Free Energy

1
Free Energy and Equilibrium
2

• When components of a reaction are mixed, they
  will proceed, rapidly or slowly (depending on
  kinetics) to the equilibrium position.
• Equilibrium position defined as the point at
  which forward and reverse reaction rates are
  equal (Chapter 13).
• Thermodynamics point of view equilibrium point
  occurs at the lowest value of free energy
  available to the reaction system.

3

• Consider the simple reaction below
• A(g) ? B(g)
• where 1.0 mole of gaseous A is placed in a
  reaction vessel at a pressure of 2.0 atm.
• Total free energy of system
• GTotal GA GB
• Figure (a) illustrates the initial free energies
  of A and B.

4

• As A changes to B, GA will decrease because PA is
  decreasing (b).
• GB will increase because PB is increasing.
• Reaction will proceed to right (products) as long
  as GTotal decreases (GB is less than GA).
• At some point the pressures of A and B reach the
  values such as GA is equal to GB.

5

• The system is at equilibrium and has reached
  minimum free energy (c).
• No longer driving force to change A to B or B to
  A, so system remains at this position and
  pressures of A and B remain constant.

6

• A(g) ? B(g) 1.0 mol A(g) at 2.0 atm
• Plot of free energy versus the mole fraction of A
  is shown in (a) below.
• Minimum free energy is reached when 75 of A has
  been changed to B.
• At this point, the pressure of A is 0.25 times
  the original pressure or
• (0.25)(2.0 atm) 0.50 atm
• Pressure of B is
• (0.75)(2.0 atm) 1.5 atm

7

• K (equilibrium constant) can be calculated using
  equilibrium pressures
• Overall free energy curve is shown in (c).
• This demonstrates any mixture of A(g) and B(g)
  containing 1.0 mol (A plus B) at a total pressure
  of 2.0 atm will react until it reaches the
  minimum in the curve.

8

• In summary,
• Reactions proceed to minimum free energy
  (equilibrium)
• Corresponds to the point where
• Gproducts Greactants
• or
• ?G Gproducts Greactants 0

9

• Remember,
• ?G ?Go RT ln(Q)
• At equilibrium, ?G equals 0 and Q K.
• So ?G 0 ?Go RT ln(K)
• Or ?Go -RT ln(K)

10

• ?Go -RT ln(K)
• Case 1 ?Go 0.
• When ?Go 0 the free energies of reactants and
  products are equal when all components are in
  standard states (1 atm for gases).
• System is at equilibrium when the pressures of
  all reactants and products are 1 atm, which means
  K 1.

11

• ?Go -RT ln(K)
• Case 2 ?Go lt 0.
• ?Go (Goproducts Goreactants) is negative
• Goproducts lt Goreactants
• System is not at equilibrium and system will
  adjust to the right to reach equilibrium.
• K will be greater than 1 (pressures of products
  at equilibrium are greater than pressures of
  reactants).

12

• ?Go -RT ln(K)
• Case 3 ?Go gt 0.
• ?Go (Goproducts Goreactants) is positive
• Goreactantslt Goproducts
• System will not be at equilibrium and system will
  adjust to the left to reach equilibrium.
• K will be less than 1(pressure of the reactants
  will be greater than 1 atm and the pressure of
  the products will be less than 1 atm).

13

• Cases 1, 2 and 3 are summarized in Table 16.6.
• The value of K for a specific reaction can be
  calculated from the equation
• ?Go -RT ln(K).

14
Free Energy and Work

• The maximum possible useful work obtainable from
  a process at constant temperature and pressure is
  equal to the change in free energy.
• Free energy is energy free to do work.
• wmax ?G

• The amount of work obtained is always less than
  the maximum.

15

• Henry Bents First Two Laws of Thermodynamics
• First law You cant win, you can only break
  even
• Second law You cant break even
• Read this section in the book!!!!!!!!!!!!