Transcript Document

Thermochemistry
Chapter 5
Energy
• The ability to do work or transfer heat.
– Work: Energy used to cause an object that
has mass to move.
– Heat: Energy used to cause the temperature
of an object to rise.
Potential Energy
Energy an object possesses by virtue of its
position or chemical composition.
Kinetic Energy
Energy an object possesses by virtue of its
motion.
1
KE =  mv2
2
Potential energy
Energy in
kinetic energy
Energy out
kinetic energy
The energy something possesses due to its motion, depending on mass and velocity.
Energy
A
C
B
Kinetic Energy – energy of motion
KE = ½ m v 2
mass
velocity (speed)
Potential Energy – stored energy
Batteries (chemical potential energy)
Spring in a watch (mechanical potential energy)
Water trapped above a dam (gravitational potential energy)
School Bus or Bullet?
Which has more kinetic energy;
a slow moving school bus or a fast moving bullet?
Recall: KE = ½ m v 2
BUS
KE = ½ m v 2
KE(bus) = ½ (10,000 lbs) (0.5 mph)2
BULLET
KE = ½ m v 2
KE(bullet) = ½ (0.002 lbs) (240 mph)2
Either may have more KE, it depends on the mass of the bus and the velocity
of the bullet.
Which is a more important factor: mass or velocity? Why?
(Velocity)2
Units of Energy
• The SI unit of energy is the joule (J).
kg m2
1 J = 1 
s2
• An older, non-SI unit is still in widespread
use: The calorie (cal).
1 cal = 4.184 J
System and Surroundings
• The system includes
the molecules we want
to study (here, the
hydrogen and oxygen
molecules).
• The surroundings are
everything else (here,
the cylinder and piston).
Work
• Energy used to move
an object over some
distance.
• w = F  d,
where w is work, F is
the force, and d is the
distance over which
the force is exerted.
Heat
• Energy can also be
transferred as heat.
• Heat flows from
warmer objects to
cooler objects.
First Law of Thermodynamics
• Energy is neither created nor destroyed.
• In other words, the total energy of the universe is
a constant; if the system loses energy, it must be
gained by the surroundings, and vice versa.
Use Fig. 5.5
Internal Energy
The internal energy of a system is the sum of all
kinetic and potential energies of all components of
the system; we call it E.
Use Fig. 5.5
Internal Energy
By definition, the change in internal energy, E,
is the final energy of the system minus the initial
energy of the system:
E = Efinal − Einitial
Use Fig. 5.5
Changes in Internal Energy
• If E > 0, Efinal > Einitial
– Therefore, the
system absorbed
energy from the
surroundings.
– This energy change
is called
endergonic.
Changes in Internal Energy
• If E < 0, Efinal < Einitial
– Therefore, the
system released
energy to the
surroundings.
– This energy change
is called exergonic.
Changes in Internal Energy
• When energy is
exchanged between
the system and the
surroundings, it is
exchanged as either
heat (q) or work (w).
• That is, E = q + w.
Exchange of Heat between
System and Surroundings
• When heat is absorbed by the system from the
surroundings, the process is endothermic.
Exchange of Heat between
System and Surroundings
• When heat is absorbed by the system from the
surroundings, the process is endothermic.
• When heat is released by the system to the
surroundings, the process is exothermic.
State Functions
Usually we have no way of knowing the
internal energy of a system; finding that value
is simply too complex a problem.
State Functions
• However, we do know that the internal energy of
a system is independent of the path by which the
system achieved that state.
– In the system below, the water could have reached
room temperature from either direction.
State Functions
• Therefore, internal energy is a state function.
• It depends only on the present state of the system, not on
the path by which the system arrived at that state.
• And so, E depends only on Einitial and Efinal.
State Functions
• However, q and w are
not state functions.
• Whether the battery is
shorted out or is
discharged by running
the fan, its E is the
same.
– But q and w are different
in the two cases.
Work
When a process
occurs in an open
container, commonly
the only work done is a
change in volume of a
gas pushing on the
surroundings (or being
pushed on by the
surroundings).
Work
We can measure the work done by the gas if
the reaction is done in a vessel that has been
fitted with a piston.
w = −PV
Enthalpy
• If a process takes place at constant pressure
(as the majority of processes we study do) and
the only work done is this pressure-volume
work, we can account for heat flow during the
process by measuring the enthalpy of the
system.
• Enthalpy is the internal energy plus the product
of pressure and volume:
H = E + PV
Enthalpy
• When the system changes at constant
pressure, the change in enthalpy, H, is
H = (E + PV)
• This can be written
H = E + PV
Enthalpy
• Since E = q + w and w = −PV, we can
substitute these into the enthalpy
expression:
H = E + PV
H = (q+w) − w
H = q
• So, at constant pressure the change in
enthalpy is the heat gained or lost.
Endothermicity and
Exothermicity
• A process is
endothermic, then,
when H is
positive.
Endothermicity and
Exothermicity
• A process is
endothermic when
H is positive.
• A process is
exothermic when
H is negative.
Enthalpies of Reaction
The change in
enthalpy, H, is the
enthalpy of the
products minus the
enthalpy of the
reactants:
H = Hproducts − Hreactants
Enthalpies of Reaction
This quantity, H, is called the enthalpy of
reaction, or the heat of reaction.
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g) H = -1604 kJ
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
2. When we reverse a reaction, we change the sign of
H:
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
2. When we reverse a reaction, we change the sign of
H:
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
H = +802 kJ
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
2. When we reverse a reaction, we change the sign of
H:
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
H = +802 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
2. When we reverse a reaction, we change the sign of
H:
3. Change in enthalpy depends on state:
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
2. When we reverse a reaction, we change the sign of
H:
3. Change in enthalpy depends on state:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
H = -802 kJ
Enthalpies of Reaction
For a reaction
1. Enthalpy is an extensive property (magnitude H is
directly proportional to amount):
2. When we reverse a reaction, we change the sign of
H:
3. Change in enthalpy depends on state:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
H = -802 kJ
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
H = -890 kJ
Enthalpies of Reaction
H reaction   H (products )   H (reactants )
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
b) Calculate the amount of heat transferred when
2.4g of Mg reacts at constant pressure.
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
b) Calculate the amount of heat transferred when
2.4g of Mg reacts at constant pressure.
moles Mg  2.4 g
24.3g / mol
 0.10 mol Mg
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
b) Calculate the amount of heat transferred when
2.4g of Mg reacts at constant pressure.
2
.
4
g
moles Mg 
 0.10 mol Mg
24.3g / mol
there is a ratio between the moles of Mg used
and the heat produced
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
b) Calculate the amount of heat transferred when
2.4g of Mg reacts at constant pressure.
moles Mg  2.4 g
24.3g / mol
- 1204 kJ
x

2 Mg
0.10 mol
 0.10 mol Mg
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
b) Calculate the amount of heat transferred when
2.4g of Mg reacts at constant pressure.
moles Mg  2.4 g
24.3g / mol
- 1204 kJ
x

2 Mg
0.10 mol
x   60kJ
 0.10 mol Mg
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
c) How many grams of MgO are produced during
an enthalpy change of 96.0 kJ?
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
c) How many grams of MgO are produced during
an enthalpy change of 96.0 kJ?
2 MgO
x

 1204kJ  96.0kJ
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
c) How many grams of MgO are produced during
an enthalpy change of 96.0 kJ?
2 MgO
x

 1204kJ  96.0kJ
x  0.16 mol MgO
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
c) How many grams of MgO are produced during
an enthalpy change of 96.0 kJ?
2 MgO
x

 1204kJ  96.0kJ
x  0.16 mol MgO
g MgO  0.16mol ( 40.3g / mol )
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
c) How many grams of MgO are produced during
an enthalpy change of 96.0 kJ?
2 MgO
x

 1204kJ  96.0kJ
x  0.16 mol MgO
g MgO  0.16mol(40.3g / mol )
 6.42 g
Enthalpies of Reaction
Problem: 5.33
2 Mg(s) + O2(g)  2 MgO(s)
H = -1204 kJ
d) How many kilojoules of heat are absorbed when
7.50g of MgO is decomposed into Mg and O2 at
constant pressure?
Enthalpies of Reaction
Problem: 5.33
2 MgO(s)  2 Mg(s) + O2(g)
H = 1204 kJ
d) How many kilojoules of heat are absorbed when
7.50g of MgO is decomposed into Mg and O2 at
constant pressure?
Enthalpies of Reaction
Problem: 5.33
2 MgO(s)  2 Mg(s) + O2(g)
H = 1204 kJ
d) How many kilojoules of heat are absorbed when
7.50g of MgO is decomposed into Mg and O2 at
constant pressure?
moles MgO  7.50 g
40.3g / mol
 0.186 mol
Enthalpies of Reaction
Problem: 5.33
2 MgO(s)  2 Mg(s) + O2(g)
H = 1204 kJ
d) How many kilojoules of heat are absorbed when
7.50g of MgO is decomposed into Mg and O2 at
constant pressure?
moles MgO  7.50 g
40.3g / mol
 0.186 mol
1204kJ
x

2 MgO 0.186 mol
Enthalpies of Reaction
Problem: 5.33
2 MgO(s)  2 Mg(s) + O2(g)
H = 1204 kJ
d) How many kilojoules of heat are absorbed when
7.50g of MgO is decomposed into Mg and O2 at
constant pressure?
moles MgO  7.50 g
40.3g / mol
 0.186 mol
1204kJ
x

2 MgO 0.186 mol
x  112kJ
Calorimetry
Calorimetry
Since we cannot
know the exact
enthalpy of the
reactants and
products, we measure
H through
calorimetry, the
measurement of heat
flow.
Calorimetry
• Use a calorimeter (measures heat flow).
• Two kinds:
– Constant pressure calorimeter (called a coffee
cup calorimeter)
– Constant volume calorimeter is called a bomb
calorimeter.
Specific Heat Capacity and Heat
Transfer
• The quantity of heat transferred depends
upon three things:
1.The quantity of material (extensive).
2.The difference in temperature.
3.The identity of the material gaining/losing heat.
Heat Capacity and Specific Heat
• The amount of energy required to raise the
temperature of a substance by 1 K (1C) is
its heat capacity.
• Molar Heat Capacity – The amount of heat
required to raise the temperature of one
mole of a substance by 1 Kelvin.
Heat Capacity and Specific Heat
Specific Heat (Specific Heat Capacity, c) – The amount
of heat required to raise 1.0 gram of a substance by 1
Kelvin.
q
c
mT 
c  specific heat
m  mass
T  change in temperature, T f  Ti
Constant Pressure Calorimetry
By carrying out a
reaction in aqueous
solution in a simple
calorimeter such as this
one, one can indirectly
measure the heat
change for the system
by measuring the heat
change for the water in
the calorimeter.
Constant Pressure Calorimetry
Because the specific
heat for water is well
known (4.184 J/mol-K),
we can measure H for
the reaction with this
equation:
q = m  c  T
Calorimetry
Constant-Pressure Calorimetry
- Atmospheric pressure is constant
H = qP
qsystem = -qsurroundings
- The surroundings are composed of the water in the
calorimeter and the calorimeter.
qsystem = - (qwater + qcalorimeter)
- For most calculations, the qcalorimeter can be ignored.
qsystem = - qwater
csystemmsystem Tsystem = - cwatermwater Twater
Bomb Calorimetry
Reactions can be
carried out in a sealed
“bomb,” such as this
one, and measure the
heat absorbed by the
water.
Bomb Calorimetry
• Because the volume in
the bomb calorimeter is
constant, what is
measured is really the
change in internal energy,
E, not H.
• For most reactions, the
difference is very small.
Calorimetry
Bomb Calorimetry (Constant-Volume Calorimetry)
- Special calorimetry for combustion reactions
- Substance of interest is placed in a “bomb” and filled to a high
pressure of oxygen
- The sealed bomb is ignited and the heat from the reaction is
transferred to the water
- This calculation must take into account the heat capacity of the
calorimeter (this is grouped together with the heat capacity of water).
qrxn = -Ccalorimeter(T)
Calorimetry
Problem 5.50
NH4NO3(s)  NH4+(aq) + NO3-(aq)
Twater = 18.4oC – 23.0oC = -4.6oC
mwater = 60.0g
cwater = 4.184J/goC
msample = 3.88g
qsample = -qwater
qsample = -cwatermwater Twater
qsample = -(4.184J/goC)(60.0g+3.88g)(-4.6oC)
qsample = 1229J
- Now calculate H in kJ/mol
Calorimetry
Problem 5.50
NH4NO3(s)  NH4+(aq) + NO3-(aq)
Twater = 18.4oC – 23.0oC = -4.6oC
mwater = 60.0g
cwater = 4.184J/goC
msample = 3.88g
qsample = 1229J
moles NH4NO3 = 3.88g/80.032g/mol = 0.04848mol
H = qsample/moles
H = 1229J/0.04848mol
H = 25.4 kJ/mol
Calorimetry
A 1.800g sample of octane, C8H18, was burned in a
bomb calorimeter whose total heat capacity is 11.66
kJ/oC. The temperature of the calorimeter plus
contents increased from 21.36oC to 28.78oC. What is
the heat of combustion per gram of octane? Per mole of
octane?
2 C8H18 + 25O2  16 CO2 + 18 H2O
Twater = 28.78oC – 21.36oC = 7.42oC
Ccal = 11.66kJ/oC
msample = 1.80g
Calorimetry
A 1.800g sample of octane, C8H18, was burned in a
bomb calorimeter whose total heat capacity is 11.66
kJ/oC. The temperature of the calorimeter plus
contents increased from 21.36oC to 28.78oC. What is
the heat of combustion per gram of octane? Per mole of
octane?
2 C8H18 + 25O2  16 CO2 + 18 H2O
Twater = 28.78oC – 21.36oC = 7.42oC
Ccal = 11.66kJ/oC
msample = 1.80g
qrxn = -Ccal (Twater)
Calorimetry
A 1.800g sample of octane, C8H18, was burned in a
bomb calorimeter whose total heat capacity is 11.66
kJ/oC. The temperature of the calorimeter plus
contents increased from 21.36oC to 28.78oC. What is
the heat of combustion per gram of octane? Per mole of
octane?
2 C8H18 + 25O2  16 CO2 + 18 H2O
Twater = 28.78oC – 21.36oC = 7.42oC
Ccal = 11.66kJ/oC
msample = 1.80g
qrxn = -Ccal (Twater)
qrxn = -11.66kJ/oC(7.42oC)
Calorimetry
A 1.800g sample of octane, C8H18, was burned in a
bomb calorimeter whose total heat capacity is 11.66
kJ/oC. The temperature of the calorimeter plus
contents increased from 21.36oC to 28.78oC. What is
the heat of combustion per gram of octane? Per mole of
octane?
2 C8H18 + 25O2  16 CO2 + 18 H2O
Twater = 28.78oC – 21.36oC = 7.42oC
Ccal = 11.66kJ/oC
msample = 1.80g
qrxn = -Ccal (Twater)
qrxn = -11.66kJ/oC(7.42oC) = -86.52kJ
Calorimetry
A 1.800g sample of octane, C8H18, was burned in a
bomb calorimeter whose total heat capacity is 11.66
kJ/oC. The temperature of the calorimeter plus
contents increased from 21.36oC to 28.78oC. What is
the heat of combustion per gram of octane? Per mole of
octane?
Hcombustion(in kJ/g)
Hcombustion = -86.52kJ/1.80g =
Calorimetry
A 1.800g sample of octane, C8H18, was burned in a
bomb calorimeter whose total heat capacity is 11.66
kJ/oC. The temperature of the calorimeter plus
contents increased from 21.36oC to 28.78oC. What is
the heat of combustion per gram of octane? Per mole of
octane?
Hcombustion(in kJ/g)
Hcombustion = -86.52kJ/1.80g = -48.1 kJ/g
Calorimetry
A 1.800g sample of octane, C8H18, was burned in a
bomb calorimeter whose total heat capacity is 11.66
kJ/oC. The temperature of the calorimeter plus
contents increased from 21.36oC to 28.78oC. What is
the heat of combustion per gram of octane? Per mole of
octane?
Hcombustion(in kJ/g)
Hcombustion = -86.52kJ/1.80g = -48.1 kJ/g
Hcombustion(in kJ/mol)
Hcombustion = -86.52kJ/0.01575mol =
Calorimetry
A 1.800g sample of octane, C8H18, was burned in a
bomb calorimeter whose total heat capacity is 11.66
kJ/oC. The temperature of the calorimeter plus
contents increased from 21.36oC to 28.78oC. What is
the heat of combustion per gram of octane? Per mole of
octane?
Hcombustion(in kJ/g)
Hcombustion = -86.52kJ/1.80g = -48.1 kJ/g
Hcombustion(in kJ/mol)
Hcombustion = -86.52kJ/0.01575mol = -5492 kJ/mol
Hess’s Law
H is well known for many reactions, and
it is inconvenient to measure H for
every reaction in which we are
interested.
• However, we can estimate H using H
values that are published and the
properties of enthalpy.
Hess’s Law
Hess’s law states that
“If a reaction is carried
out in a series of
steps, H for the
overall reaction will be
equal to the sum of
the enthalpy changes
for the individual
steps.”
Hess’s Law
Because H is a state
function, the total
enthalpy change
depends only on the
initial state of the
reactants and the final
state of the products.
Hess’s Law
– rxns in one step or multiple steps are additive
because they are state functions
• eg.
• CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
2H2O(g) 
2H2O(l)
• CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
 H = - 802 kJ
 H = - 88 kJ
 H = - 890 kJ
Practice
– Calculate  H for the conversion of graphite to
diamond:
• Cgraphite

Cdiamond
• Cgraphite + O2(g)  CO2(g)
• Cdiamond + O2(g)  CO2(g)
 H = -393.5 kJ
 H = -395.4 kJ
• Cgraphite + O2(g)  CO2(g)  H = -393.5 kJ
• CO2(g)  Cdiamond + O2(g)  H = 395.4 kJ
Cgraphite
 Cdiamond
 H = + 1.9 kJ
Enthalpies of Formation
- There are many type of H, depending on what you
want to know
Hvapor – enthalpy of vaporization (liquid  gas)
Hfusion – enthalpy of fusion (solid  liquid)
Hcombustion – enthalpy of combustion
(energy from burning a substance)
Enthalpies of Formation
- A fundamental H is the Standard Enthalpy
of Formation ( H of )
Enthalpies of Formation
- A fundamental H is the Standard Enthalpy
of Formation ( H of )
o

H
f ) – The
Standard Enthalpy of Formation (
enthalpy change that accompanies the
formation of one mole of a substance from
the most stable forms of its component
elements at 298 Kelvin and 1 atmosphere
pressure.
Enthalpies of Formation
- A fundamental H is the Standard Enthalpy of
Formation ( H of )
o

H
Standard Enthalpy of Formation (
f ) – The
enthalpy change that accompanies the formation of
one mole of a substance from the most stable
forms of its component elements at 298 Kelvin and
1 atmosphere pressure.
“The standard enthalpy of formation of the most
stable form on any element is zero”
Enthalpies of Formation
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in 3 steps:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in 3 steps:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• Imagine this as occurring
in 3 steps:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
• The sum of these
equations is:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Calculation of H
We can use Hess’s law in this way:
H = nHf(products) - mHf(reactants)


where n and m are the stoichiometric
coefficients.
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H =
=
=
=
[3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]
[(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]
(-2323.7 kJ) - (-103.85 kJ)
-2219.9 kJ
Enthalpies of Formation
Problem 5.72
a) N2O4(g) + 4 H2(g)  N2(g) + 4 H2O(g)
N2O4(g)
H2(g)
N2(g)
H2O(g)
9.66 kJ/mol
0 kJ/mol
0 kJ/mol
-241.82 kJ/mol
H = [1mol(H(N2)) + 4mol(H(H2O))] – [1mol(H(N2O4)) + 4mol(H(H2))]
H = [1mol(0kJ/mol) + 4mol(-241.82kJ/mol)] – [1mol(9.66kJ/mol) +
4mol(0kJ/mol)]
= -976 kJ
Enthalpies of Formation
Problem 5.72
b) 2 KOH(s) + CO2(g)  K2CO3(s) + H2O(g)
KOH(s)
CO2(g)
K2CO3(s)
H2O(g)
-424.7 kJ/mol
-393.5 kJ/mol
-1150.18 kJ/mol
-241.82 kJ/mol
H = [1mol(H(K2CO3)) + 1mol(H(H2O))] – [2mol(H(KOH)) + 1mol(H(CO2))]
H = [1mol(-1150.18kJ/mol) + 1mol(-241.82kJ/mol)] –
[2mol(-424.7kJ/mol) + 1mol(-393.5kJ/mol)]
= -149.1kJ
Energy in Foods
Most of the fuel in the food we eat comes
from carbohydrates and fats.
Foods and Fuels
Foods
• 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal.
• Energy in our bodies comes from carbohydrates and fats
(mostly).
• Intestines: carbohydrates converted into glucose:
C6H12O6 + 6O2  6CO2 + 6H2O, H = -2816 kJ
• Fats break down as follows:
2C57H110O6 + 163O2  114CO2 + 110H2O, H = -75,520 kJ
Fats contain more energy; are not water soluble, so are good for
energy storage.
Fuels
The vast majority
of the energy
consumed in this
country comes
from fossil fuels.
Foods and Fuels
Fuels
• Fuel value = energy released when 1 g of substance is
burned.
• Most from petroleum and natural gas.
• Remainder from coal, nuclear, and hydroelectric.
• Fossil fuels are not renewable.
• In 2000 the United States consumed 1.03  1017 kJ of
fuel.
• Hydrogen has great potential as a fuel with a fuel value
of 142 kJ/g.