Spontaneity, Entropy and Free Energy
Thermodynamics considers only the initial and final states, not time.
Kinetics Recall reaction Rates considered pathway,
concentrations, temperatures and activation energy.
Consider …
Fe will rust but never spontaneously turn back
into Fe.
A tree burns but never reverts back into a tree
Spontaneous: Occurs without outside intervention. A
reaction is spontaneous because of an increase in
disorder. There are more possible positions in a random,
disordered arrangement, more positive, than complete
order.
Entropy (S) is a measure of randomness and
disorder.
S(solid) < S(liquid) << S(gas)
Sample Ex. 16.1 and 16.2
Second Law of Thermodynamics:
In any spontaneous reaction there is always an
increase in Entropy in the Universe.
Or
S(Universe) = S(system) + S(surroundings)
Effect of Temperature:
S(surroundings) is of primary interest at constant temperature and pressure.
Exothermic reactions, -H always increase the entropy of the surroundings, +S.
But the impact of heat is more significant at low temperatures.
Ex., Giving $50 to a student increases the wealth more than giving $50 to a rich person.
Sign of S(surroundings) depends on the direction of heat flow:
Endothermic- Heat moves from surroundings to system, -S
Exothermic- Heat moves from system to surroundings, +S
Magniture of S(surroundings) depends on temperature.
Endothermic (+H) -S(surroundings) = - J/K
Exothermic (-H) +S(surroundings) = +J/K
OR
S(surroundings) =
Sample Ex. 16.4
H  J 
 
T K
H is change in enthalpy, T is Kelvin absolute temperature.
Review Table 16.3
Free Energy: (G) Named After Gibbs and looks at temperature dependence on spontaneity.
This relationship states that:
*G= H -TS
All quantities refer to the system. That is if no superscript is given we are referring to the system. Or So
Any reaction carried out at constant T and P will be spontaneous only if G is negative. Or any reaction is
spontaneous only in a direction in which Free Energy is negative. (Or the S(universe) is positive).
Review Table 16.4 and Table 16.5
Need to consider the H and S
System wants to lower it energy and therefore have more randomness
At low temperature H dominates
At high temperature S dominates
In Class: 16.5
During a phase change G=0 Assume that Ho and So are independent of temperature
At what temperature is the following process spontaneous? Endothermic boiling point.
Br2(l)  Br2(g)
o
H =31.0 kJ/mol So=93.0 J/K . mol
(BP VP of liquid = VP gas, equilibrium) Set G=0; G= H -TS or Ho= TSo; T= 333 K
Homework Practice (Set 1), P. 783 #23, 24, 25, 26, 27, 28, 29, 30, 31, 32
Recap For Reinforcement: (Day 2 Discussion with Questions –Entropy)
Nature tends to move from more ordered to more random states.
S (solid) < S (liquid) < S (gas)
Increasing the temperature increases the entropy.
2nd Law of Thermodynamics: The entropy of the universe is always increasing.
S universe = S system + S surroundings, S universe must be positive to be spontaneous.
Standard Entropy values (at 25oC and 1 atm, 1 M) are listed on tables. These are always positive values.
The 3rd Law of Thermodynamics tells us that a completely ordered pure crystalline solid has an
entropy value of 0 J/K. Unlike enthalpy, entropy values can be measured. The more complicated the
molecule the larger the S (J/mol . K) value.
Q1: Explain why H2(g) has a Hf = 0 kJ/mol but a S= 131 J/mol K? Compare that to H2O(g)
Hf=-242 kJ/mol and a S= 189 J/mol K.
Q2: Which is predicted to have the larger Standard Entropy value? CH 4 or C4H10?
So for Reactions can be calculate as: So= n(products)So - n(reactants)So (Standard Conditions)
Q3: Calculate the value for the standard entropy change So(rxn) for the reaction:
CaCO3(s) 
 CaO(s) + CO2(g)
Why is the sign value positive for this reaction?
Relation of Entropy to Spontaneity:
S universe = S system + S surroundings, S universe will be positive for any spontaneous change
The sign of S surroundings depends on the heat flow.
If heat flows out of the system into the surroundings then Ssurr will be positive. More
disorder.
If the system absorbs heat from the surroundings then Ssurr will be negative. Lower kinetic
energy, more order.
The magnitude of Ssurr depends on two factors: The amount of heat transferred and the absolute
temperature.
 S(surroundings) =
H  J 
 
T K
H is change in enthalpy, T is Kelvin absolute temperature
Q4: Show by calculation that the vaporization of water is spontaneous at 120 oC and 1atm.
H2O(l)

 H2O(g)
1st calculate Ssys with standard entropy values. (Appendix 4, A20) (+119 J//k)
2nd calculate the Ssurr with Hfor (App. 4, A20) +44 kJ/mol (endothermic) ( g > l)
S(surroundings) =
 H
T
= S(surroundings) =
 44
393
= - 0.112 kJ/K
Substituting:
S(Universe)


= S(system) + S(surroundings)
0.119kJ/mol + (- 0.112 kJ/mol) = 0.007 kJ/K > 0, reaction is spontaneous