11

16.8 Free Energy and the Equilibrium Constant

16.8 Free Energy and the Equilibrium Constant

G < 0 spontaneousG < 0 spontaneous G > 0 not spontaneous, but reverse reaction is G > 0 not spontaneous, but reverse reaction is

spontaneousspontaneous G = 0 not spontaneous in either directionG = 0 not spontaneous in either direction

• This is a state of equilibrium between reactants This is a state of equilibrium between reactants and products; they coexist in a stable condition.and products; they coexist in a stable condition.

• Many situations are not in the standard state, so Many situations are not in the standard state, so we need some way to relate we need some way to relate G to G to GGoo and the and the existing conditions.existing conditions.

22

EquilibriumEquilibrium

• We define the reaction quotient to describe We define the reaction quotient to describe the extent of reaction.the extent of reaction.

• aA + bB aA + bB ⇌⇌ cC + dD cC + dD

• Q = Q = PPCCcc P PDD

dd

PPAAaa P PBB

bb

• Reminder: Only the pressure of gases (PReminder: Only the pressure of gases (PAA) ) or the concentration ([A]) of dissolved or the concentration ([A]) of dissolved substances appear in the reaction quotient; substances appear in the reaction quotient; pure solids and pure liquids are omitted.pure solids and pure liquids are omitted.

33

Equilibrium: Reaction Quotient

Equilibrium: Reaction Quotient

• 2H2H22(g) + O(g) + O22(g) (g) ⇌⇌ 2H 2H22O(g)O(g)

• Q = PQ = PHH22OO22/P/PHH22

22PPOO22

• 2H2H22(g) + O(g) + O22(g) (g) ⇌⇌ 2H 2H22O(l)O(l)

• Q = 1/PQ = 1/PHH2222PPOO22

55

Go and EquilibriumGo and Equilibrium

• Q can be used to relate Q can be used to relate G to G to GGoo

G = G = GGoo + RT ln Q + RT ln Q

R = 8.314 J/mol KR = 8.314 J/mol K

T = absolute temperature (K)T = absolute temperature (K)

lnln is natural logarithm; find this button on is natural logarithm; find this button on your calculator because it will be useful your calculator because it will be useful throughout the semester.throughout the semester.

66

Go and EquilibriumGo and Equilibrium

G = G = GGoo + RT ln Q + RT ln Q• no products: Q = 0no products: Q = 0• no reactants: Q = no reactants: Q = • standard state: all P=1, all [X]=1, so Q = standard state: all P=1, all [X]=1, so Q =

1 and ln Q = ln 1 = 0, 1 and ln Q = ln 1 = 0, G = G = GGoo • Between the extremes, Q > 0, but < Between the extremes, Q > 0, but < and and

G changes as the reaction progresses G changes as the reaction progresses (decreases as the reaction goes forward)(decreases as the reaction goes forward)

77

Go and EquilibriumGo and Equilibrium

• CO(g) + HCO(g) + H22O(g) O(g) ⇌⇌ CO CO22 (g) + H (g) + H22(g)(g)

• Q = PQ = PCOCO22PPHH22

/P/PCOCOPPHH22OO

• Initially, Q = 0, but then it increasesInitially, Q = 0, but then it increases• As the reaction progresses, As the reaction progresses, G decreases until a G decreases until a

minimum value of G is reached; at this point, minimum value of G is reached; at this point, G = 0 and net reaction ceases (though the G = 0 and net reaction ceases (though the forward and reverse reactions still occur, but forward and reverse reactions still occur, but they offset one another).they offset one another).

88

Go and EquilibriumGo and Equilibrium

99

Go and EquilibriumGo and Equilibrium

• When When G = 0 = G = 0 = GGoo + RT ln Q, we call this + RT ln Q, we call this a state of equilibrium and Q = Qa state of equilibrium and Q = Qequilequil = K, the = K, the equilibrium constant.equilibrium constant.

• Thus, Thus, GGoo = - RT ln K or K = = - RT ln K or K = ee--GGoo/RT/RT

GGo o > 0 when K << 1 (lies toward reactants)> 0 when K << 1 (lies toward reactants)GGo o < 0 when K >>1 (lies toward products)< 0 when K >>1 (lies toward products)GGo o = 0 when K = 1 (equal amounts of = 0 when K = 1 (equal amounts of

reactants and products)reactants and products)

1010

Go and EquilibriumGo and Equilibrium

• The position of equilibrium depends on The position of equilibrium depends on the value of the value of GGoo..

< 0

1111

Group WorkGroup Work

GGoo = - RT ln = - RT ln KK• Calculate K for the following reaction at 25Calculate K for the following reaction at 25ooC.C.

HH22(g) + Br(g) + Br22(g) (g) ⇌⇌ 2HBr(g) 2HBr(g) GGoo = = -- 53.2 kJ/mol 53.2 kJ/mol

• --53,200 J/mol = 53,200 J/mol = -- 8.314 J/mol K x 298 K x ln 8.314 J/mol K x 298 K x ln KK

• ln ln KK = 21.47 = 21.47• KK = 2.12 x 10 = 2.12 x 1099 ((KK has no units) has no units)• When When KK is this large, the reaction proceeds far is this large, the reaction proceeds far

toward products to reach equilibrium.toward products to reach equilibrium.

• PPHH2 2 = P= PBrBr22

= 4.72 x 10 = 4.72 x 10--10 10 atm when Patm when PHBrHBr = 1 atm = 1 atm