→ 1 Law of Thermodynamics
→ Enthalpies of Reaction - Hess' Law, etc.
→ Bond Energies
We've discussed some physical processes - eg. dissolution of
ionic solids (salts) in water - & some chemical reactions -
precipitation, acid-base & redox ...
in terms of changes of state and changes in molecular formula
- making/breaking of bonds, transfer of electrons, etc.
Energy flow also accompanies these processes.
THERMODYNAMICS provides the framework to
understand energy flow & its role in determining whether a
process actually occurs or not.
eg. 2 Na(s) + 2 H O(l) → 2 NaOH(aq) + H (g)
occurs spontaneously, liberating a good deal of heat energy
However, the reverse reaction never happens.
25 More egs.
MgSO (s4 & NH NO (s)4 3
dissolve spontaneously in water.
When MgSO (s) d4ssolves in water, heat is liberated.
→ used in hot packs
When NH NO (4) d3ssolves in water, heat is consumed.
→ used in cold packs
Thermochemistry deals with the accounting of heat transfer.
We defer the question of what makes a process spontaneous till
later when we take up entropy & Gibbs free energy.
Thermodynamics hinges on being very precise in our
description of a "system" and its processes
eg. Types of Systems:
isolated closed open
system system system
system is sealed off system has system has permeable,
outside world impermeable walls, but non-thermally-
- no heat or matter is not thermally insulating walls
flow across boundary insulated - heat and matter flow
- rigid walls ⇒ - heat flow, but - walls can be rigid or
no mechanical work - no matter flow non-rigid
done on or by system - walls can be rigid or
26 Properties of Systems
1. Discrete labels:
- chemical species identity
- phase identity
eg. CH (4) Zn(s) Br 2l)
2. Continuous variables:
p, T, n, V
pressure, temperature, # of moles, volume
U, H, S, G
energy, enthalpy, entropy, Gibbs free energy
State Functions - they depend on the state of the system
pi= 1 atm p f 3 atm
T i= 300 K T f= 450 K
ni = 1 mole some process n f= 1 mole
V i= 10 m V f = 5 m
U i, Hi Si G i U f, Hf, Sf, Gf
initial system state final state
Processes cause changes in system state & corresponding
changes in system state functions.
Change in a state function depends only on the initial &
final state - not on how the change was achieved ⇒
∆p = p −fp = 2iatm ∆T = T − f = 15i K
∆n = n −fn = 0imol ∆V = V − f = −5im
- mechanical or electrical
We will focus on mechanical work
distance through which
spring is compressed
Apply a force to a spring to compress it
→ work is done on the spring
work = force × distance
The spring stores this energy.
Work is done on the system, which increases the energy of
Compressing a gas in a cylinder is similar:
gas has higher
V apply force on
i gas in cylinder
via piston V f
28 Temperature & Heat
Hot vs. Cold
Temperature measures the average kinetic energy per atom
T1 T2 T1< T 2
Heat flows from system 2 (on the right) to system 1 (on the
⇒ T 1ncreases & T d2creases until the two systems achieve
the same temperature – thermal equilibrium
q Heat flowing to system
T increases the temperature
q > 0
q Heat flowing from system
decreases the temperature
q < 0
Note – other things (i.e. other than heat flow) can change the
⇒ the sign of q does not always correspond to the sign of ∆T
- measuring heat transfer
29 Consider transferring 100 J to …
a) 18 g of H O(l) ⇒ ∆T = 1.33 K
b) 18 g of Cu(s) ⇒ ∆T = 14.5 K
The resulting temperature change depends on the type of
material – in addition to the amount of material.
In general, Heat Capacity
q = C ∆T
= m s ∆T
mass Specific Heat
Units of s : J g K
s = Amount of heat required to raise the temperature of 1 g
of the substance by 1 K
Note, s is always > 0
s[H2O(l)] = q = 100 J = 4.18 J g K1 −1
m ∆T (18 g) × (1.33 K)
s[Cu(s)] = q = 100 J = 0.378 J g K −1
m ∆T (18 g) × (14.5 K)
30 Calorimetry utilizes a bulk quantity of a substance with a
well known specific heat in thermal equilibrium with a
system of interest.
A physical or chemical process occurs in the system. Any
heat produced or consumed enters or is drawn from the
The heat produced or consumed can be deduced from the
change in temperature – there is no heat flow to or from entire
calorimeter, since it is thermally isolated.
However, there must be no physical or chemical processes in
the bulk, as the resulting temperature change would not
correspond to the heat.
1 mol solid ice at 1 mol liquid water
0° C & at 0° C &
1 atm pressure 1 atm pressure
q = 6008 J
Here, there is a substantial heat transfer resulting in
no temperature change
→ Heat of fusion
31 1 mol liquid water 1 1 mol liquid water
at 25.00° C & at 26.33° C &
1 atm pressure 2 1 atm pressure
Two different processes can produce the same final state
from a given initial state.
1. q1= 100 J & w1= 0 J
2. q2= 0 J & w2= 100 J
1. a flame transfers 100 J of heat to system
2. a submerged propeller delivers 100 J of work to system
q ≠ q & w ≠ w
1 2 1 2
q 1 w 1 = q 2 w 2
It turns out this is true for all processes connecting
a particular initial state to a particular final state.
⇒ q + w is the change in a state function …
∆U = q + w
The first law of
32 How does one measure ∆U ?
∆U = q + w
Arrange for either q or w to be zero & measure the other.
We’ve seen how to measure q.
- just be sure that w = 0 & you will measure ∆U
- this is accomplished by enclosing the system in rigid
walls, with no internal moving parts
We can also measure w for a process for which the system is
- we assume a constant pressure
distance surface area = A
= d thermal isolulation
q = 0
Force = p extA
Mechanical work done on system
= Force ×distance Pressure is
= (p ext A) d force per unit
= p ext (A d)
work done on system = − p ext∆V reversible
= − p ∆V process - i.e. do
33 Knowing the work done on the system & ensuring that q = 0,
∆U = w = − p ext∆V
if pextis constant
In the case of w = 0,
∆U = q
- the subscript V denotes a constant volume process
- this is required to make w = 0
A heat measurement at constant volume determines ∆U
Let’s go back to the constant pressure process - more easily
managed – eg. just vent the system to the atmosphere
(mechanical equilibrium is needed)
- a constant volume requires a bomb calorimeter which
must be very strong to withstand high internal pressures
which are possible – depending on the process being
34 For a constant pressure process, we still have
∆U = q + w
- note that we are not now considering the case of q = 0
If we measure q, we measure at constant pressure
qp = ∆U − w
w = − p ∆V
if p = p ext is constant.
q p = ∆U + p ∆V
= ∆( U + p V) p is constant
Because constant pressure processes are so important, the
H = U + p V
is given a name, enthalpy, and used extensively.
Ufinal U initial
Vf> V i
⇒ ∆V > 0
⇒ w = −p∆V < 0
i.e. work is done by the system
also, q = 0
ensuring that q = 0 So, ∆U = q + w < 0
2 HCl(aq) + Zn(s) → H (g) +
36 Exothermic Endothermic
q < 0 q > 0
constant V : constant V :
∆U = q V < 0 ∆U = q V > 0
constant p : constant p :
∆H = q < p ∆H = q > 0p
For a chemical re