CHEM 163

Chapter 20

Spring 2009

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3-minute exercise

Is each of the following a spontaneous change?• Water evaporates from a puddle• A small amount of sugar dissolves in hot

tea• Methane burns in air• A hamburger becomes uncooked

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Thermodynamics

First Law: Law of Conservation of Energy

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Second Law: Systems change towards more disorder

Internal E of a system

wqE

heat

work

surrsys EE

Limitation:

Explains change, but not direction

Spontaneous Change

• Change that occurs without continuous E input

• Change can only be spontaneous in one direction, under a given set of conditions

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Enthalpy (∆H):• heat gained or lost at constant P• Sign of ∆H

• Exothermic or endothermic• No information about spontaneity

Entropy (∆S):• Freedom of particle motion (dispersed E of

motion)

Energy Levels• Each atom or molecule has quantized E

levels– Electronic– Kinetic

• Vibrational, rotational, translational

• Microstate: combined E at any given point– each microstate is equally possible (equal E)

for a given set of conditions

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WkS lnentrop

y

Number of microstates

Boltzmann constant= R/NA = 1.38 x 10-23

J/K

(J/K)

Entropy Change: Microstates

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ifsys SSS

AN

AN

R2ln

i

f

W

Wk ln

2lnR

for 1 mol

AN2

Entropy Change: Heat Changes

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T

qS rev

sys

• Remove 1 grain of sand• Gas does work on piston

• absorbs heat to maintain E• Works for tiny changes (totally reversible)

Always Increasing EntropyAll real processes occur spontaneously in

direction that increases the entropy of the universe.

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surrsysuniv SSS 0

Perfect crystal at T = 0 K has S = 0• 1 microstate 1lnkS 0

Standard Molar Entropy (S°) • S increase from 0 to standard state

• 1 atm (gases)• 1 M (solutions)• Pure substance, most stable form

(liquids/solids)

What affects S°?

1. Temperature change↑ T ↑ S

2. Phase changeabsorb heat ↑ S

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As # of microstates (or kinetic E) increases, S increases

ofus

ovap SS

gl ls

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4. Atomic Size• heavier atoms• allotropes

5. Molecular Complexitymore complex

closer E levels

more microstateso

graphiteodiamond SS

more types of movement more

microstatesOnly applies to molecules in same physical state

3. DissolutionIons: increased S, except small, highly charged ionsMolecules (solid or liquid): ∆S ≈ 0 Gases: decreased S

3-minute practice

What is the sign of ∆Ssys?

• A pond freezes in winter

• Atmospheric CO2 dissolves in the ocean

• 2 K (s) + F2 (g) 2KF (s)

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Standard Entropy of Reaction Increasing disorder:

orxnS

0 orxnS

Predict ∆Srxn:

orxt

oprod

orxn nSmSS

Decreasing disorder:

0 orxnS

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

orxnS )2(

3

oNHS )31(

22

oH

oN SS

Change in # moles of gas

Calculate ∆Srxn:

∆Suniverse

Decrease in ∆Ssys only if greater increase ∆Ssurr

System acts as heat sink or drain1.Exothermic

2.Endothermic

0sysq 0surrq 0 surrS

0sysq 0surrq 0 surrS

T

qS sys

surr T

HS sys

surr

at constant P

Measure ∆Hsys to determine ∆Ssurr

Spontaneous at 298 K?N2 (g) + H2 (g) NH3 (g)3 21. Balance equation!

2. Calculate ∆Ssys

3. Calculate ∆Hsys

4. Calculate ∆Ssurr

5. Calculate ∆Suniv

osysS )31(

22

oH

oN SS )2(

3 oNHS

1932 )6.13035.191( KJ197

osysH 0 o

fNHH

32 0 kJ8.91

T

HS

osys

surr

K

kJ

298

8.91 KJ308

surrosysuniv SSS KJ111

Entropy at Equilibrium

Approaching equilibrium: At equilibrium:

No net change 0 univS

? sysS surrsys SS

0 univS

revfwd SS

5-minute Practice

Calculate ∆Ssys for the combustion reaction of ammonia (producing nitrogen dioxide and water vapor).

Gibbs Free Energy• Measure of spontaneity• Combines enthalpy and entropy

surrosysuniv SSS

T

HSS syso

sysuniv

sysosysuniv HSTST

osyssyssys STHG

Spontaneous if…∆Suniv > 0

univsys STG

T∆Suniv > 0

-T∆Suniv < 0 ∆Gsys < 0

∆G0G Spontaneous process

0G Nonspontaneous process

0G Process at equilibrium

Standard Free Energy Change osys

osys

osys STHG

Standard Free Energy of Formation :

of

of

orxn rxtprod

nGmGG

ofG

E change when 1 mol of compound is made from its elements in standard states

Element in standard state: 0 ofG

• Nonspontaneous process:• Process may occur if work is done to the system• How much work is needed?

∆G and work (constant T & P)

• Spontaneous process:• ∆G = maximum useful work done by the

system

w = ∆G

• wmax only if process is totally reversible

• Actually does less w < wmax

• Extra E lost as heat

Useful Work• Excludes work done by or on

atmosphere• Some free energy is always lost to heat

∆Hsys?

)(O2

25)(HC 2188 gl )( OH9)(CO 22 gg

∆Ssys?

∆Gsys?

< 0

> 0

< 0

∆Gsys = wmax that can be done by system

∆Gsys > w actually done by system

In some multistep reactions, ∆G from one reaction can cause an otherwise nonspontaneous reaction to occur. “coupling of reactions”

What about T?

• Typically ∆H > T∆S• For ∆G to be negative, need ∆H to be…

• What about at high T?

osys

osys

osys STHG

negative

• T∆S term can dominate• Sign of ∆S becomes important

1. ∆H < 0 & ∆S > 0

2. ∆H > 0 & ∆S < 0

3. ∆H > 0 & ∆S > 0

4. ∆H < 0 & ∆S < 0

4 situations:

∆G > 0 at low T; ∆G < 0 at high T

∆G < 0 at low T; ∆G > 0 at high T

∆G < 0

∆G > 0

High T v. Low T?

Spontaneous “limit” at what temperature?

STHG 0

STH

S

HT

Equilibria and ∆G

• If Q < K, reaction…

• If Q > K, reaction…

• If Q = K, at equilibrium

→

←

∆G < 0

∆G > 0

∆G = 0

Q/K

Q/K

Q/K

< 1

> 1

= 1

0ln KQ

0ln KQ

0ln KQ

ln ln

K

QRTG ln

proportional

KRTQRTG lnln

KRTGo ln

Make Q standard state (all values = 1)

0

oGQRTG ln

Homework due MONDAY, May 11th

Chap 20: #19, 26, 30, 41, 50, 52, 58, 70, 78, 106

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