ThermodynamicsThermodynamicsChapter 19Chapter 19

Liquid benzene

Production of quicklimeProduction of quicklime

Solid benzene

⇅

CaCO3 (s) CaO + CO⇌ 2

Gibbs Energy

For a constant-pressure & constant temperature process:

G = Hsys - TSsysGibbs energy

(G)

G < 0 The reaction is spontaneous in the forward direction

G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction

G = 0 The reaction is at equilibrium

Fig 19.17 Analogy between Potential Energy and Free Energy

Fig 19.18 Free Energy and Equilibrium

aA + bB cC + dD

G°rxn nG° (products)f= mG° (reactants)f-

Standard free-energy of reaction (Gorxn) ≡ free-energy

change for a reaction when it occurs under standard-state conditions.

Standard free energy of formation (G°)

• Free-energy change that occurs

when 1 mole of the compound

is formed from its elements

in their standard states.

f

Fig 19.19 Energy Conversion

What’s “Free” About Gibbs Energy?

• ΔG ≡ the theoretical maximum amount of work that can bedone by the system on the surroundings at constant P and T

• ΔG = − wmax

What’s “free” about the Gibbs energy?What’s “free” about the Gibbs energy?

• “ “Free” does not imply that the energy has no costFree” does not imply that the energy has no cost

• For a constant-temperature process, “free energy”For a constant-temperature process, “free energy”

is the amount available to do workis the amount available to do work

e.g., Human metabolism converts glucose toe.g., Human metabolism converts glucose to

COCO22 and H and H22O with a O with a ΔΔG° = -2880 kJ/molG° = -2880 kJ/mol

This energy represents approx. 688 CalThis energy represents approx. 688 Cal

or about two Snickers bars worth... or about two Snickers bars worth...

Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity

The Haber process for the production of ammonia involves the equilibrium

Assume that ΔH° and ΔS° for this reaction do not change with temperature.

(a)Predict the direction in which ΔG° for this reaction changes with increasing temperature.

(b) Calculate the values ΔG° of for the reaction at 25 °C and 500 °C.

(a) The temperature dependence of ΔG° comes from the entropy term.

• We expect ΔS° for this reaction to be negative because the number of molecules of gas is smaller in the products.

• Because ΔS° is negative, the term –T ΔS° is positive and grows larger with increasing temperature.

• As a result, ΔG° becomes less negative (or more positive) with increasing temperature.

• Thus, the driving force for the production of NH3 becomes smaller with increasing temperature.

G = Hsys - TSsys

Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity

The Haber process for the production of ammonia involves the equilibrium

Assume that ΔH° and ΔS° for this reaction do not change with temperature.

(a)Predict the direction in which ΔG° for this reaction changes with increasing temperature.

(b) Calculate the values ΔG° of for the reaction at 25 °C and 500 °C.

Go = Hsys - TSsys

• The reaction is nonspontaneous at 500 oC

• The reaction is spontaneous at 25 oC

Gibbs Free Energy and Chemical Equilibrium

• We need to distinguish between ΔG and ΔG°

• During the course of a chemical reaction, not all

products and reactants will be in their standard states

• In this case, we use ΔG

• When the system reaches equilibrium, the sign of ΔG°

tells us whether products or reactants are favored

• What is the relationship between ΔG and ΔG°?

Gibbs Free Energy and Chemical Equilibrium

ΔG = ΔG° + RT lnQ

R ≡ gas constant (8.314 J/K•mol)

T ≡ absolute temperature (K)

Q ≡ reaction quotient = [products]o / [reactants]o

At Equilibrium:

ΔG = 0 Q = K

0 = ΔG° + RT lnK

ΔG° = − RT lnK

When not all products and reactants are in their standard states:

G° = - RT lnK

Table 19.5

RToGΔ

eK

or

Calculate ΔG° for the following process at 25 °C:

BaF2 (s) ⇌ Ba2+(aq) + 2 F− (aq); Ksp = 1.7 x 10-6

Example

ΔG = 0 for any equilibrium, so:

ΔG° = − RT ln Ksp

Equilibrium lies to the left

ΔG° = − (8.314 J/mol∙K) (298 K) ln (1.7 x 10-6)

ΔG° = + 32.9 kJ/mol

ΔG° ≈ + 33 kJ/mol

Thermodynamics in living systems

• Many biochemical reactions have a positive ΔGo

• In living systems, these reactions are coupled to a

process with a negative ΔGo (coupled reactions)

• The favorable rxn drives the unfavorable rxn

CC66HH1212OO66 ((ss)) + 6O + 6O22 ((gg) ) 6CO 6CO22 ((gg)) + 6H + 6H22O O ((ll)

Metabolism of glucose in humansMetabolism of glucose in humans

ΔΔG° = -2880 kJ/molG° = -2880 kJ/mol

• Does not occur in a single step as it would in simple combustion

• Enzymes break glucose down step-wise

• Free energy released used to synthesize ATP from ADP:

Fig 19.20 Free Energy and Cell Metabolism

ADP + H3PO4 → ATP + H2O ΔG° = +31 kJ/mol

(Free energy stored)