZATION, Hvap

Definition: the enthalpy change (energy
absorbed) when one mole of the compound
is converted from a liquid to a gas without
a change in temperature.

Note: for water Hvap is 2260 J/g
LATENT HEAT OF VAPORIZATION,
Hvap
A = solid
B = melting (solid + liquid)
C = liquid
D = boiling (liquid + gas)
E = gas
Example 1: How much heat is released by 250.0 g of H2O as it cools
from 125.0C to -40.0C?
Five steps…
1. Cool the steam
2. Condense
3. Cool the liquid water
4. Freeze
5. Cool the solid ice
m∙csteam∙T
m(-Hvap)
m∙cwater∙T
m(-Hfus)
m∙cice∙T
Example 1: How much heat is released by 250.0 g of H2O as it cools
from 125.0C to -40.0C?
When substances change state, they often
have different specific heats:
 cice
= 2.09 J/goC
 cwater = 4.18 J/goC
 csteam = 2.03 J/goC
Example 1: How much heat is released by 250.0 g of H2O as it cools
from 125.0C to -40.0C?
cooling = exothermic → negative heat values
qsteam = mcT = (250.0g)(2.03J/goC)(100.0–125.0) = -12,700 J
qvap = mHvap = (250.0g)(-2260J/g) = -565,000 J
qwater = mcT = (250.0g)(4.18J/goC)(0-100) = -105,000 J
qfus = mHfus = (250.0g)(-334J/g) = -83,500 J
qice = mcT = (250.0g)(2.09J/goC)(-40.0-0) = -20,900 J
qtotal = -787,000J
-787 kJ
Now YOU try a few…
Example 2: How much heat energy is required to bring
135.5 g of water at 55.0oC to it’s boiling point (100.oC)
and then vaporize it?
Water must be heated to it’s boiling point.
q = mcT = (135.5 g)(4.18J/goC)(100-55)
q = 25.5 kJ
2. Water must be vaporized:
q = mHvap = (135.5 g)(2260 J/g)
q = 306 kJ
3.
q = 25.5 kJ + 306 kJ = 332 kJ
1.
Example 3: How much energy is required to convert 15.0
g of ice at -12.5oC to steam at 123.0oC?
1.
2.
3.
4.
5.
Heat the ice
Melt the ice
Heat the water
Vaporize the water
Heat the steam.
Example 3: How much energy is required to convert 15.0
g of ice at -12.5oC to steam at 123.0oC?
qice = mcT = (15.0g)(2.09J/goC)(0.0 - -12.5) = 392 J
qfus = mHfus = (15.0g)(334J/g) = 5,010 J
qwater = mcT = (15.0g)(4.18J/goC)(100-0) = 6,270 J
qvap = mHvap = (15.0g)(2260J/g) = 33,900 J
qsteam = mcT = (15.0g)(2.03J/goC)(123.0-100.0) = 700. J
qtotal = 392 J + 5,010 J + 6,270 J + 33,900 J + 700. J = 46,300J
46.3 kJ
Enthalpies of Formation
enthalpy
H
change
(delta)

f
standard
conditions
formation
Enthalpies of Formation
 usually
exothermic
 see table for Hf value
 enthalpy of formation of an element
in its stable state = 0
 these can be used to calculate H
for a reaction
Standard Enthalpy Change
Standard enthalpy change, H, for a
given thermochemical equation is = to the
sum of the standard enthalpies of
formation of the product – the standard
enthalpies of formation of the reactants.
H

rxn

(H

products
f
) - (H
sum of
(sigma)

reactants
f
)
Standard Enthalpy Change

elements in their standard states can be omitted:
2 Al(s) + Fe2O3(s)  2 Fe(s) + Al2O3(s)
ΔHrxn =
(ΔHfproducts) - (ΔHfreactants)
ΔHrxn = ΔHfAl2O3 - ΔHfFe2O3
ΔHrxn = (-1676.0 kJ) – (-822.1 kJ)
ΔHrxn = -853.9 kJ
Standard Enthalpy Change

the coefficient of the products and reactants in
the thermochemical equation must be taken into
account:
2 Al(s) + 3 Cu2+(aq)
ΔHrxn =
 2 Al3+(aq) + 3 Cu(s)
(ΔHfproducts) - (ΔHfreactants)
ΔHrxn = 2ΔHfAl3+ - 3ΔHfCu2+
ΔHrxn = 2(-531.0 kJ) – 3(64.8 kJ)
ΔHrxn = -1256.4 kJ
Standard Enthalpy Change

Example: Calculate H for the combustion of
one mole of propane:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
ΔHrxn =
(ΔHfproducts) - (ΔHfreactants)
ΔHrxn = [3ΔHfCO2+4ΔHfH2O] - ΔHfC3H8
ΔHrxn = [3(-393.5kJ)+4(-285.8kJ)]–(-103.8 kJ)
ΔHrxn = -2219.9 kJ
Standard Enthalpy Change

Example: The thermochemical equation for the
combustion of benzene, C6H6, is:
C6H6(l) + 15/2 O2(g)  6CO2(g) + 3H2O(l)
H = -3267.4 kJ
Calculate the standard heat of formation of benzene.
-3267.4kJ = [6(-393.5kJ)+3(-285.8kJ)]–ΔHfC6H6
-3267.4kJ = -3218.4–ΔHfC6H6
-49.0kJ = –ΔHfC6H6
ΔHfC6H6 = +49.0 kJ
Standard Enthalpy Change

Example: When hydrochloric acid is added to a
solution of sodium carbonate, carbon dioxide gas
is formed. The equation for the reaction is:
2H+(aq) + CO32-(aq)  CO2(g) + H2O(l)
Calculate H for this thermochemical equation.
ΔH = [(-393.5kJ)+(-285.8kJ)]–[2(0 kJ)+(-677.1kJ)
ΔH = (-679.3kJ)–(-677.1kJ)
ΔH = -2.2 kJ
```