Chapter 19Chemical Thermodynamics

Entropy, Enthalpy, and Free Energy

Thermodynamic Quantities

• H : Enthalpy, heat energy

• E : Total energy, q + w

• S : Entropy, disorder

• G : Free energy, measure of spontaneity, energy available to do work

• Relationship between thermodynamic quantities:

G = H – TS

• Sign of G

G < 0 Reaction is spontaneous, reaction will proceed in

the forward direction

G < 0 Reaction is not spontaneous, however the reaction

going in the reverse direction is spontaneous

G = 0 The reaction is at equilibrium, nothing will happen

The Meaning of “Spontaneous”

• Spontaneous reaction:– Product favored– Given sufficient time, a combination of

reactants will be converted to products K >> 1

• Non spontaneous reaction:– Reactant favored– Given sufficient time, nothing will happen

K << 1

Problems

a) H2 + I2 < ----- > 2HI Kc = 51

b) N2 + O2 < ----- > 2NO Kc = 1 x 10-30

• Which reaction is spontaneous in the forward direction?

• Which reaction has a negative G0?

What Drives a Reaction to Occur?

• Driving Forces:– The tendencies of concentrated energy and

matter to disperse

• Enthalpy:– Exothermic reactions disperse energy– Energy flows from hot area to cold area

• Entropy:– The tendency of concentrated matter to

disperse

Will a process be spontaneous?

• If energy and matter are both dispersed in a reaction, it is definitely spontaneous

• If only energy or matter is dispersed, then the relative effects of enthalpy and entropy determine spontaneity

• If neither matter nor energy is dispersed, then that process will not be spontaneous and reactants will remain, no matter how long we wait

The Three Laws of Thermodynamics

First Law: The total energy of the universe is a constant

Esystem = -Esurroundings and E = q+w

Second Law: The total entropy of the universe

is always increasing

Third Law: The entropy of a pure perfectly formed crystalline substance at

absolute zero is zero.

Entropy

• Boltzman’s Expression for Entropy: A quantitative measure of matter dispersal or disorder

Ludwig Boltzman (1844 – 1906)S = k log W

S = entropyk = constantW = number of ways atoms or molecules can be arranged

Entropy

• Entropy of a substance is determined by calorimetry

S = q / T

• Absolute entropy of a substance at any temperature can be determined– Absolute entropy at 0 K is 0.– Absolute entropy S0 is the entropy of a pure

substance relative to its entropy at absolute zero

S: Entropy Generalizations

1. Entropies of gases >>> entropy of liquids > entropies of solids

2. Entropies of complex molecules are > than entropies of simpler molecules

3. Entropies of ionic solids become smaller as the attractions between ions become stronger

4. Entropy usually increases when a pure liquid or solid dissolves in a solvent

5. Entropies of liquids comprised of molecules with similar structures are smaller when hydrogen bonding is possible

6. Entropy increases when a dissolved gas escapes from a solution

Problem

• Predict whether S is positive or negative for the following processes

CaCO3(s) --- > CaO(s) + CO2(g)

2CO(g) + O2 --- > 2CO2(g)

Ag+(aq) + Cl-(aq) --- > AgCl(s)

The “naught” Notation and Standard States

S0, H0, G0

Solid: Pure

Liquid: Pure

Gas: 1 atm

Solution: 1 M

Temperature: 25oC

Free Elements

• The absolute entropy S0 of a free element in its standard state is NOT zero

• S0 refers to the entropy increase in warming a pure substance from absolute zero where its entropy is zero, to 25oC

S0 = S0(at 25oC) – S0(at –273oC) = S0 – 0 = S0

Free Elements (continued)

Hf0 of a free element in its standard

state is always equal to zero

Gf0 of a free element in its standard

state is always equal to zero

Second Law of Thermodynamics

• The total entropy of the universe is constantly increasing

• Whenever anything happens, matter, energy or both become more dispersed or disordered

• For any spontaneous process S > 0

Calculating Entropy Changes for a reaction

S0reaction

= n S0products

– n S0reactants

where S0 is the absolute entropy of each compound

S0 is the entropy change that occurs when reactants in their standard states (pure, 1atm or 1M) are converted completely to products in their standard states.

• Calculate S0 for the Haber Process

N2(g) + 3H2(g) --- > 2NH3(g)

S0reaction = 2mol (195.2 J/mol K)

-1 mol (191.5 J/mol K)

-3 (130.6 J/mol K)

= -198.4 J/mol K

Enthalpies and Entropies of Formation

The enthalpy or entropy change associated with forming a compound from its free elements.

Hf0 = Hf

0(compound)

– n Hf0

(elements)

= Hf0

(compound) – 0

Look up Hf0

in a Table

Sf0 = S0

(compound) – n S0

(elements)

Look up S0(compound) and S0

(element) for each element in a Table and substract

S0reaction vs. Sf

0

S0 is the entropy of the reactionthe entropy of formation

S0 is the weighted sum of all the absolute entropies of the product minus the weighted sum of all the absolute entropies of the reactants

Look up all the S0 values for the elements or compounds involved in the reaction and calculate S0 using

S0 = n S0products

– n S0reactants

Sf0 is the entropy change for the reaction

which forms the compound from its elements in their standard states

Look up all the S0 values and calculate Sf0

using

Sf0 = S0

(compound) – n S0

(elements)

Problem

• Calculate the entropy change for the following reaction and calculate DSf0 for CH3OH (l)

• CO(g) + 2H2(g) --- > CH3OH (l)

Solution

S0reaction = S0 [CH3OH(l)] – S0[CO(g)]

-2S0 [H2(g)]

= 1mol 126.8 J/mol K – 1 mol 197.6 J/mol K

-2 mol 130.7 J/mol K = -332.2 J/K

• Formation Reaction

C(s) + 2H2(g) + ½ O2(g) --- > CH3OH (l)

Sf0 = S0 [CH3OH (l)] – S0[C(s)] – 2S0[H2(g)]

- ½ S0 [O2(g)]

= 1 mol (126.8 J/ mol K) – 5.69 J/ mol K

- 2 mol (130.58 J/mol K) – ½ (205 J/mol K)

= -242.6 J/ mol K

Notes:

• Absolute entropies are always positive

• The units for S are Joules / mol K NOT Kilojoules / mol K

• The units for G and H are in kilojoules / mol

Gibbs Free Energy

• A measure of the amount of energy involved in a reaction which is available (free) to do work

G is a quantity which tell us– Whether a reaction is spontaneous or not– Relates enthalpy and entropy– Units: kilojoules / mol

Calculating G

Greaction = Hreaction – TSreaction

Greaction = nGf products – nGf reactants

Greaction = G0reaction + RT ln Q

The sign of G

G < 0Reaction is spontaneous as written

G = 0Reaction is at equilibrium

G > 0 Reverse reaction is spontaneousWork must be done to make the reaction occur

Standard Free Energy G0

G0reaction is the free energy change when

a mixture of only reactants all in their standard states is completely converted to a mixture of all products in their standard states.

The sign of G0 and K

G0 < 0 Reaction is spontaneous as written, product favored at equilibrium K > 1

G0 = 0Very rare, at equilibrium [C]c[D]d = [A]a[B]b

G0 > 0Not spontaneous, reactant favored at equilibrium K < 1

G0 and K

At equilibrium

Greaction = 0Therefore since

Greaction = G0reaction + RT ln Q

At equilibrium

0 = G0reaction + RT ln K

G0reaction = - RT ln K

And

K = e- G0/ RT