Chemical Thermodynamics

BLB 12th Chapter 19

Chemical Reactions

1. Will the reaction occur, i.e. is it spontaneous? Ch. 5, 19

2. How fast will the reaction occur? Ch. 14

3. How far will the reaction proceed? Ch. 15

Review terms

energy heat work pathway state function system surroundings

Review terms, cont.

exothermic endothermic enthalpy enthalpy change standard state std. enthalpy of formation 1st Law of

Thermodynamics

19.1 Spontaneous Processes Spontaneous – proceeds on its own without

any outside assistance product-favored; K > 1 not necessarily fast the direction a process will take if left alone and

given enough time. Nonspontaneous

opposite direction of spontaneous reactant-favored; K < 1 not necessarily slow

Spontaneity and Energy

Examples of spontaneous systems: Brick falling Ball rolling downhill Hot objects cooling Combustion reactions

Are all spontaneous processes accompanied by a loss of heat, that is, exothermic?

Spontaneity is temperature-dependent.

Reversible & Irreversible Systems

Reversible – a change in a system for which the system can restored by exactly reversing the change – a system at equilibriumex. melting ice at 0°C

Irreversible – a process that cannot be reversed to restore the system and surroundings to their original states – a spontaneous processex. melting ice at 25°C

See p. 788-790 (last paragraph of section)

19.2 Entropy and the 2nd Law of Thermodynamics

Entropy, S – measure of randomness State function Temperature-dependent

A random (or dispersed) system is favored due to probability.

“Entropy Is Simple – If We Avoid the Briar Patches”

Frank Lambert, Occidental College, ret.

http://entropysimple.oxy.edu

Entropy Change

ΔS = Sfinal − Sinitial (a state function)

(isothermal) as for phase changes.

ΔS > 0 is favorable

T) (constant T

qS rev

Calculating ΔS for Changes of State

)(

:1

KinTT

H

T

qS

moleFor

Kmol

Jor

K

Jofunits

T

qS

syssyssys

Problem 26 The freezing point of Ga is 29.8°C and the enthalpy of fusion is 5.59 kJ/mol.

a. Is ΔS + or − for Ga(l) → Ga(s) at the freezing point?

b. Calculate the value of ΔS when 60.0 g of Ga(l) solidifies.

System & Surroundings

Dividing the universe:

1. System – dispersal of matter by reaction: reactants → products

2. Surroundings – dispersal of energy as heat

2nd Law of Thermodynamics

The entropy of the universe increases for any spontaneous process. ΔSuniv > 0

ΔSuniv = ΔSsys + ΔSsurr

For a “reversible” process: ΔSuniv = 0. For an irreversible process:

Net entropy increase ►spontaneous Net entropy decrease ► nonspontaneous

19.3 The Molecular Interpretation of Entropy

Molecules have degrees of freedom based upon their motion Translational Vibrational Rotational

Motion of water (Figure 19.8, p. 796) Lowering the temperature decreases

the entropy.

Boltzman & Microstates S = k ln W

(W = # of microstates) If # microstates ↑,

then entropy ↑. Increasing volume,

temperature, # of molecules increases the # of microstates.

Examples of systems that have increased entropy

Entropy increases for: Changes of state: solid → liquid → gas (T) Expansion of a gas (V) Dissolution: solid → solution (V) Production of more moles in a chemical

reaction (# of particles) Ionic solids: lower ionic charge

S° (J/mol·K)

Na2CO3 136

MgCO3 66

Changes of State

H2O state S° (J/mol·K)

l 69.91

g 188.83

# microstates ↑, system entropy ↑

Dissolution

Expansion of a Gas

2 NO(g) + O2(g) → 2 NO2(g)

S° + or −?

Problem 22 TNT (trinitrotoluene) Detonation

4 C3H5N3O9(l) → 6 N2(g) + 12 CO2(g) + 10 H2O(g) + O2(g)

a) Spontaneous?

b) Sign of q?

c) Can the sign of w be determined?

d) Can the sign of ΔE be determined?

3rd Law of Thermodynamics

The entropy, S, of a pure crystalline substance at absolute zero (0 K) is zero.

19.4 Entropy Changes in Chemical Reactions

Standard molar entropy values, S° (J/mol·K): increase in value as temperature increases

from 0 K have been determined for common

substances (Appendix C, pp. 1059-1061) increase with molar mass increase with # of atoms in molecule

Calculating ΔS°sys

ΔS°sys = ∑nS°(products) - ∑mS°(reactants)

(where n and m are coefficients in the chemical equation)

2 NO(g) + O2(g) → 2 NO2(g)

S° =

Problem 54(c)

Problem

Entropy Changes in the Surroundings

T

H

T

qS syssyssurr

Heat flow affects surroundings. As T increases, ΔH becomes less important. As T decreases, ΔH becomes more important.

Calculating ΔS°univ

ΔSuniv = ΔSsys + ΔSsurr

by obtaining ΔSsys and ΔSsurr

If ΔSuniv > 0, the reaction is spontaneous.

But there is a better way – one in which only the system is involved.

19.5 Gibbs Free Energy

The spontaneity of a reaction involves both enthalpy (energy) and entropy (matter).

Gibbs Free Energy, ΔG makes use of ΔHsys and ΔSsys to predict spontaneity.

ΔGsys represents the total energy change for a system.

G = H – TS or ΔG = ΔH – TΔS or, under standard conditions:

ΔG° = ΔH° – TΔS°

syssyssys

syssys

G

univ

syssysuniv

STHG

STHSTT

HSS

sys

)(

Gibbs Free Energy

If: ΔG < 0, forward reaction is spontaneous ΔG = 0, reaction is at equilibrium ΔG > 0, forward reaction is nonspontaneous

In any spontaneous process at constant temperature and pressure, the free energy always decreases.

ΔG is a state function. ΔGf° of elements in their standard state is

zero.

Calculating ΔG°sys

ΔG°sys = ΔH°sys − TΔS°sys

or

ΔG°sys = ∑nΔG°f (products) - ∑mΔG°f (reactants)

(where n and m are coefficients in the chemical equation)

19.6 Free Energy and Temperature

ΔH ΔS −TΔS ΔG Reaction

− + − always −spontaneous at all TK > 1

+ − + always +nonspontaneous at all TK < 1

− − +− @ low T+ @high T

spontaneous at low Tnonspontaneous at high T

+ + −+ @ low T− @high T

nonspontaneous at low Tspontaneous at high T

Problem 60(b) & 83

Driving force of a reaction

For a reaction where ΔG < 0: Enthalpy-driven – if ΔH < 0 and ΔS < 0; at

low temp. Entropy-driven – if ΔH > 0 and ΔS > 0; at

high temp. “cross-over point” is where ΔG = 0

S

HTSTHSTH

0

Problem

19.7 Free Energy and K

If conditions are non-standard:

ΔG = ΔG° + RT lnQ R = 8.3145 J/mol·K If at equilibrium:

ΔG = ΔG° + RT lnQ = 0

ΔG° = −RT lnK

Problem

Problem 81