Chem 14 : Physical Chemistry for Engineers 1 [Notes]
LESSON 1: PROPERTIES OF GASES
GAS


Fills up any container it occupies
Made up of molecules (or atoms) in continuous
random motion, with average speeds that increase as
the temperature is raised.
Exercise
Calculate the pressure (in pascals and atmospheres)
exerted by a mass of 1.0 kg pressing through the point of
a pink of area 1.0 x 10-2 mm2 at the surface of the Earth.
9.8 𝑚
)
𝑠2
2
1𝑚
1 × 10−2 𝑚𝑚2 (
)
1000 𝑚𝑚
= 9.8 × 108 𝑃𝑎
1 𝑎𝑡𝑚
9.8 × 108 𝑃𝑎 (
) = 9671 𝑎𝑡𝑚
101.325 × 103 𝑃𝑎
𝐹 𝑚𝑔
𝑝= =
=
𝐴
𝐴
STATE OF GASES


Physical state of a substance is defined by physical
properties.
Each substance is described by an equation of
state.
- 𝑝 = 𝑓(𝑇, 𝑉, 𝑛) – general form
- 𝑝𝑉 = 𝑛𝑅𝑇 – for ideal gas
1 𝑘𝑔 (
-
Mechanical Equilibrium
-
Measuring pressure
 Barometer – measures pressure exerted by
the atmosphere.
 Pressure gauges – measures pressure of a
gas in a container.
PHYSICAL STATES
Volume, V
- Space occupied by the gas
- SI unit: cubic meter (m 3)
Name
Cubic centimeter
Milliliter
Symbol
1 cm3
1 mL
Liter
1L
Cubic Feet
1 ft3
Value
1x10-6 m3
1x10-6 m3
1x10-3 m3
1000 mL
0.0283 m3
Amount of Substance (mass or moles), n (moles)
- Expressed in kilograms (kg) or number of moles
(mol).
- Number of moles, n
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒, 𝑔
𝑛=
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑔𝑎𝑠, 𝑔/𝑚𝑜𝑙
Pressure, p
- Force divided by the area to which force is
applied.
- Origin of force exerted by a gas is the incessant
battering of the molecules on the walls of the
container.
𝐹
𝑃=
𝐴
- SI unit: pascal (Pa); 1 Pa = 1 N/m 2
Name
Bar
Atmosphere
Torr
Milliliters of
mercury
Pound per square
inch
Symbol
1 bar
1 atm
1 Torr
Value
1x105 Pa
101.325 kPa
133.32 Pa
1 mmHg
133.32 Pa
1 psi
6.894757 Pa
Exercise
Derive an equation for the pressure at the base of a
column of liquid of mass density 𝜌 and height ℎ at the
surface of the Earth.
𝑃=
𝐹 𝑚𝑔 𝜌𝑉𝑔 𝜌𝐴ℎ𝑔
=
=
=
= 𝜌𝑔ℎ
𝐴
𝐴
𝐴
𝐴
Temperature, T
- Indicates the direction of flow of energy through a
thermally conducting, rigid
wall.
- If energy flows from A to B
when they are in contact, then
A has a higher temperature
than B.
- Types of boundaries:
 Diathermic
wall
–
thermally conducting
 Adiabatic wall – thermally insulating
- A property that indicates whether two objects would
be in “thermal equilibrium” if they were in contact
through a diathermic boundary.
1
TRANS:
-
-
Zeroth
Law
of
Thermodynamics
 If A is in thermal
equilibrium with B, and B
is in thermal equilibrium
with C, then C is also in thermal equilibrium
with A.
 It justifies the use of a thermometer: Celsius
Scale and Thermodynamic Temperature
Scale.
Temperature scales
 Celsius/Kelvin: K=°C+273.15
 Fahrenheit/Rankine: R=°F+460
 Celsius/Fahrenheit: °F=°C*(1.8)+32
 Kelvin/Rankine: K=R+460
*Note: Absolute temperature is temperature
expressed in Kelvin or Rankine
*A difference in temp of 1°C is equivalent to a
difference of 1K
LESSON 2: GAS LAWS
GAS LAWS

Boyle’s law
- 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑛𝑎𝑡 𝑛, 𝑇)
- Each hyperbola is an isotherm

Avogadro’s principle
- 𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑛 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑛𝑡𝑎𝑡 𝑝, 𝑇)
- 𝑉1 𝑛2 = 𝑉2 𝑛1
Note: gas laws are increasingly true as P approaches
zero, but reasonably reliable at normal pressures (P = 1
bar)
COMBINED GAS EQUATION
𝑃1 𝑉1 𝑃2 𝑉2
=
𝑇1
𝑇2


𝑃1 𝑉1 = 𝑃2 𝑉2
Charles’ law
- 𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑇 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛, 𝑝)
- 𝑝 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑇 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑛, 𝑉) *GL
- Lines are isobars and isochores
A real gas behaves more like a perfect as pressure is
lowered and as temperature is increased.
Exercise
In an industrial process, nitrogen is heated to 500 K in a
vessel of constant volume. If it enters the vessel at 100
atm and 300 K, what pressure would it exert at the
working temperature if it behaved as a perfect gas.
𝑃1 𝑃2
=
𝑇1 𝑇2
𝑇2
500 𝐾
(100 𝑎𝑡𝑚) = 167 𝑎𝑡𝑚
𝑃2 = 𝑃1 =
𝑇1
300 𝐾
PERFECT GAS EQUATION

-
𝑉1 𝑇2 = 𝑉2 𝑇1
𝑃1 𝑇2 = 𝑃2 𝑇1 *GL

The observations combine into a single expression
𝑝𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑛𝑇
𝑝𝑉 = 𝑛𝑅𝑇
A gas that obeys the above equation is called a
perfect gas (ideal gas).
R values
8.31447
8.20574 x 10-2
8.31447 x 10-2
1.98721
Unit
J/mol K
dm3atm/mol K
dm3bar/mol K
cal/mol K
2
TRANS:

MOLECULAR INTERPRETATION

Boyle’s law
- If a gas is compressed to half its volume, then
twice as many molecules strike the wall at a
given time.
- Average force exerted on the wall is doubled.
- Applicable to all gases at low pressure

Charles’ law
- Raising the temperature of a gas, increases the
average speed of its molecules.
- Molecules collide more frequently with the walls,
thus exert greater pressure on the walls.


KINETIC MODEL OF GASES


Assumptions
- The gas consists of molecules of mass m in
ceaseless random motion.
- The size of the molecules is negligible.
- The molecules interact only through brief,
infrequent, elastic collisions.
Pressure-volume relation
-
1
1
𝑝𝑉 = 𝑛𝑀𝑐 2 ; 𝑐 = (𝑣 2 )2
3
M = molar mass of molecules
c = root-mean square speed of the molecules

If c depends only on T
- 𝑝𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇) → 𝐵𝑜𝑦𝑙𝑒 ′ 𝑠 𝑙𝑎𝑤

For ideal gas, pV = nRT, thus
1
-
3𝑅𝑇 2
( )
𝑀
𝑐=
The higher the T, the higher the c.
At given T, heavy molecules travel more slowly.


Dalton’s law
- The
total
pressure,
P,
exerted by a
mixture
of
gases is the
sum of the
partial
pressure of the
gases (PA, PB,…)
𝑃 = 𝑃𝐴 + 𝑃𝐵 + ⋯
We can also express the partial pressure in terms of
the mole fraction, x, of the component, J, in the gas.
𝑃𝐽 = 𝑥𝐽 𝑃
Where
𝑛𝐽
𝑥𝐽 =
𝑛 = 𝑛𝐴 + 𝑛𝐵 + ⋯
𝑛
Since
𝑥𝐴 + 𝑥𝐵 + ⋯ = 1
𝑃𝐴 + 𝑃𝐵 + ⋯ = (𝑥𝐴 + 𝑥𝐵 + ⋯ )𝑃 = 𝑃
The sum of the partial pressure is equal to the total
pressure.
Exercise
The mass percentage composition of dry air at sea level
is approximately N2: 75.5, O2: 23.2, Ar: 1.3. What is the
partial pressure of each component when the total
pressure is 1.00 atm?
100 𝑔
𝑛(𝑁2 ) = 0.755 (
) = 2.69 𝑚𝑜𝑙
28.02 𝑔/𝑚𝑜𝑙
100 𝑔
𝑛(𝑂2 ) = 0.232 (
) = 0.725 𝑚𝑜𝑙
32.00 𝑔/𝑚𝑜𝑙
100 𝑔
𝑛(𝐴𝑟) = 0.013 (
) = 0.033 𝑚𝑜𝑙
39.95 𝑔/𝑚𝑜𝑙
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠: 3.45 𝑚𝑜𝑙
Mole fraction:
𝑁2 =
MIXTURE OF GASES

Partial pressure of a component gas J (PJ)
- Pressure that the gas would exert if it occupied
the container alone at the same temperature.
𝑛𝐴 𝑅𝑇
𝑛𝐵 𝑅𝑇
𝑃𝐴 =
𝑃𝐵 =
𝑉
𝑉
2.69
0.725
0.033
= 0.780 𝑂2 =
= 0.210 𝐴𝑟 =
= 0.0096
3.45
3.45
3.45
Partial pressure:
𝑁2 = 0.780(1 𝑎𝑡𝑚) = 0.780 𝑎𝑡𝑚
𝑂2 = 0.210(1 𝑎𝑡𝑚) = 0.210 𝑎𝑡𝑚
𝐴𝑟 = 0.0096(1 𝑎𝑡𝑚) = 0.0096 𝑎𝑡𝑚
*Note that we did not need to assume ideal gas
conditions. Definition of partial pressure applies to both
perfect and real gases.
LESSON 3: REAL GASES

REAL GAS

Real gases show deviation
from the perfect gas law
because molecules interact
with each other.
- Repulsive forces
 Short-range
interactions
-
Significant when molecules are almost in
contact (at high pressures)
Attractive forces
 Important when molecules are fairly close
together but not necessarily touching (at
moderate pressures)
3
TRANS:
REAL GAS BEHAVIOR

PRESSURE
Low
Gas
behaves
virtually perfectly
Moderate
- Average
separation is
few molecular
diameters.
- Attractive
forces
dominate
repulsive
forces.
- Gas is more
compressible
than ideal gas.
High
- Average
separation is
small
- Repulsive
forces
dominate
- Gas is less
compressible
than ideal gas
Ar
CO2
He
O2


Condensation
o A → B: gas pressure
rises according to
Boyle’s law.
o B → C: deviations from
Boyle’s law appear.
o C → D → E: volume
decreases without a
change in p; liquid
appears.
o At E, sample is entirely
liquid

Critical point of gas
o TC: critical temperature
o pC: critical pressure
o VC: critical molar volume
o At & above TC, a supercritical fluid (gas) exits
𝑉𝑚
𝑉𝑚0
Vm = V/n, molar volume of gas
Vmo = molar volume of ideal gas
𝑝𝑉𝑚 = 𝑅𝑇𝑍 𝑠𝑖𝑛𝑐𝑒 𝑉𝑚𝑜 = 𝑅𝑇/𝑝


Z = 1 for a perfect gas under all conditions
Deviation from Z is a measure of departure from
perfect behavior.
REAL GAS


At high pressures,
o Z > 1, Vm > Vmo
o Repulsive forces dominant
At moderate pressures,
o Z < 1, Vm < Vmo
o Attractive forces dominant
T B, K
411.5
714.8
22.64
405.9

COMPRESSIBILITY FACTOR
𝑍=
Boyle Temperature, TB
o Temperature at which
properties of a real gas
coincide with those of a
perfect gas as p
approaches o.
o At TB, B = o.
Ar
CO2
He
O2
Pc, atm
48.0
72.9
2.26
50.14
Vc, cm3/mol
75.3
94.0
57.8
78.0
Tc, K
150.7
304.2
5.2
154.8
LESSON 4: VAN DER WAALS EQUATION AND PRINCIPLE OF CORRESPONDING STATES
EQUATIONS OF STATE

Van der Waals Equation
𝑛𝑅𝑇
𝑛 2
𝑝=
−𝑎( )
𝑉 − 𝑛𝑏
𝑉
𝑅𝑇
𝑎
𝑝=
− 2
𝑉𝑚 − 𝑏 𝑉𝑚
a, b: Van der Waals’
coefficients
Exercise
Estimate the molar volume of Co2 at 500 K and 100 atm
by treating it as a Van der Waals gas. (For CO2, a = 3.5
dm6 atm mol-2 and b = 4.267 x 10-2 dm3 mol-1)
𝑅𝑇
𝑎
−
𝑉𝑚 − 𝑏 (𝑉𝑚 )2
2
(𝑉𝑚 − 𝑏)(𝑉𝑚 ) 𝑃 = 𝑅𝑇(𝑉𝑚 )2 − (𝑉𝑚 − 𝑏)𝑎
𝑅𝑇
𝑎
𝑎𝑏
(𝑉𝑚 )3 − (𝑏 + ) (𝑉𝑚 )2 + ( ) 𝑉𝑚 −
=0
𝑃
𝑃
𝑃
𝑃=
Substituting all values in the equation,
𝑉𝑚 = 0.366 𝑑𝑚3 /𝑚𝑜𝑙
4
TRANS:
VAN DER WAALS LOOPS

Berthelot
𝑝=

𝑅𝑇
𝑎
−
𝑉𝑚 − 𝑏 𝑇𝑉𝑚2
Dieterici
𝑝=
𝑅𝑇𝑒 −𝑎/𝑅𝑇𝑉𝑚
𝑉𝑚 − 𝑏
PRINCIPLE OF CORRESPONDING STATES


FEATURES OF VDW EQUATION



Perfect gas isotherms are obtained at high
temperatures and large molar volumes.
o At high temperatures, 𝑉𝑚 ≫ 𝑏, thus 𝑉𝑚 − 𝑏 ≈ 𝑉𝑚
𝑅𝑇
𝑎
𝑅𝑇
𝑃=
−
→𝑃=
2
𝑉𝑚 − 𝑏 (𝑉𝑚 )
𝑉𝑚
Liquids and gases coexist when cohesive and
dispersing effects are in balance.
The critical constants are related to the Van der
Waals Coefficients
𝑎
8𝑎
𝑉𝑥 = 3𝑏 𝑃𝑐 =
𝑇𝑐 =
2
27𝑏
27𝑅𝑏
o Critical compressibility factor
𝑃𝑐 𝑉𝑐 3
𝑍𝑐 =
=
𝑅𝑇𝑐 8

Critical constants are characteristic properties of
gases.
Reduced variables: divide actual variable by critical
constants.
𝑃
o Reduced pressure 𝑃𝑟 =
𝑃𝑐
𝑉𝑚
o
Reduced volume 𝑉𝑟 =
o
Reduced temperature 𝑇𝑟 =
𝑉𝑐
𝑇
𝑇𝑐
Real gases at the same reduced volume and reduced
temperature have the same reduced pressure.
o Works best for gases composed of spherical
molecules
LESSON 5: FIRST LAW OF THERMODYNAMICS
THE FIRST LAW: THE CONCEPTS


Universe
o System = part of the world we are interested in
o Surroundings = region outside the system
Types of system
o Open = system that allows the flow of matter and
energy from the system to the surroundings and
vice versa.
o Closed = system that only allows the flow of
energy from the system to the surroundings and
vice versa.
o Isolated = system that does not allow any flow of
matter or energy.
WORK AND ENERGY


Work – is done when an object is moved against an
opposing force.
Energy – capacity of a system to do work.


When work is done on the system, energy increases.
When the system does work, energy is reduced.

When the energy of a system changes as a result of
temperature difference, between a system and its
surroundings, energy has been transferred as heat.

Exothermic process – Process that releases energy
as heat. E.g., combustion reactions
Endothermic process – Process that absorbs energy
as heat. E.g., vaporization of water

5
TRANS:


INTERNAL ENERGY




𝑉𝑓
Heat energy transfer is random molecular motion.
Work energy transfer is uniform molecular motion.
→ Total kinetic and potential energy of the molecules
in the system.
→ A function of the properties that define the current
state of the system (i.e. state function)
→ Change in internal energy
∆𝑈 = 𝑈𝑓 − 𝑈𝑖
Internal energy, work and heat are all measured in
Joules.
o 1 J = 1 kg.m2.s-2
o 1 cal = 4.814 J (enough to raise the temp of 1g of water
𝑤 = − ∫ 𝑝𝑒𝑥 𝑑𝑉
𝑉𝑖


Free expansion
o Expansion
against
zero
opposing force (pex = 0)
o W=0
o e.g. system expands to a
vacuum
Expansion
against
constant
pressure
𝑉𝑓
𝑤 = −𝑝𝑒𝑥 ∫ 𝑑𝑉
𝑉𝑖
𝑤 = −𝑝𝑒𝑥 (𝑉𝑓 − 𝑉𝑖 )
𝑤 = −𝑝𝑒𝑥 ∆𝑉
by 1°C)
FIRST LAW OF THERMODYNAMICS


The internal energy of a system may be changed
either by doing work on the system or heating it.
First Law of Thermodynamics
o The internal energy of an isolated system is
constant.
o ∆𝑈 = 𝑞 + 𝑤
o q: work done on the system; w: energy
transferred as heat to the system
o
o


Reversible change: change that can be reversed by
an infinitesimal modification of a variable
A system is in equilibrium if an infinitesimal change in
the conditions in opposite directions results in
opposite changes in its state.
o Reversible expansion
𝑉𝑓
𝑤 = − ∫ 𝑑𝑉
o
w > 0, q > 0 energy is transferred to the system
as heat or work
w < 0, q < 0 energy is lost from system as heat
or work
Exercise
If an electric motor produced 15 kJ of energy each second
as mechanical work and lost 2 kJ of heat to the
surroundings, what is the change in the internal energy of
the motor each second?
o
𝑉𝑖
Reversible isothermal expansion of a perfect
gas
𝑉𝑓
𝑑𝑉
𝑤 = −𝑛𝑅𝑇 ∫
𝑉𝑖 𝑉
𝑉𝑓
𝑤 = −𝑛𝑅𝑇 ln
𝑉𝑖
Reversible isothermal expansion vs irreversible
expansion at constant pressure
∆𝑈 = 𝑞 + 𝑤
∆= −15 + (−2) = −17 𝑘𝐽
Suppose that when each spring is wound, 200 kJ of work
was done on it but 15 kJ escaped to the surroundings as
heat. What is the change in internal energy?
∆𝑈 = 𝑞 + 𝑤
∆= 200 + (−15) = 85 𝑘𝐽
EXPANSION WORK


For infinitesimal changes, 𝑑𝑈 = 𝑑𝑞 +
𝑑𝑤
Expansion work
o Work arising from a change in
volume (e.g., work done by a
gas as it expands and drives
back the atmosphere)
o General expression
𝑑𝑤 = −𝐹𝑑𝑧 = −𝑝𝑒𝑥 𝐴𝑑𝑧
𝑑𝑤 = −𝑝𝑒𝑥 𝑑𝑉
Exercise
Calculate work done when 50 g of iron reacts with
hydrochloric acid to produce hydrogen has in (a) closed
vessel of fixed volume, and (b) an open beaker at 25°C.
a. 𝑤 = ∫ 𝑃𝑒𝑥 𝑑𝑉 = 0
b. 𝑤 = −𝑃𝑒𝑥 (𝑉𝑓 − 𝑉𝑖 )
𝑛𝑅𝑇
= −𝑃𝑒𝑥 (
)
𝑃𝑒𝑥
= −𝑛𝑅𝑇
𝐽
𝑔 ) (8.314 𝑚𝑜𝑙 𝐾 ) (298.15 𝐾)
55.85
𝑚𝑜𝑙
= −2213.36 𝐽 or −2.2 𝑘𝐽
= −(
50 𝑔
6
TRANS:
HEAT TRANSACTIONS



In general
o 𝑑𝑈 = 𝑑𝑞 + 𝑑𝑤𝑒𝑥𝑝 + 𝑑𝑤𝑒
At constant volume, no additional work
o 𝑑𝑈 = 𝑑𝑞
o ∆𝑈 = 𝑞𝑣
Exercise
What is the change in molar enthalpy of nitrogen gas
when it is heated from 25°C to 100°C?
Calorimetry
o The study of heat transfer during a physical and
chemical process
o Uses a calorimeter
RELATION BETWEEN CP AND CV
VARIATION OF U WITH T


Internal energy of a substance increases as
temperature is increased
Heat capacity at constant volume
𝜕𝑈
o 𝐶𝑉 = ( )
𝜕𝑇 𝑉
𝐶𝑝 − 𝐶𝑉 = 𝑛𝑅
𝐶𝑝,𝑚 − 𝐶𝑉,𝑚 = 𝑅
ADIABATIC CHANGES


Change in U arises solely from
step 2
∆𝑈 = 𝐶𝑉 (𝑇𝑓 − 𝑇𝑖 ) = 𝐶𝑉 ∆𝑇
Since q = 0
𝑤𝑎𝑑 = 𝐶𝑉 ∆𝑇
1
𝑉𝑖 𝑐
𝑇𝑓 = 𝑇𝑖 ( )
𝑉𝑓

At constant volume:
o ∆= 𝑞𝑉 = 𝐶𝑉 ∆𝑇
*where 𝑐 = 𝐶𝑉,𝑚 /𝑅
ENTHALPY



𝐻 ≡ 𝑈 + 𝑝𝑉
→ state function
At constant pressure, no additional work
o 𝑑𝐻 = 𝑑𝑞
o ∆𝐻 = 𝑞𝑃
For an ideal gas,
o ∆𝐻 = ∆𝑈 + ∆𝑈 + ∆𝑛𝑔 𝑅𝑇
VARIATION OF H WITH T



Heat capacity
pressure
𝜕𝐻
o 𝐶𝑃 ≡ ( )
at
constant
Exercise
Argon undergoes an adiabatic reversible expansion,
initially at 25°C from 0.50 dm 3 to 1.00 dm3. The molar heat
capacity of argon at constant volume is 12.48 J/mol K.
Find the work done during the expansion.
𝑊𝑎𝑑 = ∆𝑈 = 𝐶𝑉 ∆𝑇
𝐽
(1 𝑚𝑜𝑙) (12.48
) (𝑇𝑓 − 25°𝐶)
𝑚𝑜𝑙 𝐾
1
𝑉𝑖 𝑐
𝑇𝑓 = 𝑇𝑖 ( )
𝑉𝑓
𝐽
𝐶𝑉,𝑚 12.48 𝑚𝑜𝑙 𝐾
𝑐=
=
= 1.5
𝐽
𝑅
8.314
𝑚𝑜𝑙
1
𝜕𝑇 𝑃
At constant pressure
o ∆𝐻 = 𝑞𝑃 = 𝐶𝑃 ∆𝑇
If CP is temp-dependent
𝑐
o 𝐶𝑝,𝑚 = 𝑎 + 𝑏𝑇 + 2
𝑇
0.5 𝑑𝑚3 1.5
𝑇𝑓 = 298.15 𝐾 (
) = 187.82 𝐾
1.0 𝑑𝑚3
𝑊𝑎𝑑 = ∆𝑉
𝐽
= (1 𝑚𝑜𝑙) (12.48
) (187.82 𝐾 − 298.15 𝐾)
𝑚𝑜𝑙 𝐾
= −1377 𝐽
7
TRANS:
LESSON 6: THERMOCHEMISTRY
THERMOCHEMISTRY





Is the study of the energy transferred as heat during the
course of chemical reactions.
It is a branch of thermodynamics because a reaction vessel
and its contest form a system, and chemical reactions result
in the exchange of energy between the system and the
surroundings.
Uses calorimetry to measure the energy supplied or
discarded as heat by a reaction.
o 𝑞𝑉 = ∆𝑈
@ constant volume
o 𝑞𝑃 = ∆𝐻
@ constant pressure
Endothermic reactions (∆𝐻 > 0)
o Absorbs energy by cooling the surroundings
Exothermic reactions (∆𝐻 < 0)
o Releases energy by heating the surroundings.
STANDARD ENTHALPY CHANGE, ∆𝑯°


The standard state of a substance at a specified
temperature is its pure form at 1 bar.
∆𝐻° for a reaction or a physical process is the difference
between the products in their standard states and the
reactants in their standard states, all at the same specified
temperature.
𝐻2 𝑂(𝑠) → 𝐻2 𝑂(𝑔)
∆𝐻°𝑠𝑢𝑏
𝐻2 𝑂(𝑠) → 𝐻2 𝑂(𝑙)
𝐻2 𝑂(𝑙) → 𝐻2 𝑂(𝑔)
∆𝐻°𝑓𝑢𝑠
∆𝐻°𝑣𝑎𝑝
Overall:
𝐻2 𝑂(𝑠) → 𝐻2 𝑂(𝑔)
∆𝐻°𝑠𝑢𝑏 = ∆𝐻°𝑓𝑢𝑠 + ∆𝐻°𝑣𝑎𝑝

Standard enthalpy changes of a
forward process and its reverse
differ in sign

∆𝐻°(𝐴 → 𝐵) = −∆𝐻°(𝐵 → 𝐴)


∆𝐻°𝑣𝑎𝑝 𝑓𝑜𝑟 𝐻2 𝑂 = +44 𝑘𝐽/𝑚𝑜𝑙
∆𝐻°𝑐𝑜𝑛𝑑 𝑓𝑜𝑟 𝐻2 𝑂 = −44 𝑘𝐽/𝑚𝑜𝑙
*Enthalpies of transition
ENTHALPIES OF CHEMICAL CHANGE


Standard enthalpy of vaporization, ∆𝑣𝑎𝑝 𝐻°
o Enthalpy change per mole when a pure liquid at 1 bar
vaporizes to gas at 1 bar
𝐻2 𝑂(𝑙) → 𝐻2 𝑂(𝑔)
o
∆𝑣𝑎𝑝 𝐻°(373𝐾) = +40.66
Conventional
temperature
for
thermodynamic data is 298.15 K (25°𝐶).


∆𝐻°𝑟 = {3𝐻°𝑚 (𝐶) + 𝐻°𝑚 (𝐷)} − {2𝐻°𝑚 (𝐴) + 𝐻°𝑚 (𝐵)}
𝐻°𝑚(𝐽) is the standard molar enthalpy of species J
𝑚𝑜𝑙
reporting
∆𝐻°𝑟 =
∑
𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
Standard enthalpy of transition, ∆𝑡𝑟𝑠 𝐻°
o Standard enthalpy change that accompanies a
change of physical state
Standard enthalpy of vaporization, ∆𝑣𝑎𝑝 𝐻°
o ∆𝐻 per mole when a pure liquid at 1 bar vaporizes to
a gas at 1 bar
𝑘𝐽
o 𝐻2 𝑂(𝑙) → 𝐻2 𝑂(𝑔)
∆𝑣𝑎𝑝 𝐻°(373𝐾) = +40.66
𝑣𝐻°𝑚 −
∑
𝑣𝐻°𝑚
𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
HESS’ LAW

The standard enthalpy of an overall reaction is the sum of
the standard enthalpies of the individual reactions into
which a reaction may be divided.
Standard enthalpy of fusion, ∆𝐻°𝑓𝑢𝑠
o ∆𝐻° accompanying the conversion of a solid to liquid
𝑘𝐽
o 𝐻2 𝑂(𝑠) → 𝐻2 𝑂(𝑙)
∆𝑓𝑢𝑠 𝐻°(273𝐾) = 6.01
Exercise
The standard reaction enthalpy for the hydrogenation of
propene CH2=CHCH3(g) + H2(g) → CH3CH2CH3(g) is -124 kJ/mol.
The reaction enthalpy for the combustion of propane
CH3CH2CH3(g) + 5O2(g) → 3CO2(g) + 4H2O(l) is -2220 kJ/mol.
Calculate the standard enthalpy of combustion of propene.
A change in enthalpy is independent of the path between
the two states.
9
𝐶3 𝐻6(𝑔) + 𝑂2(𝑔) → 3𝐶𝑂2(𝑔) + 3𝐻2 𝑂(𝑙)
2
𝑚𝑜𝑙
ENTHALPY AS A STATE FUNCTION

Chemical Equation + Standard Reaction Enthalpy
𝑘𝐽
𝐶𝐻4(𝑔) + 𝑂2(𝑔) → 𝐶𝑂2(𝑔) + 2𝐻2 𝑂(𝑙) ∆𝐻°𝑟 = −890
For reaction 2A + B → 3C + D
𝑘𝐽
𝑚𝑜𝑙

∆𝐻° = −890 𝑘𝐽
𝑚𝑜𝑙
ENTHALPIES OF PHYSICAL CHANGE

Thermochemical equation
𝐶𝐻4(𝑔) + 𝑂2(𝑔) → 𝐶𝑂2(𝑔) + 2𝐻2 𝑂(𝑙)
Pure, separate reactants in their standard states
→ pure, separate products in their standard states
EXAMPLE OF ∆𝑯°

∆𝐻°𝑓𝑢𝑠 + ∆𝐻°𝑣𝑎𝑝
The same value of ∆𝐻° will be obtained however the change
is brought about between the same initial and final states.
𝐶3 𝐻6 (𝑔) + 𝐻2 (𝑔) → 𝐶3 𝐻8 (𝑔)
𝐶3 𝐻8 (𝑔) + 5𝑂2 (𝑔) → 3𝐶𝑂2 (𝑔) + 4𝐻2 𝑂(𝑙)
𝐻2 𝑂(𝑔) → 𝐻2 (𝑔) + 1/2 𝑂2 (𝑔)
9
𝐶3 𝐻6 (𝑔) + 𝑂2 (𝑔) → 3𝐶𝑂2 (𝑔) + 3𝐻2 𝑂(𝑙)
2
∆𝑟 𝐻°/(𝑘𝐽/𝑚𝑜𝑙)
-124
-2220
+286
-2058
8
TRANS:
STANDARD ENTHALPIES OF FORMATION

∆𝐻°𝑓
The standard enthalpy of reaction for the formation of
a compound from its elements in their reference states.
Reference State: most stable at the specified
temperature and 1 bar. Example,
 Reference state of carbon at 298 K – graphite
 Reference state of mercury at 298 K – liquid Hg
 Reference state of nitrogen at 298 K – gas N2
∆𝐻°𝑓 of Benzene at 298 K
6𝐶(𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒) + 3𝐻2 (𝑔) → 𝐶6 𝐻6 (𝑙)

∆𝐻°𝑓 = 49.0 𝑘𝐽/𝑚𝑜𝑙
Reaction enthalpy in terms of enthalpies of formation
𝑇2
𝐻(𝑇2 ) = 𝐻(𝑇1 ) + ∫ 𝐶𝑝 𝑑𝑇
𝑇1
𝑇2
∆𝑟 𝐻°(𝑇2 ) = ∆𝑟 𝐻°(𝑇1 ) + ∫ ∆𝑟 𝐶°𝑃 𝑑𝑇
𝑇1
∆𝑟 𝐶°𝑃 =
∑
𝑣𝐶°𝑃,𝑚 −
𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
∑
𝑣𝐶°𝑃,𝑚
𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
Exercise
The standard enthalpy of formation of gaseous H2O at 298 K is
-241.82 kJ/mol. Estimate its value at 100°C given the following
values of the molar heat capacities at constant pressure:
H2O(g): 33.58 J/molK; H2(g): 28.84 J/molK; O2(g): 29.37 J/molK.
Assume that the heat capacities are independent of
temperature.
Main equation to use:
𝑇2
∆𝑟 𝐻°(𝑇2 ) = ∆𝑟 𝐻°(𝑇1 ) + ∫ ∆𝑟 𝐶°𝑃 𝑑𝑇
𝑇1
∆𝑟 𝐻°(𝑇2 ) = ∆𝑟 𝐻°(𝑇1 ) + ∆𝑟 𝐶°𝑃 (𝑇2 − 𝑇1 )
∆𝐻°𝑟 =
∑
𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
𝑣∆𝐻°𝑓 −
∑
𝑣∆𝐻°𝑓
𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
TEMPERATURE DEPENDENCE OF REACTION ENTHALPIES


Standard reaction enthalpies at different temperatures may
be calculated from heat capacities and the reaction
enthalpy at some other temperature
When a substance is heated from T1 to T2, its enthalpy
changes from H(T1) to
Writing the chemical equation to determine stoichiometric
coefficients:
1
𝐻2 (𝑔) + 𝑂2 (𝑔) → 𝐻2 𝑂(𝑔)
2
1
(𝐻
∆𝑟 𝐶°𝑃 = 𝐶°𝑃,𝑚 2 𝑂, 𝑔) − {𝐶°𝑃,𝑚 (𝐻2 , 𝑔) + 𝐶°𝑃,𝑚 (𝑂2 , 𝑔)}
2
= −9.94 𝐽/𝑚𝑜𝑙𝐾
𝑘𝐽
𝐽
∆𝑓 𝐻°(373𝐾) = −241.82
+ (75𝐾) (−9.94
)
𝑚𝑜𝑙
𝑚𝑜𝑙𝐾
𝑘𝐽
= −242.6
𝑚𝑜𝑙
THE FIRST LAW (CONTINUATION)


State Function
o Depend only in the current
state of the system
o Exact differential
o Example: Internal Energy
and Enthalpy

Path Function
o Process that describes the
preparation of the state
o Inexact differential
o Example: Work and Heat

For changes in both V and T
𝜕𝑈
𝜕𝑈
o 𝑑𝑈 = ( ) 𝑑𝑉 + ( ) 𝑑𝑇
o


Let U=U(V,T)
When V changes to V + dV at constant T
𝜕𝑈
o 𝑈 ′ = 𝑈 + ( ) 𝑑𝑉

When T changes to T + dT at constant V
𝜕𝑈
o 𝑈 ′ = 𝑈 + ( ) 𝑑𝑇
𝜕𝑉 𝑇
𝜕𝑇 𝑉
𝜕𝑇 𝑉
Internal Pressure and CV
𝜕𝑈
o 𝜋𝑇 = ( )
o
CHANGES IN INTERNAL ENERGY
𝜕𝑉 𝑇
In a closed system of constant
n, the infinitesimal change in U
is
proportional
to
the
infinitesimal changes in V and T
𝜕𝑉 𝑇
𝜕𝑈
𝐶𝑉 = ( )
𝜕𝑇 𝑉

Therefore,
o 𝑑𝑈 = 𝜋 𝑇 𝑑𝑉 + 𝐶𝑉 𝑑𝑇

Internal Pressure
o When
no
interactions
between molecules, U is
independent
of
their
separation,
hence
independent of the V of the
sample
o Therefore, for a perfect gas,

𝜋𝑇 = 0
9
TRANS:
JOULE EXPERIMENT



James Joule thought to measure
𝜋 𝑇 by observing the change in T
of a gas when it is allowed to
expand into a vacuum
Result
o He observed no change in
temperature
Implications
o w = 0 and q = 0
o So, ∆𝑈 = 0
o U does not change much
when gas expands isothermally, therefore 𝜋 𝑇 = 0.
→ experiment extracted an essential limiting property of a gas,
a property of a perfect gas
Derive the equation for the isothermal compressibility for an
ideal gas.
1 𝜕(𝑛𝑅𝑇)
1
1
1
1
𝜅𝑇 = − (
) = − (𝑛𝑅𝑇) (− 2 ) = −𝑝 (− 2 ) =
𝑉
𝜕𝑝
𝑉
𝑝
𝑝
𝑝
𝑇
Introducing 𝛼, ( ) = 𝛼𝜋 𝑇 𝑉 + 𝐶𝑉

For a perfect gas, 𝜋 𝑇 = 0
𝜕𝑈
( ) = 𝐶𝑉
𝜕𝑇 𝑃
Relationship between Cp and Cv
𝜕𝐻
𝜕𝑈
𝐶𝑃 − 𝐶𝑉 = ( ) − ( )
𝜕𝑇 𝑃
𝜕𝑇 𝑃
Since H = U + pV = U + nRT
𝜕𝐻
𝜕𝑈
𝐶𝑃 − 𝐶𝑉 = ( ) + 𝑛𝑅 − ( ) = 𝑛𝑅
𝜕𝑇 𝑃
𝜕𝑇 𝑃
In general,
𝛼 2 𝑇𝑉
𝐶𝑃 − 𝐶𝑉 =
𝐾𝑇


CHANGES IN U AT CONSTANT PRESSURE


𝜕𝑈
𝜕𝑉
( ) = 𝜋 𝑇 ( ) + 𝐶𝑉
𝜕𝑇 𝑃
𝜕𝑇 𝑃
Expansion Coefficient
o Fractional change in volume that accompanies a rise
in temperature
1 𝜕𝑉
o 𝛼= ( )
𝑉 𝜕𝑇 𝑃
Isothermal Compressibility
o Fractional change in volume when pressure increases
1 𝜕𝑉
o 𝜅𝑇 = − ( )
𝑉 𝜕𝑝 𝑇
Exercise
Derive the expression for the expansion coefficient of a perfect
gas.
𝛼=
1 𝜕(𝑛𝑅𝑇/𝑝)
1 𝑛𝑅 𝑑𝑇 𝑛𝑅 1
(
) = ×
=
=
𝑉
𝜕𝑇
𝑉
𝑝 𝑑𝑇 𝑝𝑉 𝑇
𝑝
𝜕𝑈


𝜕𝑇 𝑃
CHANGES IN ENTHALPY




Enthalpy, H = U + pV
At constant pressure, ∆𝐻 = 𝑞𝑃
Let 𝐻 = 𝐻(𝑝, 𝑇)
For a closed system of constant n
𝜕𝐻
𝜕𝐻
𝑑𝐻 = ( ) 𝑑𝑃 + ( ) 𝑑𝑇
𝜕𝑝 𝑇
𝜕𝑇 𝑝
𝑑𝐻 = −𝜇𝐶𝑃 𝑑𝑝 + 𝐶𝑃 𝑑𝑇
Joule-Thomson coefficient
𝜕𝑇
𝜇=( )
𝜕𝑝 𝐻
10
TRANS:
THE SECOND LAW: ITS CONCEPTS
TYPES OF PROCESSES THAT OCCUR IN NATURE


Calculate the entropy change of a sample of a perfect gas when
it expands isothermally from a volume Vi to a volume Vf.
Spontaneous
o A gas expands to fill the available volume
o Hot body cools to the temperature of surroundings
o Chemical reaction runs in one direction
o Spontaneous direction of change – direction of
change that does not require to be done to bring the
change about
Non-spontaneous
o Work is needed for change to occur
o Examples:

Confining gas to a smaller volume

Cooling an object using a refrigerator

Reactions occurring in reverse (e.g. electrolysis)
𝑓
𝑞𝑟𝑒𝑣
ENTROPY OF THE SURROUNDINGS

SECOND LAW OF THERMODYNAMICS


No process is possible in which the sole
result is the absorption of heat from a
reservoir and its complete conversion
into work.
All real engines have both a hot sink
and cold sink.
WHAT DETERMINES THE DIRECTION OF
SPONTANEOUS CHANGE?




The direction of spontaneous change for a bouncing ball on
a floor. On each bounce, some of its energy is degraded
into thermal motion of the atoms on the floor, and that
energy disperses.
Spontaneous change: a direction of change that leads to
dispersal of the total energy of the isolated system.
𝑑𝑞𝑟𝑒𝑣 1 𝑓
= ∫ 𝑑𝑞𝑟𝑒𝑣
𝑇
𝑇 𝑖
𝑖
𝑉𝑓
= −𝑤𝑟𝑒𝑣 = 𝑛𝑅𝑇 ln
𝑉𝑖
𝑉𝑓
∆𝑆 = 𝑛𝑅 ln
𝑉𝑖
∆𝑆 = ∫

For the surroundings
𝑑𝑞𝑠𝑢𝑟,𝑟𝑒𝑣 𝑑𝑞𝑠𝑢𝑟
𝑑𝑆𝑠𝑢𝑟 =
=
𝑇𝑠𝑢𝑟
𝑇𝑠𝑢𝑟
𝑞𝑠𝑢𝑟
∆𝑆𝑠𝑢𝑟 =
𝑇𝑠𝑢𝑟
True
whether
change
is
reversible
or
irreversible;
surroundings
in
internal
equilibrium
For any adiabatic change, ∆𝑆𝑠𝑢𝑟 = 0
Exercise
Calculate the entropy change in the surroundings when 1.00 mol
H2O(l) is formed from its elements under standard conditions at
298 K. ∆𝐻°(𝐻2 𝑂) = −286 𝑘𝐽. The heat released is supplied to
the surroundings.
∆𝑆𝑠𝑢𝑟 =
𝑞𝑠𝑢𝑟 2.86 × 105 𝐽
𝐽
=
= 960
𝑇𝑠𝑢𝑟
298 𝐾
𝐾
ENTROPY AS STATE FUNCTION
Entropy
o Measure of the energy dispersed in a process.
o A state function.
The 2nd law in terms of entropy
o The entropy of an isolated system increases in the
course of a spontaneous change.
∆𝑆𝑡𝑜𝑡 > 0

Stot: total entropy of the system and its surroundings
1ST LAW AND 2ND LAW


1st Law of Thermodynamics: uses internal energy to identify
permissible changes
2nd Law of Thermodynamics: uses entropy to identify the
spontaneous changes among those permissible changes
ENTROPY: DEFINITION

A change in the extent to which energy is dispersed
depends on how much energy is transferred as heat
o Heat stimulates disorderly motion

Thermodynamic definition of entropy
𝑑𝑞𝑟𝑒𝑣
𝑑𝑆 =
𝑇
𝑓
𝑑𝑞𝑟𝑒𝑣
∆𝑆 = ∫
𝑇
𝑖
Exercise
11
TRANS:
XXX
TOPIC
TOPIC
SUBTOPIC
SUBSUBSUB
●
List 1
o List 2
▪ List 3
VIDEO
SUBTOPIC2
SUBSUB2
PARAMETER
S
DESCRIPTION
12