Second Law of
Thermodynamics:
THE TOTAL
ENTROPY OF THE UNIVERSE IS CONTINUALLY
INCREASING, or DSUNIVERSE is
positive.
DSUNIVERSE = DSSYSTEM + DSSURROUNDINGS
If more order is created in
a system, (i.e., DSSYSTEM is
negative), a
greater disorder MUST occur in the
surroundings
(i.e., DSSURROUNDINGSis
positive and greater than
DSSYSTEM).
A Molecular Interpretation of Entropy:
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.
We must also consider the
motions within molecules as a
measure of
their disorder.
There are three atomic modes
of motion:
1. translation
(the movement of a molecule from one
point in space to another),
2. vibration
(the shortening and lengthening of bonds,
including
the change in bond angles),
3. rotation
(the spinning of a molecule about some axis).
Some form of energy is
required to get a molecule to
translate,
vibrate, or rotate.
The more energy stored in
translation, vibration, and
rotation,
the greater the degrees of freedom of the
molecule,
and the higher the entropy.
In
a perfect crystal at 0 K, there is no translation,
rotation, or vibration of molecules. This represents a
state of perfect order.
This is the Third Law of
Thermodynamics:
The entropy of a perfect crystal at 0 K is zero.
Calculation of Entropy
Changes
Standard molar entropy, S°:
the entropy of a substance
in its
standard state.
Similar in concept to DH°.
Units of S° are J/mol-K,
and therefore, much smaller
than
enthalpies (DH)!
Note: units of DH°
are kJ/mol.
Standard molar entropies of
elements are not zero; only
a perfect
crystal at 0 K has zero entropy!
For a chemical reaction that
produces n products from m
reactants:
DS°
= Sum nS°(products)
- Sum mS°(reactants)
For N2(g) + 3H2(g) ----> 2NH3(g):
DS°= [2S°(NH3)] - [S°(N2) + 3S°(H2)]
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Gibbs Free Energy:
For a spontaneous
reaction, the entropy of the universe
must
increase.
Reactions with large
negative DH values are spontaneous.
How do we look at DSand DHto be able to predict
whether a
reaction is spontaneous?
Gibbs free energy, G, of a state is: G = H - TS
For a process occurring at constant
temperature:
DG = DH - TDS
There are three important
conditions:
1. If DG
< 0, the forward reaction is spontaneous.
2. If DG =
0, the reaction is at equilibrium and no net
reaction will occur.
3. If DG
> 0, the forward reaction is NOT spontaneous.
(However,
the reverse reaction IS spontaneous).
If DG > 0, work must be
supplied from the
surroundings to drive the reaction BACKWARDS.
For a reaction to occur, the
free energy of the reactants
decreases to
a minimum (equilibrium) and then
increases to
the free energy of the products.
Standard Free Energy
Changes:
Just like enthalpy, standard
free energies of formation
can be
calculated for substances in their standard
states,
i.e., pure solids, liquids or gases at 1 atm
and 298 K.
So, for a reaction of m reactants going to n
products
DGo = Sum nDGfo(products) - Sum mDGfo(reactants)
DGfo values of ELEMENTS in their most stable states
are zero.
The standard free energy of
a reaction tells us in which
direction
the reaction will proceed. Thus,
If DGo > 0, the reaction moves spontaneously to
reactants.
If DGo < 0, the reaction moves spontaneously to
products.
Free Energy and Temperature:
Focus on DG = DH - TDS:
1. If DH
< 0 and DS
> 0, then DG is ALWAYS
negative.
2. If DH
> 0 and DS
< 0, then DG is ALWAYS
positive
(i.e., the reverse of 1.)
3. If DH
< 0 and DS
< 0, then DG is
negative at low
temperatures.
4. If DH
> 0 and DS
> 0, then DG is
negative at
high
temperatures.
Even though a reaction has a
negative DG
(i.e.,
SPONTANEOUS!), it may
occur too slowly to be observed.
Remember.....
Thermodynamics
gives us the direction of a spontaneous
process.
Kinetics gives us the rate of the
process.
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Free Energy and K:
Under any
conditions: DG = DGo +
RT lnQ
where Q =
[products]/[reactants]
At
equilibrium, DG
= 0 and Q = Keq, so:
0 = DGo + RT lnKeq
Therefore, DGo = - RT lnKeq
so, when DGo < 0, K > 1
and when DGo = 0, K = 1
and when DGo >0, K < 1
R is the gas constant in J/mol-K = 8.314 J/mol-K
This is consistent
with our previous determination that
when K > 1, the products predominate (equilibrium
lies to the
right), and when K < 1, reactants
predominate
(equilibrium lies to the left).
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Driving Nonspontaneous
Reactions:
If DG > 0, work must
be supplied from the
surroundings to drive the reaction.
Biological systems use
spontaneous reactions to drive
non-spontaneous reactions.
In biology, disordered
nutrients are organized into
biological
constituents.
Therefore, DS is large and negative,
and many of these
reactions
are non-spontaneous.
Metabolism of food supplies
the energy to drive these
non-spontaneous reactions.
Example, glucose oxidation:
C6H12O6(s) + 6O2 -----> 6CO2+ 6H2O(l)
DGo = - 2880 kJ/mol
The free energy
released by glucose oxidation is used
to convert
low energy ADP to high energy ATP.
When ATP is converted
back to ADP, the energy released
is used to
convert simple molecules into complex cell
constituents. The low energy ADP is again available for
conversion
to ATP by glucose oxidation.