Chapter 18: Entropy, Free Energy, Equilibrium
Spontaneous processes:
- proceed on its own without any outside assistance
- have directional character;
Chemical thermodynamics is concerned with energy
relationships in chemical reactions.
i.e.: enthalpy, entropy
A process that is spontaneous in one direction
is not spontaneous in the opposite direction.
Examples:
First law of thermodynamics: energy is conserved.
U = q + w
where:
- Gas expanding into a vacuum; it will never reverse itself.
- When two eggs are dropped they spontaneously break.
- Nail rusting; wouldn’t expect it to just “unrust.”
- Ice melting
U is the change in internal energy,
q is the heat absorbed by the system,
w is the work done
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Reversible and Irreversible Processes
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An irreversible process cannot be reversed to restore the
system and surroundings back to their original state.
A reversible process is one that can go back and forth
between states along the same path, restoring the
system to its original state.
- a different path (with different q and w) must be taken.
- there is no net change in the system or the
surroundings when this cycle is completed.
- completely reversible processes are too slow to
be attained in practice.
(infinitesimally small steps)
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When gas molecules spread out from separate 1L bulbs
into a 2 L system, there is an increase in the
randomness or disorder.
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Entropy and the Second Law of Thermodynamics
- processes in which the disorder or entropy of the system
increases tend to be spontaneous.
- reverse process would be nonspontaneous.
Entropy, S, is a thermodynamic term that reflects the
disorder, or randomness, of the system.
The more disordered the system is, the larger the value of S.
Entropy is a state function, independent of path.
For a system, S = Sfinal – Sinitial.
If S > 0 the disorder increases, if S < 0 the order increases.
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The Second Law of Thermodynamics
For a reversible process:
In spontaneous processes, entropy of the universe increases.
The change in entropy of the universe is the sum of the
change in entropy of the system and the change in entropy of
the surroundings.
Suniv = Ssystem + Ssurroundings = 0
For a spontaneous process (i.e., irreversible):
Suniv = Ssystem + Ssurrroundings > 0
Suniv = Ssystem + Ssurroundings
Entropy is not conserved: Suniv is continually increasing.
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The entropy of a system indicates its disorder.
A gas is less ordered than a liquid, which is less
ordered than a solid.
Any process that increases the number of gas molecules
leads to an increase in entropy.
Consider:
2 NO (g) + O2 (g) → 2 NO2 (g)
When NO(g) reacts with O2 (g) to form NO2 (g),
the total number of gas molecules decreases.
- entropy decreases (more order)
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Individual molecules have degrees of freedom associated
with motions within the molecule.
• Types of motion: translation, vibration, rotation
Translational: Movement of the entire molecule.
Motion across some cartesian coordinate.
Vibrational: Periodic motion of atoms within a molecule.
Shortening and lengthening of bonds.
Rotational: Rotation of the molecule on an axis or rotation
about  bonds.
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Predicting S
Consider the melting of ice.
As ice, water molecules are held rigidly in a lattice.
As the solid melts, molecules become more randomly
distributed and decrease the order of the system,
(entropy increases)
Consider KCl dissolving in water.
Solid KCl has ions in a highly ordered arrangement.
Energy is required to induce translation, vibration or rotation.
The more energy stored in translation, vibration, and rotation,
the greater the entropy.
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When the crystal dissolves, the ions are more randomly
distributed and have more freedom; S increases
But…what happened to the water?
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Water molecules are slightly more ordered than before
as they hydrate the ions.
Entropy will increase as we increase T of the perfect crystal.
(Vibrational motion)
Both ordering and disordering occur, with disorder
usually predominating for most salts.
• The entropy changes dramatically at a phase change.
When a solid melts, the molecules and atoms have a large
increase in freedom of movement. (Even more so for boiling)
In general, entropy will increase when:
- liquids or solutions are formed from solids.
- gases are formed from solids or liquids.
- the number of gas molecules increases.
The Third Law of Thermodynamics:
The entropy of a perfect pure crystal at 0 K is zero.
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Entropy Changes in Chemical Reactions
Absolute entropy are based on a reference point of zero
for a perfect crystalline solid at 0K (the 3rd law).
For chemical reactions:
So =  n So(products) -  m So(reactants)
Standard molar entropy, So is the molar entropy of a
substance in its standard state. (J/K mol)
N2 (g) + 3 H2 (g) → 2 NH3 (g)
Some observations about So values:
- Standard molar entropies of elements are not zero.
- Sogas > Soliquid > Sosolid
- So tends to increase with increasing molar mass of
the substance.
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So = {2 So (NH3) – [So (N2) + 3 So (H2)]}
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Gibbs Free Energy
Entropy Changes in the Surroundings
Processes/reactions which increase in entropy,
tend to be spontaneous.
For an isothermal process, T = 0
Ssurr = -qsys / T
Reactions with large negative H values,
tend to be spontaneous.
For a reaction at constant pressure,
Can S and H predict whether a reaction is spontaneous?
H = qsys (constant pressure)
Souniv = Sosurr + Sosys
The Gibbs free energy, G, of a state is: G = H – TS
Sosurr or Sosys may be either positive or negative,
but for an isolated system, Souniv will be either  0.
When Souniv > 0, isolated systems are spontaneous.
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At constant T and P, the free energy change is:
G = H – TS
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G = H – TS
If G < 0, forward reaction is spontaneous.
Standard Free Energy Changes
If G = 0, reaction is at equilibrium.
If G > 0, reverse reaction is spontaneous.
- work must be supplied from the surroundings
to drive the reaction in the forward direction.
The equilibrium position in a spontaneous process is given
by the minimum free energy available to the system.
The standard free energy change for a process is given by:
G decreases until it reaches this minimum value.
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Tabulated : fGo (= 0 for elements)
Go =  n fGo(products) -  m fGo(reactants)
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Free Energy and Temperature
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Consider the following reaction:
The sign of G indicates whether reaction is spontaneous.
G = H – TS
• If H < 0 and –TS < 0; G will always be < 0. (spontaneous)
H2O (s)  H2O (l)
At a temperature less than 0oC:
H > TS; therefore G > 0
• If H > 0 and –TS > 0; G will always be > 0. (not spontaneous)
The melting of ice is not spontaneous when the
temperature is less than 0oC.
• If H and –TS have different signs;
G will depend on the sign and magnitudes of H and S.
H > 0, S > 0
At a temperature greater than 0oC:
H < TS; therefore G < 0
Temperature will be an important factor.
The melting of ice is spontaneous when the
temperature is greater than 0oC.
At 0oC:
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H = TS; therefore G = 0
Ice and water are in equilibrium at 0oC
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Free Energy and the Equilibrium Constant
• Go and Keq apply to standard conditions.
• G and Q apply to any conditions.
For any chemical process, the general relationship between
the standard free energy change and the free energy change
under any other conditions, G, is given by:
G = Go + RT lnQ
• At equilibrium, Q = Keq and G = 0, so:
If Go < 0, then K > 1
If Go = 0, then K = 1
If Go > 0, then K < 1
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Go = – RT lnK
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