1. No category

Ch. 16: Energy and
Chemical Change
Sec. 16.3: Thermochemical Equations
Objectives
Write thermochemical equation for
chemical reactions and other processes.
Describe how energy is lost or gained
during changes of state.
Calculate the heat absorbed or released
in a chemical reaction.
Review
The change in energy is an important
part of chemical reactions so chemists
include ΔH as part of the chemical
equation.
Thermochemical Equations
A thermochemical equation is a
balanced chemical equation that
includes the physical states of all
reactants and products and the energy
change.
The energy change is usually expressed
as the change in enthalpy, ΔH.
Thermochemical Equations
A subscript of ΔH will often give you
information about the type of
reaction or process taking place.
For example, ΔHcomb is the change
in enthalpy for a combustion
reaction or, simply, the enthalpy
(heat) of combustion.
Thermochemical Equations
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
ΔHcomb = -2808 kJ/mol
This is the thermochemical equation for
the combustion of glucose.
The enthalpy of combustion is -2808
kJ/mol.
Thermochemical Equations
The enthalpy of combustion (ΔHcomb) of
a substance is defined as the enthalpy
change for the complete burning of one
mole of the substance.
That means, 2808 kJ of heat are
released for every mole of glucose that
is oxidized (or combusts). Two moles
would release 5616 kJ:
2 mol glucose x -2808 kJ = -5616 kJ
1 mol
Standard Enthalpies
Standard enthalpy changes have the
symbol ΔHo.
Standard conditions in
thermochemistry are 1 atm pressure
and 25 0C. **DO NOT CONFUSE THESE
WITH THE STP CONDITIONS OF THE
GAS LAWS.
Calculations
How much heat is evolved when 54.0 g
of glucose (C6H12O6) is burned
according to this equation?
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
ΔHcomb = -2808 kJ/mol

Since the ΔHcomb given is for 1 mole of
glucose, you must first determine the # of
moles you have.
54.0 g x 1 mole = 0.300 moles glucose
180 g
Calculations (cont.)

You can now use the ΔHcomb value as a
conversion factor:
0.300 mol glucose x -2808 kJ = -842 kJ
1 mole
Practice Problems
How much heat will be released when 6.44 g
of sulfur reacts with O2 according to this
equation:
2S + 3O2  2SO3 ΔH0 = -395.7 kJ/mol
How much heat will be released when 11.8 g
of iron react with O2 according to the
equation:
3Fe + 2O2  Fe3O4 ΔH0 = -373.49 kJ/mol
Changes of State
The heat required to vaporize one mole
of a liquid is called its molar enthalpy
(heat) of vaporization (ΔHvap).
The heat required to melt one mole of a
solid substance is called its molar
enthalpy (heat) of fusion (ΔHfus).
Both phase changes are endothermic &
ΔH has a positive value.
Changes of State
The vaporization of water and the
melting of ice can be described by the
following equations:
H2O(l) → H2O(g) ΔHvap = 40.7 kJ/mol

One mole of water requires 40.7 kJ to
vaporize.
H2O(s) → H2O(l) ΔHfus = 6.01

kJ/mol
One mole of ice requires 6.01 kJ to
melt.
Changes of State
The same amounts of energy are released in
the reverse processes (condensation and
solidification (freezing)) as are absorbed in
the processes of vaporization and melting.
Therefore, they have the same numerical
values but are opposite in sign.
ΔHcond = - ΔHvap = -40.7 kJ/mol
ΔHsolid = - ΔHfus = -6.01 kJ/mol
Practice Problems
How much heat is released when 275 g
of ammonia gas condenses at its boiling
point? (ΔHvap = 23.3 kJ/mol)
If water at 00 C releases 52.9 J as it
freezes, what is the mass of the water?
(ΔHfus= 6.01 kJ/mol)
How much heat is required to melt 25 g
of ice at its MP? (ΔHfus= 6.01 kJ/mol)
Combination Problems
At times, you will need to calculate the amount
of heat that is absorbed or released when a
temperature change AND a phase change occur
in sequence.


Recall, the heat involved in a temperature change
is calculated by using: ΔH = mCΔT. NOTE: In this
expression, ΔH is found in joules or cal.
Heat involved in a phase change is calculated using
dimensional analysis. NOTE: ΔH will be in kJ.
Example
How much heat is released when 37.5 g
of water that is at 20.0 0C freezes?
(ΔHfus= 6.01 kJ/mol)
Water cannot freeze at 20.0 0C. It must be
at 0 0C. Therefore, we must first determine
how much energy is released when it is
cooled from 20.0 0C to 0 0C.
ΔH = mCΔT = (37.5 g)(4.184 J/g 0C)(- 20.0 0C)
= - 3138 J = -3.14 kJ

Example
How much heat is released when 37.5 g
of water that is at 20.0 0C freezes?
(ΔHfus= 6.01 kJ/mol)
-3.14 kJ of heat are released when the
sample is cooled to 0 0C.
 Now we must calculate the energy
released when the entire sample freezes.
Recall that ΔHsolid= -ΔHfus= 6.01kJ/mol
37.5 g x 1 mole x -6.01 kJ = -12.5 kJ
18 g
1 mole

Example
How much heat is released when 37.5 g
of water that is at 20.0 0C freezes?
(ΔHfus= 6.01 kJ/mol)



-3.14 kJ of heat are released when the
sample is cooled to 0 0C.
-12.5 kJ of heat are released when the
sample freezes.
The total heat released is -3.14 + -12.5 or
-15.6 kJ.
Practice Problems
Use Cw = 4.184 J/g0C; ΔHfus = 6.01
kJ/mol; ΔHvap = 40.7 kJ/mol.
How much heat is needed to melt 8 g
of ice at 0 0C to water at 15 0C?
How much heat is needed to change
28.0 g of water at 60.0 0C to steam?