•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.

•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: G Total = G A + G B •Figure (a) illustrates the initial free energies of A and B.

•As A changes to B, G A will decrease because P A is decreasing (b).

•G B will increase because P B is increasing.

•Reaction will proceed to right (products) as long as G Total decreases (G B less than G A ). •At some point the is pressures of A and B reach the values such as G A is equal to G B.

•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.

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

•K ( equilibrium constant ) can be calculated using equilibrium pressures: K  P e B P e A  1 .

5 atm 0 .

50 atm  3 .

0 •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.

•In summary, •Reactions proceed to minimum free energy (equilibrium) •Corresponds to the point where G products = G reactants or ΔG = G products – G reactants = 0

•Remember, ΔG = ΔG o •At equilibrium , ΔG equals 0 and Q = K.

•So ΔG = 0 = ΔG o + RT ln(K) •Or ΔG o + RT ln(Q) = -RT ln(K)

ΔG o = -RT ln(K)

Case 1

: ΔG o = 0.

•When ΔG o = 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.

ΔG o = -RT ln(K)

Case 2:

ΔG o < 0. •ΔG o (G o products – G o reactants ) is negative •G o products < G o reactants •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).

ΔG o = -RT ln(K) •

Case 3:

ΔG o > 0.

•ΔG o (G o products – G o reactants ) is positive •G o reactants < G o products •System will not be at equilibrium and system will adjust to the left •K will be less to reach equilibrium.

than 1(pressure of the reactants will be greater than 1 atm and the pressure of the products will be less than 1 atm).

•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 ΔG o = -RT ln(K).

•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.

w max = ΔG •The amount of work obtained is always less than the maximum .

•Henry Bent’s First Two Laws of Thermodynamics •First law: You can’t win, you can only break even •Second law: You can’t break even •Read this section in the book!!!!!!!!!!!!