Energy and Driving Forces:
The Laws of Thermodynamics
Spontaneous Reaction:
• given the required Ea, activation energy,
the reaction continues to proceed by
itself
• may be slow or fast
Thermodynamics:
• Study of energy transformations.
• There are 3 Laws of Thermodynamics
– used to predict reaction spontaneity
1st Law of Thermodynamics
• Law of Conservation of Energy
• Total energy of the universe is
constant
• Energy cannot be created or
destroyed, just transferred from
one form to another
For a chemical reaction
∆Hºuniverse = ∆Hºsystem + ∆Hºsurroundings = 0
In a chemical reaction, the PE of the reactants
and products results in the transfer of energy
from the:
1) surroundings to the chemical system (ENDO)
2) chemical system to the surroundings (EXO)
Enthalpy changes and Spontaneity
• Bond energy (BE): the minimum
energy required to break 1 mol of
bonds between 2 atoms, +ve BE
• Equals the energy released when 1 mol
of bonds are formed, -ve BE
• The greater the value, the more stable
the bond
SO, the Enthalpy of reaction can also be
calculated by:
∆Hrxn = BE break reactant bonds – BE form product bonds
eg. C2H4 + 3 O2  2 CO2 + 2 H2O (g)
H
H
O O
C C
H
O C O
H O H
H
∆Hcomb = [BE(C=C) + 4 BE(C-H) + 3 BE(O=O)]
– [ (2+2) BE(C=O) + (2+2) BE(O-H)]
= [1mol(614 kJmol-1) + 4mol(413 kJmol-1) + 3mol(495
kJmol-1)] - [4mol(745 kJmol-1) + 4 mol(467 kJmol-1 )]
= - 1097 kJ
Using Hess’ Law Summation Formula:
H comb= 

nH f prod
-

nH f reac
∆Hcomb = [2Hf° CO2 + 2Hf° H2O] - [Hf° C2H4 ]
= [2mol(-393.5 kJmol-1) + 2mol(-241.8 kJmol-1)]
- [1mol(+52.5 kJmol-1)]
= -1323.1 kJ
The values from the methods differ as the BE values
are averaged from several compounds with that bond
type.
∆Hreaction -ve (< 0)
• This indicates the formation of stronger
bonds and more stable compounds
• is most likely to be a spontaneous
reaction
Entropy Changes (∆S) and Spontaneity
• the greater the # of ways particles can arrange
themselves, the less ordered they are
• the greater the # of ways a particular state can
be achieved, the more likely that state is going
to exist
Entropy , Sº
• The measure of disorder or randomness
• increased disorder,entropy favors spontaneity
∆S > 0
2nd Law of Thermodynamics
∆Sºuniverse = ∆Sºsystem + ∆Sºsurroundings > 0
disorder increases, since ∆S > 0
•The following increases entropy:
1.The volume of a gas increases
2.The temperature of the system increases
3.Physical state solid to liquid to gas
4.Increase in the number of moles produced
5.Breaking complex molecules into smaller ones
3rd Law of Thermodynamics
• At 0K all motion ceases, the forces of
attraction have reduced entropy to a
minimum
• S = 0 at T = 0 K
• As a result as T increases, S must
increase
 S is a measure of the energy needed to
achieve a level of disorder by overcoming
the FA
 This is dependent on the substance and
the temperature reached
 J/molK
 ∆Sº =  nSº products -  nSº reactants
 ∆Sº +ve (>0) favors spontaneity
Entropy Calculations
eg. Calculate ∆Sº for
2CO(g) + O2(g) 
2CO2(g)
∆Srxnº =  nSº products -  nSº reactants
∆Srxnº = 2mol( 213.78 J•mol-1•K-1)
- (2mol(197.66 J•mol-1•K-1) +1mol(205.10 J•mol-1•K-1))
= -172.86 J/K
Consistent as fewer # of product molecules, ∆Sº is < 0.
This would mean it is nonspontaneous, but we need
to consider the ∆Hrxnº .
∆Hrxnº =  n∆Hº products -  n∆H º reactants
∆Hrxnº = 2mol(-393.5 kJ•mol-1)
- 2mol(-110.5 kJ•mol-1)
= -566.0 kJ
Enthalpy, ∆Hº, is –ve  spontaneous
But, Entropy, ∆Sº, is –ve  non-spontaneous
So the reaction is ?????!