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Chemistry 200
Focus 4
Thermodynamics: The First Law
Thermodynamics, is the study of Work
and Heat are Energy
Work
Energy
Heat
Heat & Work
Heat was thought of as a fluid, flowing from
a hot to cold substance.
Sadi Carnot, a French engineer, believed
heat flowed to produce work, just as water
flowed to turn a water wheel .
James Joule, English physicist,
showed that both heat and work
are forms of energy; now know
as the two fundamental concepts
of thermodynamics.
Systems verses the Surroundings
Thermodynamics studies how energy is transformed from one form into
another and transferred. Electricity generated in a power station is used a
great distance away.
The System
(transformed)
The Surroundings
(transferred)
To keep track of changes in energy, we always separate these two
components.
Systems verses the Surroundings
Everything around the system, such as the water bath in a reaction is
immersed, is called the surroundings. The surroundings are where
we make observations on energy transfer into or out of the system.
Systems verses the Surroundings
Universe
The system and the surroundings jointly
make up the universe.
You can “Open” the System to the
surroundings (P, V, T all change)
You can “Close” the System to the
surroundings (put a lid on it-constant V).
You can “Isolate” the System by preventing
temperature from escaping (constant
Temp=isothermal).
Systems verses the Surroundings
An open system
can exchange
both matter and
energy with the
surroundings, i.e.
automobile
engines and the
human body.
Isolated systems
have no contact
with its
surroundings. A
sealed, insulated,
rigid container; i.e.
a thermos.
A closed system has a fixed amount of matter, but it can
exchange energy with the surroundings, i.e. a cold pack
used for treating athletic injuries.
Work & Energy
Work is motion against an opposing force.
Work = opposing force × distance moved
The SI unit for work (energy), is joule, J,
where 1 J = 1 kg·m2·s-2
Work & Energy
Work is done when:
• Raising a weight against the pull of gravity;
• A chemical reaction in a battery does work when it
pushes an electric current through a circuit;
• A gas in a cylinder does work when it expands and
pushes back a piston.
Internal energy, U
In thermodynamics, the capacity of a system to do work is called its
internal energy, U.
Internal energy (U): sum of the kinetic and potential energies.
Absolute internal energy is not measurable because it includes the
energies of all the atoms, their electrons, and the components of their
nuclei .
The best we can do is to measure changes in internal energy or ΔX:
ΔX = Xfinal - Xinitial
Internal energy, U
If a system does 15 J of work, it used up some of its store of energy, so we
say its internal energy has fallen by 15 J and write ΔX = -15 J (when your body
burns food).
A negative value of ΔX, or ΔU - 15 J, means X has decreased.
If we do work on the system, lets say 15 J of work, we say its internal energy
has increased by 15 J and write ΔX = +15 J (winding a spring increases U of
the system, or we eat a sandwich).
A positive value of ΔX, or ΔU + 15 J, means X has increased.
Work: Expansive and nonexpansive
Expansion work is a change in volume of a system.
• Expansion work is when a gas expands in a cylinder pushing back a
piston which pushes out against the atmosphere.
Nonexpansion work is the flow of electrical current.
• chemical reactions do nonexpansive work by causing electrical current to
flow.
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Expansive Work
We now introduce two forms of Expansive Work (W = - PΔV):
1. Expansion at constant pressure;
2. Expansion at constant temperature with changing Pexternal.
Work
Energy
Heat
Expansive Work
Work = opposing force × distance moved
The opposing force is the external
pressure pushing against the outer
face of the piston.
How much expansion work is done is
when the system expands, ΔV,
against the external pressure Pex.
Work is done when the gas expands, the piston pushes against the
opposing force (atmospheric pressure).
Expansive Work
ΔU = “-”
work = PexA × Δ d
A × Δ d = ΔV

w = -Pex ΔV
Volume
The system does expansive work by pushing against the opposing pressure
(force), so the system losses energy as work: ΔU = “-.”
“-” because it is assumed work will be done so internal energy will decrease.
• so if ΔV is positive (an expansion), w is negative;
• so if ΔV is negative (contraction), w is positive.
SI units for Pressure
External pressure is expressed in Pascal's:
1 Pa = 1 kg·m-1·s-2
Change in External Pressure, by a cubic meter (m3) is:
w = -Pex ΔV
1 Pa · 1 m3
or 1 kg·m-1·s-2·m3 = 1 J
101.323 J = 1 L·atm,
Work: Free Expansion
If the external pressure is zero (a vacuum), than w = 0
No expansion work is done in a vacuum because there is no opposing force,
there is nothing to push against.
Expansion against zero pressure is called free expansion.
Practice 1: Suppose a gas expands by 500. mL (0.500 L) against a
pressure of 1.20 atm. How much work is done in the expansion in Joules?
w = -Pex ΔV
w = - 1.20 atm × 0.500 L = - 0.600 L·atm
Since 101.323 J = 1 L · atm
- 0.600 L · atm ×
w = -60.8 J
ΔU = -60.8 J
101.323 J
= - 60.8 J
1 L · atm
Changing external pressure
Work is a measure of P and V changes.
Measuring work where V changes in either direction as a "reversible"
expansion must be done in "infinitesimal" changes.
Of course this means that pressure increases or decreases infinitesimally
as well.
These reversible processes are of the greatest importance in
thermodynamics, and we shall encounter them many times.
Measuring isothermal work with changing external pressure
The simplest reversible change is an
isothermal (constant temperature)
expansion of an ideal gas.
The criteria for isothermal expansion is
the temperature remains unchanged.
The reason is that temperature implies
change in motion of atoms so an
"infinitesimal" change in P or V is
impossible if the atoms speed up or
slow down during the expansion
(contraction).
An ice bath can maintain
constant temperature.
Measuring isothermal work with changing external pressure
Reversible, isothermal work (constant temperature)
vfinal
wiso = -nRT ln
vinitial
Practice 2: A piston confines 0.100 mol Ar (g) in a volume of 1.00 L at
25°C. Two experiments are performed.
1.
2.
The piston is allowed to expand through 1.00 L against a constant
pressure of 1.00 atm.
The piston is allowed to expand reversibly and isothermally to the
same final volume.
Which process does more work?
We know how to calculate the amount of work two ways:
1. For expansion against constant external pressure:
w = -Pex ΔV
2. For reversible, isothermal expansion: w = -nRT ln
vfinal
vinitial
Practice 2: A piston confines 0.100 mol Ar (g) in a volume of 1.00 L at
25°C. Two experiments are performed. In one, the piston is allowed to
expand through 1.00 L against a constant pressure of 1.00 atm. In the
second, it is allowed to expand reversibly and isothermally to the same
final volume. Which process does more work?
1. we use w = -Pex ΔV = 1.00 atm × 1.00 L ×
2. isothermal expansion: w = -nRT ln
101.323 J
= - 101 J
1 L·atm
vfinal
vinitial
= 0.100 mol × 8.3145 J·K-1·mol-1 × 298 K × ln
2.00 L
1.00 L
= -172 J
The reversible process does more work (yellow) than
constant pressure (green).
Heat and Temperature
Is there a difference between heat and temperature?
Heat is what Physically happens when two
objects at different temperatures touch.
Energetic molecules in higher temperature
regions vigorously stimulate slower moving
molecules in the lower temperature region into
higher energetic states.
Heat: Flow of energy due to a T difference.
Heat and Temperature
This results in a decrease for higher energy molecules (they become
cooler) and an increase the lower energetic molecules (they become
warmer).
Temperature is measured with a thermometer, and it must come into
contact with an object. Fluid inside the thermometer, often mercury or
alcohol, reacts to changes in the energetic molecules it makes contact with.
Cold
Water
Hot
Water
Heat & Work
U = q + w
Energy flows into system via heat (endothermic): q = +x
Energy flows out of system via heat (exothermic): q = -x
Surroundings
Surroundings
Energy
Energy
System
System
U  0
U  0
Endothermic
Exothermic
System does work on surroundings: w = -x  U < 0
Surroundings do work on the system: w = +x  U > 0
Heat
units of heat:
calorie (cal)
joule (J)
English system
SI system
Joule: Energy (heat) required to raise T of one gram of water by 1C.
1 cal = 4.184 J
Food energy is measured in Calories (note the capital C).
1 Cal = 1 kcal = 1000 cal
Heat
Amount of heat = specific heat capacity × mass × change in temperature
q = Cp × m × (Tfinal – Tinitial)
Cp = Specific heat capacity (cal/g °C or J/g °C)
Tfinal = final temperature (ºC)
m = mass (g)
Tinitial = initial temperature(ºC)
Heat
• Specific heat capacity is
the energy required to
change the temperature of
a mass of one gram of a
substance by one Celsius
degree.
Two forms of heat capacity
specific heat, Cp =
C
m
where m = molar mass, and
molar heat capacity, Cpm =
Cp
where n = moles.
n
For Water the specific heat is 4.18 J·(°C)-1·g-1 or 4.18 J·K-1·g-1
and the molar heat capacity is 75 J·K-1·mol-1 .
Also commonly used in the following form:
specific heat capacity: q = m Cp ∆T
molar heat capacity: q = n Cpm ∆T
Heat
Practice 3:
• A silver-gray metal weighing 15.0 g requires 133.5 J to
raise the temperature by 10.°C. Find the heat capacity.
q = Cp x m x T
(133.5 J) = Cp x (15.0 g) x (10.°C)
Cp = 0.89 J/g°C
Al
Can you determine the
identity of the metal using
Table 10.1?
Practice 4: Calculating the heat needed to bring about a rise in
temperature Calculate the heat necessary to increase the temperature of
2.00 mol H2O(l) by 20.°C from room temperature. (Cpm = 75 J.K-1.mol-1)
q = Cp ∆T = n Cpm ∆T
q = 2.00 mol × 75 J·K-1·mol-1 × 20. K = + 3.0 kJ (“+” adding heat)
Common student error:
∆T (°C) = ∆T (K)
When working with ∆T, it’s important for the
student to remember that
0.0 °C → 100.0 °C
∆T = 100.0 °C
∆T (°C) = ∆T (K).
In (a) above, ∆T= 20.°C, is the same as
∆T= 20. K
273.0 K → 373.0 K
∆T = 100.0 K
Calorimeter
A calorimeter is a common device to measure energy transferred as heat.
1. A calorimeter has a known Ccal (calorimeter constant).
2. When a sample is placed in the calorimeter, heat
from the experiment goes into the calorimeter.
3. ∆T of the calorimeter, is the heat lost by the
experiment into the calorimeter.
4. By knowing information about the calorimeter (1)
and temperature changes (3) we can calculate,
q = Ccal ∆T, the heat form our experiment.
Calorimeter
Heat from the Experiment = Heat of the Calorimeter.
The sample generates heat. Then, that
heat is absorbed the calorimeter
(qcal = -qexp).
Calorimeters are calibrated with a
known specific heat, usually water.
Calibration means finding the
calorimeter constant Ccal.
A more sophisticated version made of metal,
called a bomb calorimeter.
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Practice 5: A constant-volume calorimeter was calibrated by carrying out a
reaction known to release 1.78 kJ of heat in 0.100 L of solution in the
calorimeter, resulting in a temperature rise of 3.65°C. Next, 50. mL of 0.20 M
HCl(aq) and 50. mL of 0.20 M NaOH(aq) were mixed in the same
calorimeter and the temperature rose by 1.26°C. What is the change in the
internal energy of the neutralization reaction?
First, we calibrate or find the amount of heat absorbed by the calorimeter.
qcal = Ccal ∆T or Ccal =
Ccal =
qcal
plug in our data,
∆T
1.78 kJ
= 0.488 kJ°C−1
3.65°C
Second, we find the heat the sample generated, we do this by measuring the
amount of heat was absorbed by the calorimeter.
Here we notice the calorimeter’s temperature rose by 1.26°C
qcal = Ccal ∆T plugging in our data, 0.488 kJ°C−1 × 1.26°C = 0.614 kJ
For our reaction, heat is leaving the reaction so q = ΔU = - 0.614 kJ
The first law of thermodynamics
Law of conservation of energy: energy of the universe is constant.
So far, we have considered the transfer of energy as work or heat
separately.
Many processes, involve changes in internal energy that are both work
and heat.
When a spark ignites the mixture of gasoline vapor and air in the engine
of a moving automobile, the vapor burns and expands, transferring
energy to its surroundings as both work and heat.
ΔU = q + w
Practice 6: A system was heated by using 300. J of heat, yet it was found
that its internal energy decreased by 150. J (the system loses it’s internal
energy). Calculate was work done on the system or did the system do work?
ΔU = q + w
ΔU = -150. J
lost
q = +300. J
added to system
w = ΔU - q
or
-150. J - (+300. J) = - 450. J
The “-” means work lost by the system or the system did – 450 J of
work.
State functions
The first law of thermodynamics: The internal energy of an isolated system
is constant.
If we were to prepare a second system consisting of exactly the same
amount of substance in exactly the same state as the first, and inspect that
system, we would find that the second system has the same internal energy
as the first system.
A property that depends only on the current state of the system and is
independent of how that state was prepared, is called a State Function.
State functions
Pressure, volume, temperature, mass and altitude of a system are also
state functions, and so is internal energy.
Work is not a state function, the physical exertion of a climber climbing
path A or B requires a wildly different amount (and kind) of “work” by the
climber; work = P∆V
If in one step we raised
its temperature, its
internal energy would
change by a certain
amount.
In multiple steps of raising
and lowering the
temperature, we use a
different amount of
energy, but the change is
the same.
The net change (A to B) in the internal energy in each step is exactly the same.
State functions
In the next example internal energy, ΔU = q + w, is calculated by two
different paths, both over the same range.
1. One path calculates internal energy only reversibly,
2. The other path is a combination of both reversible and irreversible.
The biggest difference between the two is for a reversible process the
temperature remains constant (isothermal).
Also remember that all isothermal expansions are the most efficient
(though not common).
Practice 7: Suppose that 1.00 mol of ideal gas molecules at 292 K at
3.00 atm expands from 8.00 L to 20.00 L and has a final pressure of 1.20
atm. Find the amount of work by two different paths.
(a) Path A is an isothermal, reversible expansion.
(b) Path B has two parts. In step 1, the gas is cooled at constant volume
until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed
to expand against a constant pressure of 1.20 atm until its volume is
20.00 L and T = 292 K. Determine for each path the work done, the heat
transferred, and the change in internal energy (w, q, and ΔU).
Isothermal: w = -nRT ln
vfinal
vinitial
Practice 7: Path B has two parts. In step 1, the gas is cooled at constant
volume until its pressure has fallen to 1.20 atm. In step 2, it is heated and
allowed to expand against a constant pressure of 1.20 atm until its
volume is 20.00 L and T = 292 K. Determine for each path the work done,
the heat transferred, and the change in internal energy (w, q, and ΔU).
Step 1, cool, keeping V constant. P↓T↓
Step 2, let V change V↑T↑,
keeping P constant
Practice 7: Suppose that 1.00 mol of ideal gas molecules at 292 K and
3.00 atm expands from 8.00 L to 20.00 L and a final pressure of 1.20 atm
by two different paths. (a) Path A is an isothermal, reversible expansion.
Part A: ΔU = q + w
For reversible, isothermal expansion:
w = -nRT ln
vfinal
vinitial
1.00 mol × 8.3145 J·K-1·mol-1 × 292 K × ln
20.00
8.00
= -2.22 kJ
ΔU = q + w, note for an isothermal expansion ΔU = 0
(this is just a gas expanding, so it is not a chemical
reaction losing energy).
0 = q + w or q = -w so here
= +2.22 kJ
q = - (-2.22 kJ)
Practice 7: Path B has two parts. In step 1, the gas is cooled at constant volume
until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed to
expand against a constant pressure of 1.20 atm until its volume is 20.00 L and T
= 292 K. Determine for each path the work done, the heat transferred, and the
change in internal energy (w, q, and ΔU).
Part B: ΔU = q + w
Step 1 work: cooling at constant V, no work or w = 0
1
Step 2 work: w = -Pex ΔV, ΔV = (20.00 – 8.00 )L = 12.00 L
101.323 J
w = -Pex ΔV = 1.20 atm × 12.00 L ×
= - 1.46 kJ
1 L · atm
Step 1 + Step 2 = 0 + -1.46 kJ = - 1.46 kJ
ΔU = q + w, note for an isothermal expansion ΔU = 0 (same as (a)).
0 = q + w or q = -w so here
q = - (- 1.46 kJ) = +1.46 kJ
2
Practice 7: Suppose that 1.00 mol of ideal gas molecules at 292 K and 3.00 atm expands
from 8.00 L to 20.00 L and a final pressure of 1.20 atm by two different paths. (a) Path A is
an isothermal, reversible expansion. (b) Path B has two parts. In step 1, the gas is cooled at
constant volume until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed
to expand against a constant pressure of 1.20 atm until its volume is 20.00 L and T = 292 K.
Determine for each path the work done, the heat transferred, and the change in internal
energy (w, q, and ΔU).
Nonreversible does
less work, +1.46 kJ.
Isothermal, reversible does
more work, +2.22 kJ.
ΔU = 0 for both
Enthalpy
In a chemistry lab, many reactions are done in open containers, under
nonexpansive work (ΔV = 0) conditions at a constant pressure of 1 atm.
ΔU = q + w , and for nonexpansion work w = 0
So at constant volume: ΔU = q
In this case energy supplied is strictly heat, q only.
Enthalpy (Thermochemistry): heat of chemical reactions.
For a reaction in constant pressure,
the change of enthalpy is equal to energy that flows as heat.
Hp = heat
Constant pressure
Enthalpy
Heat Transfers at Constant Pressure:
The new state function is called enthalpy, H:
H = U + PV
U, P, and V are the internal energy, pressure, and volume of the system.
Enthalpy is a state function because U (from the first law), P, and V are also
state functions.
Constant pressure means the system is open, and the work is pushing
against the atmosphere; that is the constant pressure is the constant
pressure of the atmosphere.
Enthalpy
It follows from enthalpy, H = U + PV, that changes at constant pressure:
ΔH = ΔU + PΔV
We substitute for ΔU = q + w , to get, ΔH = q + w + PΔV
A system doing expansion work only, we substitute for
w = -Pex ΔV
to get:
ΔH = q + -Pex ΔV + PΔV
Now, if we leave our system open to the atmosphere, the pressure of the
system is the same as the external pressure; so Pex = P, and ΔH = q + -PΔV
+ PΔV, the last two terms cancel to give us:
ΔH = q (constant pressure)
Previously:
ΔU = q (constant volume)
Enthalpy
Surroundings
Energy
System
Exothermic: H is negative.
(heat flows out of the system).
Exothermic
Surroundings
Energy
Endothermic: H is positive.
(heat flows into the system).
System
Endothermic
Remember that ΔH is a constant pressure (open to the atmosphere) scenario.
Enthalpy
The enthalpy of a system is like a measure of the
height of water in a reservoir. When a reaction
releases 208 kJ of heat at constant pressure, the
"reservoir" falls by 208 kJ and ΔH = -208 kJ.
Exothermic (burning)
Surrounding
Reactant
(PE)
Energy released to the surroundings as heat
Product
Exothermic = exit!
Practice 8: In an endothermic reaction at constant pressure, 30. kJ of
energy entered the system as heat. The products took up less volume than
the reactants, and 40. kJ of energy entered the system as work as the
outside atmosphere pressed down on it. What are the values of (a) ΔH and
(b) ΔU for this process?
(a) ΔH = q
at constant pressure
The reaction absorbs energy so ΔH = + 30.0 kJ
(b) ΔU = q + w, where w = -Pex ΔV ,
since ΔV is Vfinal – V initial, and here Vfinal < V initial so ΔV is “-”
then w = - 40.0 kJ × “-” = + 40.0 kJ
ΔU = + 30.0 kJ + (+ 40.0 kJ) = + 70.0 kJ (an increase in U)
The Enthalpy of Physical Change
Phase changes approximate how far apart molecules are.
Vaporization, requires adding energy to break and separate
intermolecular attractions, an endothermic process.
Freezing, is exothermic because energy is given off to allow new
intermolecular attractions to form between molecules.
Phase changes usually take place at constant pressure, the heat transfer
is due to changes in enthalpy.
Enthalpy of vaporization
Phase change: liquid to a gas or vaporizing is called the enthalpy of
vaporization, ΔHvap:
Enthalpy of vaporization,
ΔHvap = Hm(vapor) - Hm(liquid)
Where ΔHm is the molar heat
For water at its boiling point, 100°C, ΔHvap = 40.7 kJ·mol-1, and at 25°C
the value is ΔHvap = 44.0 kJ·mol-1.
This means to vaporize 1.00 mol H2O(l) (18.02 g of water) at 25°C and
constant pressure, requires 44.0 kJ of energy as heat.
All enthalpies of vaporization are positive (endo).
Stronger intermolecular forces, such as hydrogen bonds, have the
highest enthalpies of vaporization.
Practice 9: Heating a sample of ethanol, C2H5OH, of mass 23 g to its
boiling point, it was found that 22 kJ was required to vaporize all the ethanol.
What is the enthalpy of vaporization of ethanol at its boiling point?
Enthalpy of vaporization, ΔHvap = Hm(vapor)
Where Hm is the molar heat, so we need to find kJ·mole-1.
46.069 g·mol-1 C2H5OH.
23 g C2H5OH ×
1 mol C2H5OH
= 0.50 mole C2H5OH
46.069 g C2H5OH1
22 kJ C2H5OH
= 44 kJmol-1 per 1 mole C2H5OH
0.50 mol C2H5OH
Enthalpy of fusion
Phase change: solid to a liquid or melting (fusion) is called the enthalpy of
fusion,
ΔHfus = Hm(liquid) - Hm(solid)
The enthalpy of fusion of water at 0.0 °C is 6.0 kJ·mole-1: to melt 1.0 mol
H2O(s) (18 g of ice) at 0.0 °C, we have to supply 6.0 kJ of heat.
Vaporizing water takes much more energy (more than 40 kJ) because to
vaporize a gas, its molecules are separated completely, and KE increases
dramatically. In melting, the molecules stay close together, and so the
forces of attraction and repulsion are nearly as strong as in the solid.
Enthalpy of freezing
Freezing is the change from liquid to solid.
Enthalpy is a state property:
ΔHreverse process = - ΔHforward process
Since the enthalpy of fusion of water at 0.0°C is 6.0 kJ·mole-1 the enthalpy of
freezing for water is at 0.0°C is - 6.0 kJ·mole-1
Enthalpy of condensation
Condensation is the change from gas to liquid. So this process is the
reverse of ΔHvap. For water at boiling, 100°C, ΔHvap = 40.7 kJ·mol-1 ,
therefore
ΔHcondensation = - 40.7 kJ·mol-1
Enthalpy of sublimation
Sublimation is the direct conversion of a solid
into its vapor.
ΔHsublimation = Hm(vapor) - Hm(solid)
Frost disappears on a cold, dry morning as the
ice sublimes directly into water vapor. Solid
carbon dioxide also sublimes, which is why it is
called dry ice.
Enthalpy of deposition
Deposition is the reverse of sublimation a vapor into its solid.
ΔHdeposition = Hm(solid) - Hm(vapor)
Heating Curves
A heating curve shows the
variation in the temperature of
a heated sample.
2
As a sample of ice is heated
the temperature rises steadily.
1
1. In a solid the molecules are still locked together, as they oscillate
vigorously around their mean positions.
2. As the temperature rises reaching the melting point, the molecules
have enough energy to move past one another, overcoming the
attractive forces between molecules.
Heating Curves
3. At this temperature, all
the added energy is used
break all the attractive
forces, the temperature
remains constant at the
melting point until all the ice
has melted.
5
2
6
4
3
1
4. Only then does the temperature rise again, and the rise continues right up
to the boiling point.
5. At the boiling point, the temperature rise again comes to a halt.
6. Now the water molecules have enough energy to escape into the vapor
state, and all the heat supplied is used to form the vapor.
Heating Curves
A
7. After the sample has
evaporated and heating
continues, the temperature
of the vapor rises again.
B
5
6
A
A. Steeper slopes mean lower the heat capacity. Both the solid and vapor
phases have steeper slopes so the liquid has a greater heat capacity.
B. The liquids high heat capacity is due largely to loose intermolecular
hydrogen bonds that can take up more energy than stiff chemical bonds
between atoms found in the solid.
7
 Temperature (°C) 
Practice 10: Use the following information to construct a heating curve for
bromine from -7.2°C to 70.0°C. The molar heat capacity of liquid bromine
is 75.69 kJ·mole-1 and that of bromine vapor is 36.02 kJ·mole-1. For
bromine the specific heat of vaporization 0.225 J·g-1·K-1 and of liquid is
0.473 J·g-1·K-1. Bromine melts at -7.2°C and boils at 58.78°C. Also calculate
the energy to melt 10.0 g of bromine, Br2.
70.0

ΔHvap
58.78

Hg(gas)


Hg(liquid)
ΔHfus
-7.2
Htotal = ΔHfus+Hg(liq) + ΔHvap+ Hg(gas)
 Heat supplied 
10.0 g Br2
ΔHfus = 75.69 kJ·mole-1
Cg(liq) = 0.473 J·g-1·K-1
ΔHvap = 36.02 kJ·mole-1
Cg(vap) = 0.225 J·g-1·K-1
q = C∆T
Some H’s are in mole, so 10.0 g Br2 ×
Step 1 (melt/fus): 0.0626 mol ×
1 mol Br2
= 0.0626 mol Br2
159.8 g Br2
75.69 kJ 1000 J
×
= 4740 J
1 mol Br2 1 kJ
Step 2 (heat liq): gCΔT: 10.0 g × 0.473 J·g-1·K-1 × 66.0 K = 312 J
Step 3 (vap): 0.0626 mol ×
36.02 kJ 1000 J
×
= 2250 J
1 mol Br2
1 kJ
Step 4 (heat gas): 10.0 g × 0.225 J·g-1·K-1 × 11.2 K = 25.2 J
Practice 10: Use the following information to construct a heating curve for bromine from -7.2°C
to 70.0°C. The molar heat capacity of liquid bromine is 75.69 kJ·mole-1 and that of bromine
vapor is 36.02 kJ·mole-1. For bromine the specific heat of vaporization 0.225 J·g-1·K-1 and of
liquid is 0.473 J·g-1·K-1. Bromine melts at -7.2°C and boils at 58.78°C. Also calculate the energy
to melt 10.0 g of bromine, Br2.
 Temperature (°C) 
70.0

ΔHvap
58.78

-7.2

Hg(gas)

Hg(liquid)
ΔHfus
Htotal = ΔHfus+Hg(liq) + ΔHvap+ Hg(gas)
 Heat supplied 
Total: 4740 J + 312 J + 2250 J + 25.2 J = 7330 J
Enthalpy of chemical change (reaction)
Knowing enthalpies of chemical change help us determine which are good
fuels, or to study of biochemical processes.
Every chemical reaction is accompanied by energy transfer as heat.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(I) ΔH = - 890. kJ
1 mol CH4 (g) produces 890. kJ heat (energy).
The stoichiometric coefficients corresponding to an enthalpy change.
Enthalpy
Practice 11:
S(s) + O2(g)  SO2(g)
ΔH = –296 kJ
• Calculate the quantity of heat released when 2.10 g of sulfur is
burned in oxygen at constant pressure.
1 mol S(s) produces 296 kJ heat (energy).
2.10 g S x
1 mol S
32.26 g S
0.0655 mol S x
= 0.0655 mol S
– 296 kJ
1 mol S
= – 19.4 kJ
Use the H value like a conversion factor.
Practice 12: When 0.113 g of benzene, C6H6 (78.12 g·mole-1) burns in
excess O2 in a calibrated calorimeter with a heat capacity of 551 J·(°C)-1, the
temperature of the calorimeter rises by 8.60°C. Write the thermochemical
equation and determine the reaction enthalpy.
2 C6H6(g) + 15 O2(g) → 12 CO2(g) + 6 H2O(I)
H = ?
Total heat from the calorimeter using qrxn = - qcal = CcalΔT
551 J·(°C)-1 × 8.60°C = - 4.74 kJ
This is from 0.113 g, we need to know per mole:
0.113 g C6H6 ×
1 mol C6H6
= 0.00145 mol C6H6
78.12 g C6H6
−4.75 kJ
= - 3.28 × 103 kJ per mole C6H6
0.00145 mol C6H6
Our final answer is kJ per mol, though we only write it as kJ, mole is implied.
Hess’s Law
State function: a property of system that changes independently of its pathways.
Enthalpy is a state function.
In a chemical reaction, change of enthalpy is the same
whether the reaction takes place in one step or in a series of steps.
1 Step
N2(g) + 2O2(g)  2NO2(g)
H1 = 68 kJ
2 Steps
N2(g) + O2(g)  2NO(g)
2NO(g) + O2(g)  2NO2(g)
H2 = 180 kJ
H3 = -112 kJ
N2(g) + 2O2(g)  2NO2(g)
H2 + H3 = 68 kJ
Two rules about enthalpy
1. If a reaction is reversed, the sign of H is also reversed.
N2(g) + 2O2(g)  2NO2(g)
H1 = 68 kJ
2NO2(g)  N2(g) + 2O2(g)
H1 = -68 kJ
2. If the coefficients in a balanced reaction are multiplied by an integer,
the value of H is also multiplied by the same integer.
2
N2(g) + 2O2(g)  2NO2(g)
H1 = 68 kJ
2N2(g) + 4O2(g)  4NO2(g)
H1 = 2  68 kJ = 136 kJ
Combining Reaction Enthalpies: Hess's Law
Hess's law is a way to find unknown enthalpies.
• Step 1: Select one of the reactants in the overall reaction and write down
a chemical equation in which it also appears as a reactant;
• Step 2: Select one of the products in the overall reaction and write down
a chemical equation in which it also appears as a product;
• Step 3: Cancel unwanted species in the sum;
• Step 4: Once the sequence is complete, combine the standard reaction
enthalpies;
Note: Reversing the equation, changes the sign of enthalpy;
Note: Multiplying by a factor and so multiply the reaction enthalpy by the
same factor.
Practice 13: Using Hess's law, calculate the standard enthalpy for
3 C(gr) + 4 H2(g) → C3H8(g) from the following experimental data:
(a) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH° = -2220. kJ
(b) C(gr) + O2(g) → CO2(g)
ΔH° = -394 kJ
(c) H2(g) + ½ O2(g) → H2O(l)
ΔH° = -286 kJ
Step 1: Treat carbon as a reactant, (b) and multiply it through by 3.
3 C(gr) + 3 O2(g) → 3 CO2(g) ΔH° = 3(-394) kJ = -1182 kJ
Step 2: Reverse equation (a), changing the sign of H.
3 CO2(g) + 4 H2O(l) → C3H8(g) + 5 O2(g) ΔH° = +2220. kJ
Simplify by combining the two previous equations.
3 C(gr) + 4 H2O(l) → C3H8(g) + 2 O2(g) ΔH° = +1038. kJ
Continued: Overall reaction 3 C(gr) + 4 H2(g) → C3H8(g)
(a) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH° = -2220. kJ
(b) C(gr) + O2(g) → CO2(g)
ΔH° = -394 kJ
(c) H2(g) + ½ O2(g) → H2O(l)
ΔH° = -286 kJ
Previous:
3 C(gr) + 4 H2O(l) → C3H8(g) + 2 O2(g) ΔH° = +1038. kJ
Step 3: To cancel unwanted H2O and O2 , multiply equation (c) by 4.
4 H2(g) + 2 O2(g) → 4 H2O(l)
ΔH° = 4(-286) kJ = -1144 kJ
Step 4: Combine and simplify
3 C(gr) + 4 H2(g) → C3H8(g)
ΔH° = -106 kJ
The Relation Between ΔH and ΔU
Previous:
ΔH = ΔU + PΔV
For reactions in which no gas is generated or consumed, little expansion
work is done the difference between ΔH and ΔU is negligible; so we can
set ΔH = ΔU.
However, if a gas is formed in the reaction and expansion work is done
the difference can be significant.
We can use the ideal gas law to relate the values of ΔH and ΔU for gases
that behave ideally.
The relation between ΔH and ΔU, is H = U + PV, or H = U + nRT
ΔH = ΔU + ΔngasRT
Practice 14: constant-volume calorimeter showed that the heat generated
by the combustion of 1.000 mol glucose molecules in the reaction
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(I) is 2559 kJ at 298K, and so
ΔU = -2559 kJ. What is the change in enthalpy for the same reaction?
ΔH = ΔU + ΔngasRT
We need Δngas = 12 mol final – 6 mol initial = + 6 mol
R = 8.314 J·K-1·mol-1
ΔH = -2559 kJ + ( 6 mol × 8.314 J·K-1·mol-1 × 298 K)
= -2559 kJ + 14.9 kJ = - 2544 kJ
Notice that ΔH is less negative (more positive) than ΔU for reactions that
generate gases: less energy is obtained because some of the energy is used
to expand to make room for the reaction products.
Standard Reaction Enthalpies
The standard reaction enthalpy, ΔH°, is the reaction enthalpy when reactants
are in their standard states.
Standard state: pure form of a substance at exactly 1 bar.
Combining standard enthalpies of formation to calculate a standard
reaction enthalpy.
First calculate the reaction enthalpy for the formation
of all products.
Second, calculate the reaction enthalpy for all
reactants.
The difference between these two totals is the
standard enthalpy of the reaction.
𝜟𝑯° = 𝜮n 𝜟𝑯𝒇 ° 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 − 𝜮n 𝜟𝑯𝒇 ° 𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔
n is the amounts in moles, Σ (sigma) means sum.
Practice 15: You have an inspiration: maybe diamonds would make a great
fuel! Calculate the standard enthalpy of combustion of diamonds.
C(s, dia) + O2(g) →
Δ𝐻𝑓 ° (kJ·mol-1) =
1.895
0.0
CO2(g)
-393.51
Δ𝐻° = Σn Δ𝐻𝑓 ° 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − Σn Δ𝐻𝑓 ° 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
= (1 mol ) Δ𝐻𝑓 °(CO2(g)) - [(1 mol) Δ𝐻𝑓 °(C(s, dia)) + (1 mol ) Δ𝐻𝑓 °(O2(g)]
= (1 mol ) (-393.51 kJ·mol-1) – [(1 mol) (1.895 kJ·mol-1) + (1 mol) (0)]
= -393.51 - 1.895
= -395.41 kJ (a little better than ethanol made from corn)
The Born-Haber Cycle
In focus 2 we calculated energy changes in an ionic system to find lattice
energy.
The difference in molar enthalpy of solid and gas is called the lattice
enthalpy of the solid, ΔHL.
The lattice enthalpy of a solid cannot be measured directly.
Instead we measure it indirectly by using an application of Hess's law.
The procedure uses a Born-Haber cycle, a closed path of steps.
The Born-Haber Cycle
In a Born-Haber cycle, we
a) break apart the bulk elements into atoms,
b) ionize the atoms,
b
a
c) combine the gaseous ions to form the
ionic solid,
d
d) then form the elements again from the
ionic solid,
The sum of a complete Born-Haber cycle is zero, because the enthalpy of
the system must be the same at the start and finish.
Note: Lattice energies are “-”, though when calculated with the Born-Haber
cycle the value will be positive.
c
Practice 16: Devise and use a Born-Haber cycle to calculate the lattice
enthalpy of potassium chloride.
1.
2.
3.
4.
form gaseous atoms,
ionize the atoms to form gaseous ions,
allow the ions to form an ionic solid,
and then convert the solid back into the
pure elements.
2
2
1
1
4
(s)
3
Begin with the pure elements K(s) + ½ Cl2(g)
1. atomize them to form gaseous atoms,
K(s) → K(g)
Δ𝐻𝑓 = + 89 kJ·mol-1
½ Cl2(g) → Cl(g)
Δ𝐻𝑓 = + 122 kJ·mol-1
2. ionize the atoms to form gaseous ions,
K(g) → K+ (g) (1st ionization)
Δ𝐻𝑓 = + 418 kJ·mol-1
Cl(g) → Cl- (g) (electron affinity) Δ𝐻𝑓 = - 349 kJ·mol-1
3. allow the ions to form an ionic solid,
K+ (g) + Cl- (g) → KCl(s)
Δ𝐻𝑓 = -ΔHL
4. then convert the solid back into the pure elements.
KCl(s) → K(s) + ½ Cl2(g)
Δ𝐻𝑓 = -(-437 kJ·mol-1)
note sign is “-” because the reaction is reversed.
All the steps together sum to 0.
+ 89 kJ·mol-1 + 122 kJ·mol-1 + 418 kJ·mol-1 - 349 kJ·mol-1 + ΔHL -(-437 kJ·mol-1) = 0
Solve for -ΔHL = + 717 kJ·mol-1
(s)
81