Entropy and the Driving Force

• Each substance has its own entropy value. This is an absolute scale, because a perfectly ordered substance at 0 K has an entropy of 0.

• Compare this to enthalpy, where only changes, ΔH, can be measured.

Some entropy values

Substance (1 mole) So J/mol.K

C (diamond) 2.44

B (s) 5.86

Hg (l) 76.0

O2 (g) 205

Temperature effects on entropy

• S of water • (J/mol.K):

• Increasing temperature increases entropy because random motion of molecules has increased

0o(solid)

43.2

100o(liquid)

86.8

0o(liquid)

65.2

100o(gas)

196.9

Why isn’t everything a gas?

• Gases have more entropy than the equivalent solids, but energy factors – such as intermolecular forces – can make it favorable to exist as a solid

• Other things being equal, reactions that produce gases are favored

Entropy change of reactions

• ∆S can be calculated just like ΔH:

• ∆So = ΣSo(prod) - ΣSo(react)

• For: ½N2(g) + O2 (g) → NO2(g)

• So: 191.6 205.1 240.1 J/mol.K

• ΔSo = So(NO2) – [½So(N2) + So(O2)]

• = 240.1 - [½(191.6) + 205.1)]

• = -60.8 J J/mol.K

½N2(g) + O2 (g) → NO2(g)

• ΔSo = -60.8 J/mol.K

• Entropy decreases in this process, consistent with the fewer moles of gas in the products than in the reactants

Entropy, enthalpy and spontaneity:

“The Universe”

Surroundings

SystemΔH

An increase in temperature also causes an increase in entropy

The “Second Law” of Thermodynamics

• In a spontaneous process, the total entropy change of the universe must be positive (i.e., ∆S > 0)

• The S of the system can decrease, but only if S of the surroundings increases to a greater extent.

• Exothermic reactions always raise the entropy of the surroundings, which is why they are intrinsically more favorable.

The four cases:

• ΔH < 0; ΔSsys > 0: reaction is spontaneous

• ΔH < 0; ΔSsys < 0: depends on values

• ΔH > 0; ΔSsys > 0: depends on values

• ΔH > 0; ΔSsys < 0: reaction not spontaneous

• There ought to be just one function that tells us whether the reaction is spontaneous [“will tend to occur”] or not!

A quick derivation

• ΔSsurroundings = - ΔH/T• Negative sign, because if reaction is

exothermic, ΔH is negative, but if it makes the temperature of the surroundings increase, the entropy of the surroundings should increase (positive ΔS)

• 1/T because effect of increased random molecular motion is more pronounced at lower temperatures than higher

A quick derivation

S S S

SH

TT

T S T S H

define T S G

then G H T S

to tal s y s tem surround ings

sy s tem

to tal s y s tem

to tal

. . .

:

G H T S

G

Reactants

Products

For a spontaneous process, ΔG is always negative

F ind G o for: 1

2N 2 + O 2 N O 2 at 25 o C

W e found S o = - 60 .8 J / m ol K

H = H fo (prod) - H f

o (react)

= H fo (N O 2 ) - [

1

2H f

o (N 2 ) H fo (O 2 )]

(from tab les): 33 .2 k J / m ol 0

(recall H fo for elem ents is 0 )

So G o = 33 .2 k J / m ol - (298 K )(-60 .8 J / m ol K )

A fter converting - 60 .8 J to - 0 .0608 k J,

G o = 51 .4 k J

o

[ ( ) ]1

20

/ m ol

G = 51. 4 k J / m o lo

• The positive sign indicates that this reaction is not spontaneous at 25oC

• The o indicates standard conditions (P = 1 atm; concentrations = 1 M)

• ΔGo = 33.2 kJ/mol – T(-60.8 J/mol.K)

• Because ΔH is positive and ΔS is negative, this reaction can never be spontaneous.

Another way to calculate ΔGo

• ΔGfo values are also tabulated

• ΔGfo = Gibbs free energy change when

one mole of a substance is formed from its elements in their standard states

• ΔGo = ΣΔGfo(prod) – ΣΔGf

o(reactants)

So is ΔGo related to Keq?And what can ΔG tell us about systems

not at equilibrium?

• Consider: -ln(K/Q)

• Recall: ln(1) = 0

• ln (>1) = positive number

• ln (<1) = negative number

• If Q < K, reaction goes to right, -ln(K/Q) is -

• If Q> K, reaction goes to left, -ln(K/Q) is +

If Q < K, reaction goes to right, -ln(K/Q) is -If Q> K, reaction goes to left, -ln(K/Q) is +

• So we can say ∆G α -ln(K/Q)

• Specifically: ∆G = -RTln(K/Q)

• At standard conditions, Q = 1 (!) . . . so . . .

• ∆Go = -RT lnK a marvelous equation!

• R = 8.31 J/mol.K

• This allows us to calculate an equilibrium constant from basic thermodynamic information