Second Law of Thermodynamics:
THE TOTAL ENTROPY OF THE UNIVERSE IS CONTINUALLY
    INCREASING,   or    DS_UNIVERSE is positive.

     DS_UNIVERSE =  DS_SYSTEM +  DS_SURROUNDINGS

If more order is created in a system, (i.e., DS_SYSTEM is
    negative), a greater disorder MUST occur in the
    surroundings

(i.e.,  DS_SURROUNDINGS is positive and greater than
        DS_SYSTEM).
 

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.
 

This is the Third Law of Thermodynamics:
      In a perfect crystal at 0 K, there is no translation,
    rotation, or vibration of molecules.  This represents a
    state of perfect order.

    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° = S nS°(products) - S mS°(reactants)
 

For N₂(g) + 3H₂(g) ----> 2NH₃(g):

18.4 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 DS and DH to 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
   
      DG^o = S nDG_f^o(products) - S mDG_f^o(reactants)

DG_f^o 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 DG^o > 0, the reaction moves spontaneously to
                                reactants.
  If DG^o < 0, the reaction moves spontaneously to
                                products.
 

 1.  If DH < 0 and DS > 0, then DG is ALWAYS negative
      and the reaction is spontaneous as written.
 2.  If DH > 0 and DS < 0, then DG is ALWAYS positive
      and the reaction is non-spontaneous as written.
      (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, kinetics!).

    Under any conditions:         DG = DG^o + RT lnQ
             where Q = [products]/[reactants]

    At equilibrium, DG = 0 and Q = K_eq, so:
                                0 = DG^o + RT lnK_eq
          Therefore,  DG^o = - RT lnK_eq

     so, when DG^o < 0, K > 1              at equilibrium       
    and when DG^o = 0, K = 1
    and when DG^o >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 that when K < 1,
    reactants predominate (equilibrium lies to the left).
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18.6 THERMODYNAMICS IN LIVING SYSTEMS

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.

The metabolism of food supplies the energy to drive
    these non-spontaneous reactions.

            Example, glucose oxidation:

    C₆H₁₂O₆(s) + 6O₂ -----> 6CO₂+ 6H₂O(l)
                     DG^o = - 2880 kJ/mol

 The free energy released by glucose oxidation is used
    to convert low energy ADP (adenosine diphosphate)
    to high energy ATP (adenosine triphosphate).

 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.

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