Entropy and the 2nd Law of
Thermodynamics
Thermodynamic laws
1st Law – Energy is always conserved. It
cannot be created or destroyed
2nd Law – The Entropy, (randomness or
disorder), of the universe is always
increasing.
3rd Law – States that the entropy of a pure
crystal at 0 Kelvin is zero.
Entropy
Defined as randomness or disorder. Units
are J/K mole
The more disorganization in the molecules
or atoms means higher entropy.
The entropy of a gas is much greater than
a solid or liquid
There is a natural tendency to increasing
entropy.
Entropy
So, any reaction or system in which
disorder is increased will tend to be
favored.
Symbol for entropy is “S”, and we look at
the change in entropy which is, ∆S.
If ∆S is (+) then entropy is increasing –
getting more random or disorder.
If ∆S is (-) then entropy is decreasing –
getting more ordered, or less random
Entropy
The total entropy of the universe can be
described as the sum or the entropies of a
system and its surroundings.
∆Suniverse = ∆Ssystem + ∆Ssurroundings
∆Ssystem depends on positional entropy, for
example, how many gas molecules form in a
reaction. More gas molecules is more
entropy.
∆Ssurroundings depends on heat. Is the process
endothermic or exothermic.
– Exothermic process – heat is added to the
surroundings so ∆Ssurroundings is (+)
– The magnitude of ∆S depends on the amount of heat
that flows into the surroundings.
– Endothermic process – heat is take from the
surroundings, so ∆Ssurroundings is (-) negative.
– Again, the magnitude of ∆S depends on the amount
of heat that flows away from the surroundings.
∆Ssurroundings =
-∆H
T
The negative sign in front of ∆H is included
because ∆H is determined from the system’s
point of view, whereas ∆S is determined from the
surroundings point of view.
Reactions with a ∆Suniverse that have a (+)
value , or where entropy is increasing will be
considered spontaneous. This means that will
naturally proceed to make the products
∆Ssystem
∆Ssurroundings ∆Suniverse
Spontaneous
?
(+)
(+)
(+)
Yes, always
(-)
(-)
(-)
(+)
(-)
?
No, reverse
is
Yes, at high
temps.
Yes, at low
temps.
(endothermic)
(-)
(+)
(exothermic)
?
#1 For CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
What will the ∆Suniv be for the above reaction
and will it be spontaneous given:
∆Ssys = -38.5 J/K mole
∆H = -802.3 kJ/mole at 298K
∆Suniv = ∆Ssys + ∆Ssurr
∆Suniv = ∆Ssys + -(∆H)
T
= (-38.5 J/Kmol)+ -(-802,300 J/mol)
298K
= (-38.5) + (2692) =
+ 2654 J/Kmol
It is spontaneous because ∆Suniverse is
increasing (+).
2) Is entropy increasing of decreasing in the
following reactions?
a. 2KClO3(s)  2KCl(s) + 3O2(g)
Increasing – gas is formed and gases have
higher entropy (more disorder)
b. 2Ag(s) + Cl2(g)  2AgCl(s)
Decreasing – product is all solid – and
solids are more ordered and organized.
c. 2C8H18(l) + 25O2(g)  16CO2 (g) + 18H2O(g)
Increasing – product side has more gas
molecules (25 vs. 34 moles)
**More complex molecules will have more
entropy. Aqueous solutions with more ions
also have more entropy
Calculating Entropy from a
Reaction.
We can calculate entropy like we calculate
the enthalpy of a reaction, using the
standard entropies from the table in the
appendix
∆S = ∑ S products - ∑ S reactants