Lecture 9 : G, Q, and KThe Meaning of G

• Reading: Zumdahl 10.10, 10.11, 10.12

• Outline– Relating G to Q– Relating G to K– A descriptive example of G

The Second Law

• The Second Law: there is always an increase in the entropy of the universe.

• From our definitions of system and surroundings:

Suniverse = Ssystem + Ssurroundings

Defining G

• Recall, the second law of thermodynamics:

Suniv = Stotal = Ssystem + Ssurr

• Also recall:

Ssurr = -Hsys/T

• ThenStotal = Ssystem -Hsys/T

Stotal = -TSsystem+ Hsys

Defining G (cont.)

• We then define:

G = -TStotal

• Then:

• OrG = H - TS

G = -TSsys+ Hsys

G = The Gibbs Free Energy w/ Pconst

Relating G to Q

• Recall from Lecture 6:

S = R ln (final/initial)

• For the expansion of a gas

final Volume

Relating G to Q (cont.)

• Given this relationship

S = R ln (Vfinal/Vinitial)

S R ln

nRTPfinal

nRTPinitial

R ln

PinitialPfinal

R ln

PfinalPinitial

Relating G to Q (cont.)

• This equation tells us what the change in entropy will be for a change in concentration away from standard state.

Sfi Sfi R lnPfinalPinitial

Entropy change for processoccurring under standardconditions

Additional term for changein concentration.

(1 atm, 298 K) (P ≠ 1 atm)

Relating G to Q (cont.)

• How does this relate to G?

Grxn H rxn TSrxn

Grxn H orxn TSorxn RT ln

PfinalPinitial

Grxn Grxno RT ln

PfinalPinitial

Relating G to Q (cont.)

Grxn Grxno RT ln

PfinalPinitial

Grxn Grxno RT ln Q

• Generalizing to a multicomponent reaction:

Q

C C reference

coeff

prod .

C C reference

coeff

react .

• Where

Relating G to Q (cont.)

Grxn Grxno RT ln Q

aA bB cC dD K

C c D d

A a B b

K Q (at equilibrium)

K Q (away from equilibrium)

An Example

• Determine Grxn at 298 K for:

C2H4(g) + H2O(l) C2H5OH(l)

where PC2H4 = 0.5 atm (others at standard state)

Grxn Grxno RT ln Q

G°rxn = -6 kJ/mol (from Lecture 9)

An Example (cont.)

C2H4(g) + H2O(l) C2H5OH(l)

Grxn Grxno RT ln Q

Grxn = -6 kJ/mol + (8.314 J/mol.K)(298K)ln(2)

Q 1PC2H4

1atm

1

0.5atm

1atm

2

= -4.3 kJ/mol

G and K

• The Reaction Quotient (Q) corresponds to a situation where the concentrations of reactants and products are not those at equilibrium.

• At equilibrium, we have K.

• What is the relationship between G and K?

G and K (cont.)

• At equilibrium, Grxn = 0

Grxn Grxno RT ln Q

0 K

0 = G°rxn +RTln(K)

G°rxn = -RTln(K)

G and K (cont.)• Let’s look at the interaction between G° and K

G°rxn = -RTln(K)

If G° < 0

then > 1

Products are favored over reactants

reactants

products

Grxn

G and K (cont.)• Let’s look at the interaction between G° and K

G°rxn = -RTln(K)

If G° = 0

then = 1

Products and reactants are equally favored

reactants

products

Grxn = 0

G and K (cont.)• Let’s look at the interaction between G° and K

G°rxn = -RTln(K)

If G° > 0

then < 1

Reactants are favored over products

reactantsGrxn

Temperature Dependence of K

• We now have two definitions for G°

G°rxn = -RTln(K) = H° - TS°

• Rearranging:

y = m x + b

ln(K ) H R

1

T

SR

• Plot of ln(K) vs 1/T is a straight line

T Dependence of K (cont.)

• If we measure K as a function of T, we can determine H° by determining the slope of the line

ln(K ) H R

1

T

SR

slope

intercept

T Dependence of K (cont.)

• Once we know the T dependence of K, we can predict K at another temperature:

ln(K2 ) H R

1

T2

SR

-

ln(K1) H R

1

T1

SR

lnK2

K1

H R

1

T2

1

T1

the van’t Hoff equation.

An Example• For the following reaction:

CO(g) + 2H2(g) CH3OH(l) G° = -29 kJ/mol

What is K at 340 K?

• First, what is Keq when T = 298 K?

G°rxn = -RTln(K) = -29 kJ/mol

ln(K298) = (-29 kJ/mol)

-(8.314 J/mol.K)(298K) = 11.7

K298 = 1.2 x 105

An Example (cont.)• Next, to use the van’t Hoff Eq., we need H°

CO(g) + 2H2(g) CH3OH(l)

Hf°(CO(g)) = -110.5 kJ/mol

Hf°(H2(g)) = 0 Hf°(CH3OH(l)) = -239 kJ/mol

H°rxn = H°f (products) - H° f (reactants)

= H°f(CH3OH(l)) - H°f(CO(g)) = -239 kJ - (-110.5 kJ) = -128.5 kJ

An Example (cont.)• With H°, we’re ready for the van’t Hoff Eq.

lnK2

K1

H R

1

T2

1

T1

K340 = 2.0 x 103

Why is K reduced?

Reaction is Exothermic.

Increase T, Shift Eq. To React.

Keq will then decrease

lnK340

1.2x106

( 128500 J)

(8.314 J /K)

1

340K

1

298K

lnK340

1.2x106

6.4

An Example

• For the following reaction at 298 K:

HBrO(aq) + H2O(l) BrO-(aq) + H3O+(aq)

w/ Ka = 2.3 x 10-9 What is G°rxn?

G°rxn = -RTln(K) = -RTln(2.3 x 10-9)

= 49.3 kJ/mol

An Example (cont.)

• What is Grxn when pH = 5, [BrO-] = 0.1 M, and [HBrO] = 0.2 M ?

HBrO(aq) + H2O(l) BrO-(aq) + H3O+(aq)

Q H 3O

BrO HBrO

10 5 0.1

(0.2)5x10 6

An Example (cont.)• Then:

Grxn Grxno RT ln Q

= 49.3 kJ/mol + (8.314 J/mol.K)(298 K)ln(5 x 10-6)

= 19.1 kJ/mol

Grxn < G°rxn “shifting” reaction towards products

A Descriptive Example• Consider the following gas-phase

equilibrium:

OClO (g) ClOO (g)

• Now,Grxn = G°rxn +RTln(Q)

= -RTln(K) + RTln(Q)

= RTln(Q/K)

A Descriptive Example (cont.)

OClO (g) ClOO (g)

• From inspection,Grxn = RTln(Q/K)

If Q/K < 1 Grxn < 0

If Q/K > 1 Grxn > 0

OClO (g) ClOO (g)

OClO (g) ClOO (g)

If Q/K = 1 Grxn = 0 equilibrium

A Descriptive Example (cont.)• K = 0.005

Grxn = RTln(Q/K)

G°rxn = -RTln(K) = 13.1 kJ/mol

• Let’s start the rxn off with 5 atm of OClO and 0.005 atm of ClOO.

= RTln(.001/.005) = RTln(.2) = -1.6RT < 0

Grxn < 0, reaction is proceeding towards products

Q = PClOO/POClO = .001

A Descriptive Example (cont.)

Grxn = RTln(Q/K)

• Let’s start the rxn off with 5 atm. of ClOO and

0.5 atm of OClO.

= RTln(10/0.005)

= RTln(2000) >> 0

Grxn > 0, reaction is proceeding towards reactants

Q = PClOO/POClO = 10

A Descriptive Example (cont.)

• Where is equilibrium? Back to Chem 142

OClO (g) ClOO (g)Pinit (atm) 5 0.005

Pequil. (atm) 5 - x 0.005 + x

x = 0.02

POClO = 5 - .02 = 4.98 atm PClOO = 0.025 atm

K PClOOPOClO

0.005 x

(5 x).005

A Descriptive Example (cont.)

Grxn = RTln(Q/K)

K

= RTln(1) = 0 equilibrium.

Pictorally:

“Amount” of OClO Present