Entropy, Free Energy, and the Direction of Chemical Reactions
• Limitations of the First Law
∆E = q + w
E universe = E system + E surroundings
∆E system = -∆E surroundings
∆E system + ∆E surroundings = 0 = ∆E universe
The total energy of the universe is constant!
However, this does not tell us anything about the direction of change in the universe
It is noteworthy!
When gasoline burns in your car’s engine the ∆(PE) between the chemical bonds in the
fuel mixture and those in the exhaust gases is converted to the KE of the moving car
plus the heat released to the environment. If you measure q and w you will find that
energy is conserved. Now! Why doesn’t the heat released in the car’s engine convert
exhaust fumes back into gasoline and oxygen? This change doesn’t violate the First Law
but it never happens! So, we look elsewhere to predict the direction of the change.
• Spontaneous Change
A chemical or physical change in the system can occur by itself without an input of
energy from the surroundings, i.e., freezing of water at – 5oC and 1 atm pressure
• Non - spontaneous Change
The surroundings must support the system with a continuous input of energy
Spontaneous ≠ instantaneous
(Ripening, aging)
If a change is spontaneous in one direction it is not spontaneous in the opposite direction.
A chemical reaction proceeding towards equilibrium is an example of a spontaneous
change.
• Can the sign of ∆H predict spontaneous change?
A few examples
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) ∆H orxn = - 802 kJ spontaneous + exothermic
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2 Fe (s) + 3/2 O2 (g) → Fe2O3 (s) ∆H orxn = - 826 kJ spontaneous + exothermic
Na (s) + ½ Cl2 (g) → NaCl (s) ∆H orxn = - 411 kJ spontaneous + exothermic
H2O (l) → H2O (s) ∆H orxn = - 6.02 kJ spontaneous + exothermic at temperatures < 0
H2O (s) → H2O (l) ∆H orxn = + 6.02 kJ spontaneous + endothermic at temperatures > 0
H2O (l) → H2O (g) ∆H orxn = + 44.0 kJ spontaneous + endothermic at P = 1 atm., T =
298K and dry air
Recall from Chemistry 112 that most water-soluble salts have a positive ∆H oso ln and yet
dissolve spontaneously
NaCl (s) → Na+ (aq) + Cl − (aq) ∆H oso ln = + 3.9 kJ
NH4NO3 (s) → NH 4+ (aq) + NO 3− (aq) ∆H oso ln = + 25.7 kJ
And some more endothermic processes also spontaneous
N2O5 (s) → 2 NO2 (g) + ½ O2 (g) ∆H orxn = + 109.5 kJ
Ba(OH)2•8 H2O (s) + 2NH4NO3 (s) → Ba2+ (aq) + 2NO 3− (aq) + 2NH3 (aq) + 10H2O (l)
∆H orxn = + 62.3 kJ
Conclusion: The sign of ∆H cannot predict spontaneous change
A new state property is needed to establish a criterion for spontaneity that is valid at all
temperatures. This property is called Entropy.
Entropy refers to the state of order
A change in order is a change in the number of ways of arranging the particles and
dispersing their energy of motion, and it is a key factor in determining the direction of a
spontaneous process.
localized energy of motion
a) Case of a phase change
More order
Solid
→ Liquid
b) Case of dissolving a salt
More order
→
Crystalline solid + liquid
dispersed energy of motion
→
less order
Gas
less order
ions in solution
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c) Case of a chemical change
More order
→
Crystal + crystal
less order
gases + ions in solution
• How freedom of motion and dispersal of energy relate to spontaneous change
Modes of motion: translational, rotational and vibrational
Quantized energy states: discrete translational, rotational, vibrational and electronic
states
Microstate: each quantized state
With each microstate equally possible for the system, the laws of probability say that
over time all microstates are equally occupied. We focus on microstates associated with
thermal energy. Then the number of microstates, W of the system is the number of ways
it can disperse its thermal energy among the various modes of motion of all its
molecules.
1877: Ludwig Boltzmann (Austrian mathematician and physicist)
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S = k 1n W
S, entropy (JK )
k, Boltzmann’s constant, k = R / NA (R is universal gas constant, NA = Avogadro’s
number)
W, number of microstates
Conclusions
• A system with relatively few equivalent ways to arrange its components (smaller
number of microstates, W) has relatively less disorder and low entropy.
• A system with many equivalent ways to arrange its components (larger number of
microstates, W) has relatively more disorder and high entropy.
• Spontaneous expansion o a gas
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• Expansion of a gas and the increase in number of microstates
For 1 mol:
W final
Winitial
= 2 N A and
∆S system = S final − Sinitial = k ln W final − k ln Winitial = k ln
W final
Winitial
Since k = R / NA
R
R
∆S system =
ln 2 N A =
ln 2 = R ln 2 = (8.314 Jmol −1K −1 )(0.693) = 5.76 Jmol −1K −1
NA
NA
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Another approach to find ∆Ssystem is based on heat changes
∆S system =
qrev
(rev: reversible)
T
A reversible process occurs slowly enough for equilibrium to be maintained
continuously, so that direction of the change can be reversed by an infinitesimal reversal
of conditions
For our Ne example qrev = 1718 J . If we place Ne (g) in a 10-mL flask and we expand
reversibly to 20 mL at 298 K
q
1718 J
∆S system = rev =
= 5.76 JK −1
T
298K
So far,
• Spontaneous change in an isolated system proceeds with an increase in the
entropy of the system
• Entropy is a measure of disorder
The Second Law of Thermodynamics
• It sets the criterion for the direction of a spontaneous change
• We consider changes in both the system and surroundings
All real processes occur spontaneously in the direction that increases the entropy of
the universe
∆S universe = ∆S system + ∆S surroundings > 0
The Third Law of Thermodynamics
A perfect crystal at 0K has zero entropy
Recall, S = k 1n W
For a perfect crystal at 0K, W = 1, ln W = 1, so S = 0
When we warm the crystal, its total energy increases and the particles’ energy is
dispersed over more microstates, W > 1, ln W > 1, so S > 0
Conclusion: The entropy of the substance at a given temperature is an absolute value
that is equal to the entropy increase when the substance is heated from 0K to that
temperature.
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Standard Molar Entropies (So)
Standard state: gases at 1 bar and solutions at 1 M
We cannot measure absolute internal energy, E and enthalpy, H but we can measure
absolute entropies.
The units of So are J mol-1 K-1
Predicting So of a system
1. Temperature changes
So increases as temperature increases
2. Physical states and phase changes
So increases as a more ordered phase changes to a less ordered phase
∆S ovap > ∆S ofus
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3. Dissolving a gas
A gas becomes more ordered when it dissolves in a liquid or solid but when a gas
dissolves in another gas the entropy increases because of the mixing of the
molecules
4. Dissolving a solid or liquid
When we dissolve a solid or liquid things are getting complex. The entropy of a
dissolved solid or liquid is usually greater than the entropy of the pure solute. But
the nature of the solute and solvent and the dissolving process affect the overall
entropy change
NaCl AlCl3 CH3OH
S (s,l) 72.1(s) 167(s) 127(l)
So(aq) 115.1 -148 132
o
When an ionic solid dissolves into water, the crystal breaks down and the ions
experience a great increase in freedom of motion as they get hydrated and
separate. Their energy is dispersed over more microstates. Therefore, we expect
the entropy of the ions themselves to be greater in the solution than in the crystal
(case of NaCl (s)).
The small increase in entropy when ethanol dissolves in water.
↓
The entropy change accompanying the dissolution of a salt
→
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In some cases water molecules become more organized around the ions and their
motions are restricted and their restriction imposes a negative contribution to the
overall entropy change.
For Al3+ (aq) So = - 313 J mol-1 K-1 and even though So for Cl − (aq) is positive the
overall So <0.
In EtOH in H2O small increase in So because of random mixing of the molecules
5. Entropy increases down the group
Li
Atomic radius (pm) 152
Molar mass (g mol-1) 6.941
So (s)
29.1
Na
186
22.99
51.4
K
127
39.10
64.7
Rb
248
85.47
69.5
Cs
265
132.9
85.2
The same trend of increasing entropy down a group holds for similar compounds
HF
HCl HBr HI
Molar mass (g mol ) 20.01 36.46 80.91 127.9
So (g)
173.7 186.8 198.6 206.3
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For element that occurs in different allotropic forms the entropy is higher in the form
that allows the atoms more freedom of motion ( ⇒ disperses energy in more
microstates)
So graphite (2-dim): 5.69 J mol-1 K-1 (more motion)
So diamond (3-dim): 2.44 J mol-1 K-1 (less motion)
For compounds at the same phase entropy increases with chemical complexity
When gases are compared to liquids, the effect of the physical state usually
dominates that of molecular complexity.
Example
Predict relative entropy values
a) 1 mol of SO2 (g) or 1 mol of SO3 (g)
b) 1 mol of CO2 (s) or 1 mol of CO2 (g)
c) 3 mol of O2 (g) or 2 mol of O3 (g)
d) 1 mol CF4 (g) or 1 mol CCl4 (g)
e) seawater at 25oC or seawater at 35oC
f) 1 mol CaF2 (s) or 1 mol BaCl2 (s)
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• Entropy and vibrational motion
• Change in Entropy in a Chemical Reaction
∆S orxn - The entropy change that occurs when all reactants and products are in their
standard states.
∆S orxn = Σm∆So (products) - Σn∆So(reactants)
Calculate the ∆S orxn for the combustion of 1 mol of propane at 25oC
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
∆S orxn = [(3 mol CO2)(So CO2) + (4 mol H2O)(So H2O)] - [(1 mol C3H8)(So C3H8)
+ (5 mol O2)(So O2)]
∆S orxn = [(3 mol)(213.7 J/mol•K) + (4 mol)(69.9 J/mol•K)] - [(1 mol)(269.9
J/mol•K) + (5 mol)(205.0 J/mol•K)] = -374 J/K
Note: It is obvious that entropy is being lost because the reaction goes from 6 mol of
gas to 3 mol of gas.
• Entropy changes in the surroundings
The essential role of the surroundings is to either add heat to the system or to remove
heat from it. In essence the surroundings function as an enormous sink , so large that its
temperature remains constant even though its entropy changes through the loss or gain
of heat.
1) Exothermic Change: heat lost by the system is gained by surroundings. This heat gain
increases the freedom of motion of the particles in the surroundings ⇒ disperse their
energy ⇒ ∆Ssurroundings increases.
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qsystem < 0, qsurroundings > 0, ∆Ssurroundings >0
2) Endothermic Change: heat is gained by the system and it is lost form the
surroundings. The heat loss reduces the freedom of motion of particles in the
surroundings ⇒ ∆Ssurroundings decreases.
qsystem > 0, qsurroundings < 0, ∆Ssurroundings < 0
Note: the temperature of the surroundings at which the heat is transferred affects
∆Ssurroundings.
The change in entropy of the surroundings is directly related to an opposite change in
the heat of the system and inversely related to the temperature at which the heat is
transferred.
∆S surroundings =
− ∆H system
− qsystem
S
∆
=
and at constant pressure:
surroundings
T
T
Recall, a reaction is spontaneous when ∆Suniverse > 0
Example (to determine reaction spontaneity)
At 298 K the formation of NH3 (g) has a negative ∆S osystem . Is this reaction spontaneous at
298 K?
N2 (g) + 3H2 (g) → 2NH3 (g)
∆S osystem = - 197 J/K
∆H osystem = ∆H orxn = [(2 mol NH3)(-45.9 kJ/mol)] - [(3 mol H2)(0 kJ/mol) + (1 mol N2)(0
kJ/mol)] = -91.8 kJ
− ∆H system − ( −91.8kJ )(1000 J / kJ )
=
= 308 JK −1
T
298 K
o
o
= ∆S system + ∆S surroundings = −197 JK −1 + 308 JK −1 = 111JK −1 > 0
o
∆S surroundin
gs =
o
∆S universe
Therefore, reaction occurs spontaneously!
Comment: ∆Ho units are given usually in kJ but ∆So has units in J K-1. Don’t forget to
convert kJ to J
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Components of ∆Suniverse for spontaneous reactions
Entropy Change and the Equilibrium State
At equilibrium: ∆S universe = ∆S system + ∆S surroundings= 0
∆S system = - ∆S surroundings
Let us calculate ∆S universe for a phase change
H2O (l, 373 K) • H2O (g, 373 K)
o
∆S system
= 195.9 JK −1 − 86.8 JK −1 = 109.1JK −1
As we expect the entropy of the system increases as the liquid absorbs heat and changes
into a gas
∆S surroundings
− ∆H system − ∆H vaporization − 40.7 x103 J
=
=
=
= −109 JK −1
T
T
373K
∆S universe = 109.1 – 109.1 = 0
When a system reaches equilibrium neither the forward nor the reverse reaction is
spontaneous. NO NET REACTION IN EITHER DIRECTON
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