Chem 1310:
Introduction to physical chemistry
Part 1: Thermodynamics 1
Peter H.M. Budzelaar
About Energy
What happens in chemistry and physics is determined by
energy.
Things tend to move to a state of "lowest energy", but there
might be "barriers" in the way of certain changes.
Thermodynamics
"where a system
would like to be"
Kinetics
vs
"how easy it is for a
system to get there"
To work with these ideas, you need to understand what
"energy" is.
Energy, work
Energy is defined in MSJ as "capacity to do work".
Work is done "when a force acts through a distance":
w = F  Dx
The definition of energy is a bit vague, descriptive,
non-quantitative.
More practical: sum of known forms of energy:
potential: associated with forces on an object
(water in a hydro dam lake; gravity)
kinetic: macroscopic: mechanical energy (car in motion)
microscopic: thermal energy (hot water)
Quantifying energy
Energy is interesting because it is conserved
(if defined correctly!)
SI unit: Joule, J ( = kg m2 s-2)
1 kJ = 1,000 J; 1 MJ = 1,000,000 J
Older unit: calorie, cal
original definition: energy required to heat 1 cm3 liquid
water from 14.5 to 15.5 °C.
now defined as 4.184 J. Also: kcal, etc
In nutrition: 1 Cal = 1 kcal = 1,000 cal
energy that would be freed by metabolizing
(burning) the food
Quantifying energy
Use your UNITS !!!!!
Power
Concept: power = energy per time unit
1 Watt = 1W = 1 J s-1 ( = 1 kg m2 s-3 )
Light bulbs: refers to power consumption, not to
actual light energy produced
(most electrical energy becomes heat)
Capacity of an energy plant is expressed in MW
e.g. Wuskwatim ca 200 MW, Conawapa ca 1200 MW
Production limited by amount of water coming in
at a given height difference
Kinetic energy
Ekin = ½ m v2
m in kg, v in m s-1  E in kg m2 s-2 = J
A car going 3 times as fast has 9 times as much
potential to cause damage...
If we have a large collection of molecules, they all have
different energies and move in random directions, so we
don't notice any macroscopic kinetic energy.
Instead, we feel pressure (molecules hitting the wall) and
possibly heat (transfer of kinetic energy to/from
surroundings)
Potential energy
Various forms, depending on types of forces
involved.
Most common ones:
m1 m2
or on Earth : Vgrav  m g h
gravitation: Vgrav  r
12
electrostatic:
q1 q2
Velst 
r12
(g = 9.80665 m s-2)
Conservation of energy
(first law of thermodynamics)
Kinetic energy can "disappear":
shoot a bullet upwards: it will eventually slow down due
to gravitational pull of earth
Potential energy can "disappear":
water flows from a higher to a lower level
If properly defined, "total energy" is constant:
Etot = Epotential + Ekinetic is conserved
bullet: kinetic energy converted in potential energy
water: potential energy converted into heat
Conversion between kinetic
and potential energy
How far up can a good tennis player shoot a tennis
ball?
A tennis ball weighs approx 58 g = 0.058 kg
Initial speed (good player!) 190 km/h = 52.8 m s-1
Kinetic energy = 0.5* 0.058*(52.8)2 J = 80.8 J
Ignore air resistance !!!
Everything converted to potential energy:
m g h = 0.058*9.80665*h = 80.8 J  h = 142 m
Conservation of energy
Most processes involve "doing work", transferring or
converting energy.
If we know that total energy is conserved, you can
say something about how much work can be done,
or how much of one kind of energy can be
transferred into how much of another type.
Heat
Temperature is a measure of the average kinetic
energy of the components (molecules) of a system.
Heat (thermal energy) flows from hot to cold:
kinetic energy is transferred between molecules of
hot and cold regions until both have the same
temperature.
System and surroundings
The "system" is the part of the universe you are
looking at in an experiment.
The "surroundings" is the rest of the universe.
Energy is conserved for "system+surroundings", not
for the "system" separately.
System and surroundings
Heat a beaker of water in a microwave:
Electrical energy gets converted into microwave energy
(electromagnetic radiation)
Microwave energy flows into the system (beaker)
Gets absorbed there, i.e. converted into heat
The water cools: heat gets transferred from system to
surroundings, partly through contact, partly through
infrared radiation
Conservation of energy
DEsystem = qsystem + wsystem + ...
Energy used for heating the system.
If system cools, qsystem < 0.
(radiation energy,
potential energy, ...)
Work done on the system.
If work done by system, wsystem < 0.
To put heating on a scale with other forms of energy, you
need to know how much energy it costs to heat something.
Unfortunately, this is not the same for every compound.
Quantifying heat (1)
Heat capacity: the amount of energy needed to warm
up a system by 1 degree.
Remember as: the capacity of a system to absorb heat.
Measure? Relative to another, known system.
Units: J K-1 = kg m2 s-2 K-1
Does it matter whether we use K-1 or °C-1 ?
Quantifying heat (2)
Specific heat (capacity): the amount of energy
needed to warm up 1 g of a compound
by 1 degree.
Units: J K-1 g-1 = 1000 m2 s-2 K-1
Property of a compound under given conditions
(temperature, pressure, state)
Tabulated (MSJ p 225). Use this to calculate actual
heating.
Calculating heat transfer (1)
What is more likely to burn your finger:
• 1 g of water of 50 °C
• 2 g of aluminum of 60 °C
Depends on several factors: rate of heat transfer to
finger, rate of heat dissipation in body, etc.
Assuming we transfer all heat to 1 g of "finger"
quickly. Then what causes most heating will be
most likely to burn you.
Calculating heat transfer (2)
Start with:
• 1 g of finger (mostly water) at 37 °C, specific heat 4.184,
will be heated
• 1 g of water of 50 °C , specific heat 4.184, will cool
Final states "meet in the middle".
Final temperature: 0.5*(37+50) = 43.5 °C
Calculating heat transfer (3)
Start with:
• 1 g of finger (mostly water) at 37 °C, specific heat 4.184,
will be heated
• 2 g of aluminum of 60 °C , specific heat 0.902, will cool
Assume Al cools by x degrees, so final temperature
of finger and Al is (60-x) °C.
Transfer from Al = 2*0.902*x = 1.804*x J
Transfer to finger = 1*4.184*(60-x-37) = 4.184*(23-x) J
Must be equal; solve: x = 16.1, finger temperature 43.9 °C.
About water
Water has a very high heat capacity,
especially as a liquid.
 good way to "store" heat or cold
 moderates climate
 is very important in determining weather
(warm water is an important factor in creating hurricanes)
Molar heat capacity
Some properties are best expressed "per mole"
instead of "per gram". Then we call them "molar".
Problem: not everything consists of well-defined,
identical molecules.
H2O: OK
NaCl: probably OK
Diamond: maybe
Wood: ????????
Energy changes
due to internal changes
Phase change / state change
• Phase changes (solid/liquid, liquid/gas) normally happen at
a constant temperature.
• Adding heat causes more solid to melt, etc, but does not
change the temperature until all solid has melted (etc).
• Quantity of energy needed for melting:
heat of fusion
(J/g, at a given temperature).
• Also: heat of evaporation/sublimation.
Fig. 6-11, p.229
Calculating heat transfer (4)
We put 20 g of ice of 0 °C in a beaker containing 180
g of water of 20 °C, mix, and wait for heat
exchange within the beaker. What will be the final
temperature? Will there be any ice left?
The heat of fusion of ice is 333 J/g at 0 °C (MSJ p
228). Melting the ice absorbs 333*20 J = 6.66 kJ.
Cooling the remaining 180 g of water by 1 °C costs
180*4.184 J = 0.753 kJ, so the water gets cooled
by 6.66/0.753 °C = 8.8 °C to 11.2 °C....
Calculating heat transfer (5)
Now we are left with 20 g water at 0 °C and 180 g of
water of 11.2 °C, both with the same heat capacity,
so the final temperature will be a weighted
average:
(20*0+180*11.2)/200 = 10.1 °C.
In other words, the melting gives us about 10 ° of
cooling, the further cooling effect of the added
water only about 1 more degree. Ice is an effective
coolant!
Why does melting/boiling cost energy?
• Many interactions present in the well-ordered
solid state are diminished on melting.
• This is an abrupt transition because intermediate
situations would be worse: they don't have the
favourable well-ordered arrangement of the solid,
nor the freedom of movement of the liquid.
• Melting keeps a big part of the short-range
interactions present in the solid. On evaporation,
these are completely lost. Evaporation costs a lot
of energy!
Enthalpy:
the relevant energy at constant pressure
For a constant-volume process, it makes sense to talk
about DE of a system.
We can measure heat being transferred in to / out of
a system: this says something about its internal
energy (change).
Nothing moves, no work is done, so
DE = q+w = q = qV
subscript V indicates
heat transfer
at constant volume
Enthalpy:
the relevant energy at constant pressure
In the real world we often work at constant pressure
(1 bar). Now, it makes more sense to talk about heat
transfer at constant pressure, where work is being
done (expansion/compression) to keep the pressure
constant.
Heat 10 g of water from 0°C to 100°C (to measure its heat
capacity, an easy measurement): it will expand a bit as
well (say, 10%) against the surrounding atmosphere of 1
bar. To keep it at constant volume would require a very
high pressure, and this is an impractical measurement.
Enthalpy:
the relevant energy at constant pressure
At constant pressure, it is convenient to work with
DH = q P
heat transfer for processes
at constant pressure,
H is called enthalpy.
no other work being done
Processes that absorb heat at constant pressure (like
boiling): endothermic, DH < 0.
Processes that produce heat at constant pressure (like
condensation): exothermic, DH > 0.
Indicating conditions
Enthalpy change for a given process:
DH
Enthalpy change at "standard pressure" (1 bar):
DH °
Enthalpy change at 1 bar, 0°C:
DH 273°
Thermodynamic calculations
• Usually "per mole", e.g. per mole of reactant.
• Changes correspond to a particular equation, e.g.:
S(s)  ½ S2 (g)
If we would know that S(s) actually consisted of S8
molecules, we could also write
S8 (s)  4 S2 (g)
DH is 8 times as much for this. Compare to a
market: if you don't get a discount, the price per
dozen of apples is 12 times the price per apple.
Thermodynamic calculations
Remember:
At exams, you will be asked about a specific
equation; your answer should fit that equation.
Intermezzo: balancing equations
You should know how to do this!
But let's go over it again:
□Fe + □O2  □FeO
□Fe + □O2  □Fe2O3
□Fe + □O2  □Fe3O4
□C3H8 + □O2  □CO2 + □H2O
Now try for yourself, at home:
□MnO + □Mn2O7  □MnO2
Thermodynamic calculations
How much heat does burning of 1 mole of hydrogen
produce? Does it matter whether we get liquid or
gaseous water?
DH = qP
H2 (g) + ½O2 (g)  H2O(l) DH = -285.8 kJ/mol (MSJ p 237)
H2 (g) + ½O2 (g)  H2O(g) DH = -241.8 kJ/mol (MSJ p 236)
subtract:
H2O(l)  H2O(g)
DH = +44.0 kJ/mol
Chemical change:
more energy involved than in physical change
Thermodynamic calculations
Try this at home:
How much natural gas (in L) do you need to burn to
heat all the air in your house from 19°C to 20°C?
Assume:
• natural gas is pure propane (C3H8).
• the house is 150 m2 with an average height of 2.5 m
(or assume more accurate values from a house you
know).
• the heat capacity of air is 1.006 J g-1 K-1.
• air is 80% N2, 20% O2.
Interpreting energy changes:
chemistry + physics
Bond energies tend to be similar for similar
compounds.
The C-H bond has a similar strength in CH4 and in
C12H22O11 (sugar).
We can estimate the energy released in a chemical
reaction by subtracting bond energies of products
and reactants (see MSJ p352).
When tabulated DHf values are available (MSJ p250,
App J), they are more accurate and are preferred!
Interpreting energy changes:
chemistry + physics
Estimate the energy change for the reaction
of H2 with I2 to give HI.
2 H  H2
DHest = -436 kJ/mol
2 I  I2
DHest = -151 kJ/mol
H + I  HI
DHest = -299 kJ/mol
H2 + I2  2 H + 2 I  2 HI
DHest = -11 kJ/mol
+436
+151
- 2*299
Hess's law
We can add or subtract reactions, and then do the
same with the associated enthalpies etc.
Idea behind this: molecules have no memory. It
doesn't matter for the strength of a C-H bond of CH4
whether it has been produced from Ca2C + H2O or
from heating carbon in a stream of hydrogen. All
molecules of CH4 are identical. This makes it
possible to talk about the molar properties of a
compound without worrying about which mole it is,
who has made it or where.
Heats of formation
Because of Hess's law, we need only tabulate heats
of formation for all compounds to allow prediction
of enthalpy changes of arbitrary reactions.
DHf °: energy change in forming 1 mole of the
compound (at 1 bar) from its elements in their
standard states.
Heats of formation of elements in their standard
state are zero by definition.
Heats of formation (2)
For any reaction, we can imagine taking apart the
reactants to their elements, then converting these
back to the products.
Total enthalpy change:
S DHf°(products) - S DHf°(reactants)
Looks trivial, but beware:
•Choose the correct standard states!
•Use the right coefficients for every component!
Using heats of formation
Calculate the energy change for the reaction
of H2 with I2 to give HI.
H2 + I2  2 HI
Calculation:
DH = 2*DHf °(HI)-DHf °(H2)-DHf °(I2)
= 2*26.5-0-0 = 53 kJ/mol
Large difference with the bond enthalpy estimate!
Expect an error of 10-20 kJ/mol per bond enthalpy.
Reaction enthalpies
and standard states
Do you need to know what the elements are in their standard
states?
Not unless you are explicitly using non-standard states!
H2S + 2 O2  H2SO4
DH = DHf °(H2SO4)-DHf °(H2S)-2*DHf °(O2)
S(s) + O2  SO2
DH = DHf °(SO2)-DHf °(S(s))-DHf °(O2)
You probably happen to know that S(s) is S8, but that is
irrelevant. However, you will need DHf °(S(l)) for
S(l) + O2  SO2
Fuel value
Fuel can be burned to produce heat and/or do work.
Important criteria:
• Fuel value: energy released burning 1 gram of fuel
• Energy density: energy released burning a fixed
volume (1L)
Fossil fuels and fatty acids have a high fuel value;
carbohydrates less so (they are already partly oxidized).
Gases have a much lower energy density than liquids and
solids even for the same fuel value; also, lower MW 
lower energy density (this is not true for solids/liquids!).
Fuel value (2)
Estimate the fuel value of ammonia (NH3) using
bond strengths.
□NH3 + □O2  □N2 + □H2O
balance:
NH3 + ¾ O2  ½ N2 + 1½ H2O
lose: 3*N-H + ¾*O=O = 3*391+0.75*498 = 1546.5 kJ/mol
gain: ½*NN + 3*O-H = 0.5*946+3*467 = 1874.0 kJ/mol
estimate: DH  -327.5 kJ/mol, -19.3 kJ/g
Ammonia should burn well! (NN is strong, O=O is weaker)
Fuel value (3)
Calculate the fuel value of ammonia (NH3)
using heats of formation.
NH3(g) + ¾ O2  ½ N2 + 1½ H2O(l)
Enthalpy of reaction (Hess's Law):
0.5*DHf °(N2)+1.5*DHf °(H2O)-DHf °(NH3)-0.75*DHf °(O2)
= 0.5*0+1.5*-285.8+46.1-0.75*0 = -474.8 kJ/mol
Much more negative than estimate based on bond
enthalpies! (bond enthalpies are rough estimates only)
Contains corrections for states of reactants and
products.
Measuring heats of reaction
(calorimetry)
• Put reactants in separate compartments of
insulated container
• Remove separation
• Start reaction (if not spontaneous)
• Measure change in temperature
• Use available heat capacity data etc to correct
back to "standard state"
(water as a reaction product or heat absorber
is very convenient)
Fig. 6-16, p.243
Calorimetry (1)
We burn 1 g of methane in a calorimeter containing an excess
of oxygen and 200 mL of water at 25°C. What will be the
final temperature of the water (ignoring heating of the
calorimeter itself and other things)?
CH4 + 2 O2  CO2 + 2 H2O(l)
DH = DHf °(CO2)+2*DHf °(H2O)-DHf °(CH4)-2*DHf °(O2)
= -393.5-2*285.8+74.8-2*0 = -890.3 kJ/mol
Calorimetry (2)
We reacted 1 g = 0.0625 mol; the reaction produces 0.125
mol (2.25 g) H2O, so we end up with 202.25 g H2O which
will be heated by 0.0625*890.3 kJ = 55.6 kJ.
Heating: 55,600/(202.25*4.184) = 65.8°C.
Final temperature: 90.8°C
In chemical vs physical energy,
chemical energy clearly wins out!
Calorimetry (3)
Exercise 6.10: burning 1 g of Fritos chips in a
calorimeter (known heat capacity 877 J/°C)
containing 832 g water and excess oxygen, we
find a temperature change of 4.79°C. What is the
fuel value of the chips?
Total heat capacity of calorimeter+water:
877+832*4.184 = 4358 J/°C.
Total heat produced per gram (= fuel value):
4358*4.79 = 20.9 kJ/g = 5.0 kcal/g = 5.0 Cal/g
We are ignoring water formed by the reaction!