AP Chemistry : Ch. 16 Notes
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A. Spontaneity, Entropy, and Free Energy
1. Spontaneous Processes and Entropy
a. Spontaneous reactions occur under the given conditions without external stimuli.
- spontaneous processes may be fast or slow
- thermodynamics tell us only if a reaction will occur, not how fast (kinetics tells us how fast)
- thermodynamics considers only the initial and final states, the pathway is relevant only to
kinetics
- thermodynamics and kinetics are needed to fully describe a reaction
- examples - ice melting at T>0 ºC, NaCl dissolving in water, iron rusting
b. Entropy (S) - a measure of disorder
- a state function (independent of pathway)
- the driving force behind spontaneous reactions
- entropy is a thermodynamic function that describes the number of possible arrangements that are
available to a system
- nature proceeds spontaneously towards the states that have the highest probability of existing
(greatest positional probability)
- positional probability increases from solid to liquid to gas (i.e. gases have greatest entropy)
- a solution of a solid has more positional probability than the solid sitting in a beaker of water
- the more complex the molecule, the greater its entropy
- due to greater degrees of freedom or forms of motion
- vibrational motion or rotational motion (spinning of atoms on their axes)
2. Entropy and the Second Law of Thermodynamics
a. Second Law : In any spontaneous process there is always an increase in the entropy of the universe.
- the entropy change of the universe equals the sum of the changes of entropy in the system and
the surroundings
∆Suniv = ∆Ssys + ∆Ssurr
- if ∆Suniv is positive (increasing entropy) the process is spontaneous (and irreversible)
- if ∆Suniv is negative (decreasing entropy) the process is nonspontaneous
3. The Effect of Temperature on Spontaneity
a. both ∆Ssys and ∆Ssurr need to be considered
- if both ∆Ssys and ∆Ssurr are positive, then ∆Suniv is positive (spontaneous)
- if both ∆Ssys and ∆Ssurr are negative, then ∆Suniv is negative (nonspontaneous)
- in an exothermic process heat flows from the system to the surroundings and ∆Ssurr
increases(positive) and ∆Ssys decreases (negative)
- in an endothermic process heat flows from the surroundings to the system and ∆Ssys increases
(positive) and ∆Ssurr decreases (negative)
- in both of the above cases one entropy change favors an increase in ∆Suniv and one entropy
change is unfavorable towards and increase in ∆Suniv.
- In these cases the temperature is the factor which determines if ∆Suniv is positive or not.
- Consider phase changes in water. Above 100 ºC, at 1 atm, the change from the liquid state to
the gaseous state is spontaneous. Below 100 ºC at 1 atm, condensation is the spontaneous
process.
- Entropy changes in the surroundings are primarily determined by heat flow.
- The magnitude of ∆Ssurr is temperature dependent.
- The significance of exothermicity is temperature dependent.
- The impact of heat transfer to or from a system to the surroundings will be greater at lower
temperatures.
To Summarize :
- The sign of ∆Ssurr depends on the direction of heart transfer.
- The magnitude of ∆Ssurr depends on the temperature.
Signs of Entropy Changes
Spontaneous Process?
∆Ssys
∆Ssurr
∆Suniv
+
+
+
Yes
No (will be spontaneous in
AP Chemistry : Ch. 16 Notes
+
-
?
-
+
?
- At constant temperature :
∆S surr = −
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opposite direction)
Yes, if ∆Ssys has a greater
magnitude than ∆Ssurr
Yes, if ∆Ssurr has a greater
magnitude than ∆Ssys
∆H
T
- where : T is Kelvin temperature
∆H is the enthalpy of system (negative if exothermic)
- ∆Ssys will be equal in magnitude and opposite in sign of ∆Ssurr
4. Free Energy (G) - can be used to determine if a reaction is spontaneous or not at constant T and P
∆G = ∆H − T ∆S
- if ∆G is negative, the reaction is spontaneous (at constant T and P)
- if ∆G is positive, the reaction is nonspontaneous (at constant T and P)
- if ∆G = 0, the reaction is at equilibrium
- Note : A negative enthalpy value and a positive entropy value favor a negative free energy value and
a spontaneous reaction.
Case
Result
∆S positive
∆H negative
Spontaneous at all temperatures
∆S positive
∆H positive
Spontaneous at high temperatures
(exothermicity is relatively unimportant)
∆S negative
∆H negative
Spontaneous at low temperatures (where
exothermicity is dominant)
∆S negative
∆H positive
Nonspontaneous at all temperatures
(reverse process is spontaneous at all
temperatures)
a. Also, at constant T and P :
∆S univ = −
∆G
T
5. Entropy Changes in Chemical Reactions
a. Fewer molecules in products indicates a decrease in entropy.
b. A change of state also indicates a change in entropy. The change in positional entropy is dominated
by gaseous reactants or products.
c. Third Law of Thermodynamics : The entropy of a perfect crystal at 0 K is zero.
AP Chemistry : Ch. 16 Notes
d. Calculation of entropy from standard entropy values :
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∆Sº reaction = ∑ Sº products − ∑ Sº reactants
- generally, the more complex the molecule, the higher the standard entropy (more rotational and
vibrational positions)
- the standard enthalpies of elements in their standard states are not zero (unlike standard
enthalpies and standard free energies)
6. Free Energy Change and Chemical Reactions
a. Standard free energy change (∆Gº) - the free energy change that occurs in a chemical reaction when
all reactants and products are in their standard states.
b. Calculation of standard free energy change :
- free energy is a state function
∆Gº reaction = ∑ Gº products − ∑ Gº reactants
or :
∆Gº = ∆Hº-T∆Sº
or by adding ∆Gº values from chemical reactions as in Hess's law for ∆H
7. The Dependence of Free Energy on Pressure
a. The equilibrium position of a reaction system represents the lowest free energy value available.
- reason why some reactions will not go to completion
b. For an ideal gas :
- enthalpy is independent of pressure
- entropy is pressure-dependent (change in volume changes positional probabilities)
∆G = ∆G º + RT ln(Q)
- where : R = universal gas constant (8.3145J/K·mol)
T = Kelvin temperature
Q = reaction quotient
8. Free Energy and Equilibrium
a. The equilibrium point is at the lowest free energy value available to the system.
- therefore, Gprod = Greact, or ∆G = Gprod - Greact = 0
and
∆G º = − RT ln( K )
b. Relationship between ∆Gº and K :
∆Gº
K
∆Gº = 0
K=1
∆Gº < 0
K>1
∆Gº > 0
K<1
9. Free Energy and Work
a. The change in free energy, at constant temperature and pressure, is equal to the maximum possible
amount of useful work obtainable from that system.
-
wmax = ∆G
- According to second law, this amount of work is not really available - some energy will be lost
in any real pathway
- reversible processes are really hypothetical since energy is lost in the changes (it really takes
more work to restore the original condition)
- all real processes are therefore irreversible
b. Summary of first two laws of thermodynamics from Henry Bent :
- First law : You can't win, you can only break even.
- Second Law : You can't break even.