qV = rU
Constant Volume Calorimetry,
Measuring U
Heats of combustion
(r
U
o
combustion
) are
measured using a device
called a
Bomb Calorimeter.
A combustible sample is
reacted with excess O2
The heat capacity of the
bomb is constant.
The heat of reaction is found
by:
Constant Volume Calorimetry,
Measuring U
rxn
cal
cal
water
bomb
q
C
T
C
C
C
= 
×∆
=
+
Octane, the primary component of gasoline combusts by the
reaction:
C8H18(
l
) + 25/2 O2(g) 8 CO2(g) + 9H2O(
l
)
A 1.00 g sample of octane is burned in a bomb calorimeter that
contains 1.20 kg of water surrounding the bomb.
The temperature of the water rises to 33.20 °C from 25.00 °C
when the octane is reacted.
If the heat capacity of the bomb is 837 J/°C, calculate the heat of
reaction per mole of octane.
–
q
rxn =
q
water +
q
bomb
Since the temperature of the water rose, the reaction
must have been exothermic:
Therefore one can write:
Calculating Heat in an Exothermic
Reaction
–
q
rxn = mwaterCwaterTwater +
q
bombTwater
q
water
q
bomb
q
RXN = – 4803 J or – 48.0 kJ
Heat transferred per mole qV:
Calculating Heat in an Exothermic
Reaction

=
RXN
q
4.184J
g C
°
×
×
3
1.20
10 g
(
29
33.20
25.00
C
×

°
J
837
C
+
°
(
29
33.20
25.00
C
×

°
3
48.0 kJ
kJ
5.48
10
mol
mol
1.00g
114.2g

= 
×
×
The overall enthalpy change for a reaction is equal to the sum of the
enthalpy changes for the individual steps in the reaction.
Reactants
Products
rH
=
?
unknown!
Intermediate
Reaction
rH1
known
rH2
k
nown
rH1
+
rH2
=
rH
The sum of the H’s in one direction must equal the sum
in the other direction.
What if the enthalpy
changes through
another path are know?
Why?
Because enthalpy is a state function…
Path independent
!
so we can write…
Hess’s Law
Forming CO2 can
occur in a single
step or in a two
steps.
∆rHtotal is the same
no matter which
path is followed.
Hess’s law & Energy Level
Diagrams
Notice that the path from reactants to products in the desired reaction goes through
an
“intermediate compound”
in the given reactions.
This means that the path for hydrogen and nitrogen to produce ammonia goes
through hydrazine (N2H4).
Therefore, the path to the enthalpy of the reaction must be a sum of the two
given reactions!
Hess’s Law Problem:
Example:
Determine the rH for the reaction:
3H2(g) + N2(g)
2NH3(g)
rHo = ???
Given:
(1)
2 H2(
g
) + N2(
g
) N2H4(
g
)
∆rH°1 = +95.4 kJ
(2)
N2H4(
g
) + H2(
g
)
2NH3(
g
)
∆rH°2 = –187.6 kJ
adding equations (1) & (2) yields:
2 H2(
g
) + N2(
g
)
+ N2H4(
g
) +H2(
g
)
N2H4(
g
)
+
2NH3(
g
)
Hess’s Law Problem:
Example:
Determine the rH for the reaction:
3H2(g) + N2(g)
2NH3(g)
rHo = ???
Given:
(1)
2 H2(
g
) + N2(
g
) N2H4(
g
)
∆rH°1 = +95.4
kJ
(2)
N2H4(
g
) + H2(
g
)
2NH3(
g
)
∆rH°2 = –187.6 kJ
Hess’s Law Problem:
Example:
Determine the rH for the reaction:
3H2(g) + N2(g)
2NH3(g)
rHo = ???
Given:
(1)
2 H2(
g
) + N2(
g
) N2H4(
g
)
∆rH°1 = +95.4
kJ
(2)
N2H4(
g
) + H2(
g
)
2NH3(
g
)
∆rH°2 = –187.6 kJ
adding equations (1) & (2) yields:
2 H2(
g
) + N2(
g
)+ N2H4(
g
) + H2(
g
) N2H4(
g
)+
2NH3(
g
)
Look what happens…
/
/
3H2(g) + N2(g) 2NH3(g)
Hess’s Law Problem:
Example:
Determine the rH for the reaction:
3H2(g) + N2(g)
2NH3(g)
rHo = ???
You've reached the end of your free preview.
Want to read all 73 pages?
 Fall '07
 Moran
 Chemistry, Thermodynamics, Enthalpy, pH, Reaction, Energy, Kinetic Energy, °C