CHE 115
Thermochemistry
III. Thermochemistry
Thermodynamics, of which thermochemistry is one area, is the study of the heat change in a system and
work done on the system by the surroundings.
A) Heat
Cold
Block
Hot
Block
T h = initial temperature
mh = mass of hot block
Tc = initial temperature
mc = mass of cold block
c h = specific heat of hot block
cc = specific heat of cold block
When a hot block is placed next to a cold block, the temperature of the cold block rises and the
temperature of the hot block drops. At thermal equilibrium the temperatures of the two block
are the same value. The change in temperature of each block is related by the equation
mh ch (Th & Tf )
'
mc cc (Tf & Tc )
where Tf is the temperature of each block at thermal equilibrium and c is the specific heat. The
specific heat is the amount of heat required to raise the temperature of 1.00 g of a substance one
degree. The right side of the equation is the change in heat of the hot block and the left side is the
change in heat of the cold block.
heat gained by cold block:
qc
' mc cc (Tf & Tc)
heat lost by hot block:
qh
' mh ch (Tf & Th)
Since qc is a positive number (heat absorbed) and qh is a negative number (heat evolved), then
qh
% qc ' 0
q is defined as the heat absorbed by a substance and
q
' mc T
where
T
' (Tfinal & Tinitial)
or
q
' nC T
where n is the number of moles of the substance and C is the molar heat capacity of the
substance. The molar heat capacity is the amount of heat required to raise the temperature of one
mole of substance one degree.
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CHE 115
Thermochemistry
B) Work
Work is the result of an action against an opposing force over some distance.
work
' force ×distance
Although there are many kinds of work, we are interested primarily in the work done by a
frictionless piston on an ideal gas.
Pext
frictionless
piston
ideal
Pint gas
cylinder
Pext is the external pressure and Pint is the internal pressure of the gas
When Pext < Pint, then the gas expands and work is done against Pext; when Pext > Pint, then the gas
is compressed and work is done against Pint; and when Pext = Pint, then the gas is in equilibrium
with the piston.
Thermodynamic terms:
System - any piece of matter under consideration
Examples: ideal gas, chemical reaction
Surroundings - any matter that can interact with the system
Examples: cylinder, piston, reaction flask
w is defined as the work done on the system by the surroundings.
For pressure-volume work (PV work), w is defined by the equation
V
!2
w ' ! " PdV ' ! (area under PV curve)
V
1
Consider the expansion of an ideal gas from P1V1 to P2V2.
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CHE 115
Thermochemistry
Expansion
P = 20 atm
1
P2 = 4.8 atm
V2 = 4.2 L
Ideal
Gas
V1 = 1.0 L
State #1
State #2
A plot of the pressures and volumes of the ideal gas during the reversible expansion.
Since V2 > V1, the (area under the PV curve) > 0 and w = - (area under the PV curve) < 0.
Note: number > 0 is a positive number and number < 0 is a negative number.
The reverse of the definition of w occurred i.e. the gas (the system) did work on the piston (the
surroundings).
Consider the compression of an ideal gas from P1V1 to P2V2.
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CHE 115
Thermochemistry
P = 1.8 atm
1
Ideal
Gas
Compression
V1 = 4.2 L
V2 = 1.0 L
P2 = 20 atm
State #1
State #2
A plot of the pressures and volumes of the ideal gas during the reversible compression.
Since V2 < V1, the (area under the PV curve) < 0 and w = - (area under the PV curve) > 0. The
piston (the surroundings) did work on the ideal gas (the system).
The magnitude of w is dependent on the path taken from the initial state (State #1) to the final
state(State #2). The magnitude of w is path dependent.
Example: Consider the expansion of an ideal gas by two different pathways in which the initial
state (P1V1) and the final state (P2V2) are the same in both pathways.
1) Expansion into a vacuum: Pext = 0
V
! 2
w ' ! " Pext dV ' !
V
1
4
V
! 2
0 dV ' 0
"V
1
CHE 115
Thermochemistry
2) Expansion against a constant external pressure: Pext = constant
w ' !
V
V
1
1
! 2
! 2
P dV ' ! Pext dV ' Pext (V2 & V1)
"V ext
"V
w ' & Pext V
Note: Under the condition of constant pressure w = - P V for an ideal gas.
C) First Law of Thermodynamics
Consider a cyclic process for an ideal gas in a cylinder with a frictionless piston.
P2 = 20 atm
a
Ideal Gas
P1 = 20 atm
State #1
State #2
b
d
c
P3 = 10 atm
Ideal Gas
P4 = 10 atm
V2 = 20 L
V1 = 1.0 L
V4 = 1.0 L
State #3
State #4
5
V3 = 20 L
CHE 115
Thermochemistry
The net work, wnet, done in one cyclic is the sum of the work associated each step.
wnet
' wa % wb % wc % wd ' & P1 (V2 & V1) & P2(V1 & V2)
Payment for the net work, wnet, is in the form of heat, qnet. In Step a heat must be added in order
to increase the volume of the gas at constant P and n. To lower the pressure of the gas under the
condition of constant V and n in Step b, heat must be removed. In Step c the gas must lose heat
and in Step d the gas must gain heat.
qnet
' qa % qb % qc % qd
For a cyclic process
' & qnet
wnet
qnet
% wnet ' 0
In those cases where q > w, the excess heat is absorbed by the system and the internal energy of
the system will increase. When q < w, some of the internal energy of the system will used to do
the work and there will be a reduction in the internal energy of the system.
E is the symbol for internal energy.
Example of Internal Energy: When an ideal monatomic gas is the thermodynamic
system, then E, internal energy, is the kinetic energy of translation, ET = 3RT/2.
The change in the internal energy of the system is
'
E
q
% w
where q is the heat absorbed by the system and w is the work done on the system by the
surroundings.
The First Law of Thermodynamics states
i) E = q + w
ii) E is a state function.
A state function is any function in which the change in going from the initial state (state #1) to
the final state (state #2) depends only on the “conditions” of the initial and final states but not on
the path taken between the two states. State functions are path independent.
Example: Consider the expansion of one mole of an ideal monatomic gas from the same
initial state, State #1, to the same final state, State #2, by two different paths. As can be
seen from the PV curves below, the amount of work done in the process associated with
Path #1 is different from the work associated with Path #2. However, since the initial and
final states are the same in both paths, E for both paths must have the same value.
Path # 1:
E
' q % w ' 6481 J & 6481 J ' 0
Path # 2:
E
' q % w ' 2906 J & 2960 J ' 0
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CHE 115
Thermochemistry
D) Enthalpy, H
qv is the heat absorbed by the system in a constant volume process.
qp is the heat absorbed by the system in a constant pressure process.
1) A Constant Volume Process
' qv % w
E
at constant volume
w
' 0 and
E
' qv
E
' qv % 0
Thus qv is a state function.
2) A Constant Pressure Process
at constant pressure
E
' qp % w
w
' & P V and
Rearranging
7
E
' qp & P V
CHE 115
Thermochemistry
qp
qp
'
% P V ' (E2 & E1) % P(V2 & V1)
E
' (E2 % PV2) & (E1 % PV1) ' H2 & H1 '
'
qp
H
H
Since the magnitude of qp depends only on the “conditions” of the initial and final states,
qp is a state function.
H is called enthalpy and is defined as H = E + PV. Enthalpy, H, is a state function.
At Constant Pressure:
' qp
H
3) When a Chemical Reaction is the Thermodynamic System - the Relationship between E
and H
a) Reactants and/or Products are Gases
Consider the reaction at constant T and P
N2 (g)
Ÿ
% 3 H2 (g)
P1 = 1 atm
V1 at
25oC
{
P2 = 1 atm
qp = -93.38 kJ
H2
H2
N2 H2
Initial and Final States
have the same T & P
State #1
qp
'
H'& 93.38 kJ
2 NH3 (g)
NH3
NH3
}
V2 at
25oC
State #2
'
H
E
%P V
(1)
Since T and P are the same in the initial and final states, then a change in volume,
V, will result from a change in the total moles of the gases, n. Assuming ideal
gas behavior
State #1: P1 V1
' (nN2 % nH2 ) R T
P1
P V
and
State #2: P2 V2
' P2 ' P
' P(V2 & V1) ' [nNH3 & (nN2 % nH2 )] R T '
Substituting eq 2 into eq 1
H
'
8
' nNH3 R T
E
% n RT
nR T
(2)
CHE 115
Thermochemistry
where n is the TOTAL moles of GASEOUS products minus the TOTAL moles
of GASEOUS reactants.
b) Reactants and/or Products are Liquids or Solids
For solids and liquids
and thus
H
V
'
. 0 and P V . 0
%0 '
E
E
Note: Physical states: g = gas, l = liquid, s = solid, aq = aqueous solution
Example: Calculate H at 25EC for the reaction below.
6
C6H6 (l) + 15/2 O2 (g)
6 CO2 (g) + 3 H2O (l)
H '
n ' nCO & nO
2
H ' & 3264kJ %
ð
2
E %
nR T
' 6 & 15 ' & 3 mole
2
& 3 mole 8.314 moleJ &K
2
ëð
E = -3264 kJ
2
kJ
(298K) ' & 3268kJ
ëð 1000 J ë
E) Heats of Reaction
Heat of reaction under the condition of constant volume: qv = E
Heat of reaction under the condition of constant pressure: qp = H
1) Heat of Combustion
The heat change associated with the burning of a compound.
Example:
2 C4H10 (g) + 13 O 2 (g)
8 CO2 (g) + 10 H2O (l)
Thermochemical Equation
)Hocomb = -5314 kJ
Heat Change for
Reaction as Written
The heat of combustion is measured in a bomb calorimeter.
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CHE 115
Thermochemistry
Thermometer
Insulation
Water
Bomb
Bucket
Bomb Calorimeter
A weighed sample of a compound is placed in the bomb. The bomb is charged with 25
atm of O2 and ignited. The heat given up by the reaction (qreaction < 0) is absorbed by the
calorimeter (qcal > O).
qreaction
' & qcal
The change in temperature of the calorimeter is measured and the heat absorbed (qcal = Ccal
T) by the calorimeter is calculated. Since the volume of the bomb in which the reaction
takes place is constant,
qreaction
' qv '
E
' & (Ccalor % Cwater ) T ' & Ccal T ' & qcal
qreaction
'
E
' & Ccal T
where Ccalor is the heat capacity of the calorimeter components (bomb, bucket,
thermometer, etc.) and Cwater is the heat capacity of the water in the bucket.
The heat of reaction under the condition of constant pressure is calculated with the the
following equation.
H
'
E
% nR T
2) Heat of Formation
The heat change associated with the formation of a compound from its elements in their
most stable form.
Example:
N2 (g) + 3 H2 (g)
6
2 NH3 (g)
HEf = -93.38 kJ
The numeric value for the heat of a reaction is dependent on the temperature and pressure
at which the reaction is carried out. If we are to compare heats of reaction, then these
heats must be measured under the same conditions. For this reason a standard set of
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CHE 115
Thermochemistry
conditions has been chosen for reporting heats of reaction. These conditions are that the
reactants and products are in their standard states.
Standard State of:
Physical State:
Pressure:
Temperature:
gas
pure gas
partial pressure
of 1 atm
25EC
liquid
pure liquid
1 atm
solid
pure, crystalline solid
1 atm
25EC
25EC
The heat change associated with the formation of a compound in its standard state from its
elements (most stable form) in their standard states is called the standard enthalpy of
formation, HEf.
denotes standard state (1 atm and 25oC)
) Hof
denotes formation
By definition the HEf for an element is its most stable form is zero.
A table of heats of formation for compounds can be found in the text.
Example: Table of Standard Enthalpies of Formation
Compound
HEf (kJ/mole)
H2O (l)
- 285.84
NH3 (g)
- 46.69
CO2 (g)
- 393.51
H2O (g)
- 241.8
H2 (g)
0.0
H (g)
217.94
The thermochemical equations for these HEf are known by definition.
Example: Write the thermochemical equation for HEf (CO2 (g)) = -393.51 kJ/mole.
Since -393.51 kJ is the heat change for the formation of one mole of CO2 from the most stable
forms of elemental carbon and hydrogen, then
C (s) + O2 (g)
6
CO2 (g)
HEf = -393.51 kJ
3) Hess’s Law
The heat change at constant temperature and pressure for a particular chemical reaction is
constant and independent of the path taken from reactants to products.
Consider the formation of CO2 by two different pathways.
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CHE 115
Thermochemistry
CO (g)
1/2 O2 (g)
1/2 O2 (g)
O2 (g)
CO2 (g)
C (s)
The heat change is the same for both pathways to CO2.
These thermochemical
equations add to give
the summary equation.
Summary equation
{
C (s) + 1/2 O2 (g)
CO (g)
CO (g) + 1/2 O2 (g)
CO2 (g)
C (s) +
CO2 (g)
O2 (g)
)Hof = -110.52 kJ
)Hocomb = -282.99 kJ
}
These enthalpies add
to give the enthalpy
for the summary
equation.
)Hof = -393.51 kJ
4) Calculation of enthalpies that cannot be measured from enthalpies that have been measured
Procedure:
Step #1:
Write down the thermochemcial equation for which you wish to
calculate the H (or other thermodynamic function) as the summary
equation.
Step #2:
Arrange given thermochemical equations (equations with known values
for H or other thermodynamic function) so that the sum of the
equations is the summary equation.
Step #3:
Write down an appropriate value for the H (or other thermodynamic
function) associated with each thermochemical equation in Step #2. If
the thermochemical equation is the reverse of the given equation,
change the sign of the H.
Step #4:
Add up the values for the Hs ( or other thermodynamic functions) in
Step #3.
Example: Calculate the enthalpy of formation, HEf , for benzene, C6H6, at 25EC given
C6H6 (l) + 15/2 O2 (g)
6
6 CO2 (g) + 3 H2O (l)
HEcomb = -3267.7 kJ
and HEf (CO2 (g)) = - 393.51 kJ/mole ; HEf (H2O (l)) = - 285.84 kJ/mole.
Note: The thermochemical equations associated with the enthalpies of formation for CO2 and
H2O are known by definition.
C (s) + O2 (g)
H2 (g) + ½ O2 (g)
6 CO (g)
6 H O (l)
2
2
12
HEf = - 393.51 kJ
HEf = - 285.84 kJ
CHE 115
Thermochemistry
3 H2 (g) + 3/2 O2 (g)
6 C (s) + 6 O2 (g)
Step #2
3 H2O (l)
)Ho = (3 mole)(- 285.84 kJ/mole)
6 CO2 (g)
)Ho = (6 mole)(- 393.51 kJ/mole)
C6H6 (l) + 15/2 O2 (g)
)Ho = (-)(- 3267.7 kJ)
6 CO2 (g) + 3 H2O (l)
Step #1
6 C (s) + 3 H2 (g)
Step #3
)Ho = ?
C6H6 (l)
HEf (C6H6 (l)) = 49.12 kJ/mole
Alternate Method for Calculating H
HEreaction
'
j
HEf (product)
product
&
j
reactant
HEf (reactant)
where , called the mole number, is the stoichiometric coefficient in the thermochemical
equation.
Example: Calculate the enthalpy of formation, HEf , for benzene, C6H6, at 25EC given
C6H6 (l) + 15/2 O2 (g)
6
6 CO2 (g) + 3 H2O (l)
HEcomb = -3267.7 kJ
and HEf (CO2 (g)) = - 393.51 kJ/mole ; HEf (H2O (l)) = - 285.84 kJ/mole
HEcomb '
CO2
HEf (CO2 (g)) %
H2O
HEf (H2O (l )) &
C6H6
HEf (C6H6 (l )) &
O2
HEf (O2 (g ))
15
kJ
&3267.7kJ ' (6mole) (& 393.51 mole
) % (3 mole)(& 285.84 kJ ) & (1 mole)( HEf (C6H6 (l))) & ( mole)(0.0 kJ )
mole
mole
2
HEf (C6H6 (l)) ' 49.12
kJ
mole
F) Molar Heat Capacity
1) Cv: Heat Capacity for a Constant Volume Process
Consider increasing the temperature of one mole of an ideal monoatomic gas by T
degrees in a constant volume process.
E2 = 3/2 R(T + )T)
- (E1 =
3/2 RT)
)E = E2 - E1 = 3/2 R)T = qv
13
The amount of heat required
to raise the temperature of
the gas by )T degrees in a
constant volume process.
CHE 115
Thermochemistry
qV
'
Cv
T
E
T
'
(3)
Cv for an ideal monatomic gas is
' 3 R ' 12.47 moleJ &K
Cv
2
Values of Cv for some other gases
Compound:
Cv (J/mole-K):
H2
20.33
Cl2
25.10
HCl
21.00
Rearranging eq 3
' Cv T
E
(4)
Eq 4 is valid for one mole of a real gas in a constant volume process and for one mole of
an ideal gas under all conditions.
2) Cp: Heat Capacity for a Constant Pressure Process
For a constant pressure process the change in volume of one mole of an ideal gas is the
result of a change in temperature.
P V
' P(V2 & V1) ' R (T2 & T1) ' R T
(5)
The heat absorbed by one mole of an ideal gas in a constant pressure process
'
qp
H
'
E
%P V
(6)
and
Cp
'
qp
H
' Cp T
T
H
T
'
(7)
Rearranging eq 7
(8)
Eq 8 is valid for one mole of a real gas in a constant pressure process and for one mole of
an ideal gas under all conditions.
Substituting eqs 4, 5, and 8 into eq 6
Cp
T
' Cv T % R T
and dividing by T
Cp
' Cv % R
14
(9)
CHE 115
Thermochemistry
Eq 9 is valid only for an ideal gas.
Example: Calculate Cp for an ideal monatomic gas if Cv is 12.47 J/mole-K.
CP ' Cv % R ' 12.47
J
mole&K
% 8.314 moleJ &K ' 20.78 moleJ &K
SUMMARY:
• When an ideal gas is the thermodynamic system,
E
E
' n Cv T
and
' q%w
and
H
Cp
' nCp T
' Cv % R
where n is the number of moles of the ideal gas.
• When a chemical reaction is the thermodynamic system,
qreaction
'
E
' & Ccal T
and
and
E
15
' q%w
H
'
E
% nR T