1 Week 14 CHEM 1310 - Sections L and M 1 Ch. 10: Spontaneity & Entropy Spontaneous chemical reactions occur by themselves, given time, without outside intervention. 2 C 8 H 18 (l) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O(l) Octane fuel Why doesn’t the reverse reaction occur spontaneously? Equilibrium is far to the right for this reaction. Week 14 CHEM 1310 - Sections L and M 2 Ch. 10: Spontaneity & Entropy Spontaneous chemical reactions occur by themselves, given time, without outside intervention. Air Bromine Vapor Spontaneous Change More intuitive illustration… Week 14 CHEM 1310 - Sections L and M 3 Some chemical reactions are not spontaneous. Why? Air Bromine Vapor Mixture of Air and Bromine Vapor X Spontaneity
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Week 14 CHEM 1310 - Sections L and M 1
Ch. 10: Spontaneity & Entropy
Spontaneous chemical reactions occur bythemselves, given time, without outside
intervention.
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(l)
Octanefuel
Why doesn’t the reverse reaction occur spontaneously? Equilibrium is far to the right for this reaction.
Week 14 CHEM 1310 - Sections L and M 2
Ch. 10: Spontaneity & Entropy
Spontaneous chemical reactions occur bythemselves, given time, without outside
intervention.
Air Bromine Vapor Spontaneous Change
More intuitive illustration…
Week 14 CHEM 1310 - Sections L and M 3
Some chemical reactions are not spontaneous.Why?
Air Bromine VaporMixture of Air and
Bromine Vapor X
Spontaneity
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Week 14 CHEM 1310 - Sections L and M 4
How can one predict whether or not achemical reaction will be spontaneous?
THEORY: Consider 1st Law of Thermodynamics
Spontaneity
Energy can neither be created nor destroyed
Addresses questions such as:• How much energy is involved in a change?• Does energy flow into or out of a system?• What form does the energy finally assume?
Not spontaneity…
Week 14 CHEM 1310 - Sections L and M 5
How can one predict whether or not achemical reaction will be spontaneous?
THEORY: Consider Enthalpy
Δ H > 0 Endothermic Non-spontaneousΔ H < 0 Exothermic Spontaneous
NO!
NH4Cl (s) NH4+ (aq) + Cl- (aq) Δ Hro = +14.8 kJ/molENDOTHERMIC but SPONTANEOUS!!!
Spontaneity
Week 14 CHEM 1310 - Sections L and M 6
Many phase changes also disobey the theory ofenthalpy as an indicator of spontaneity…
H2O (g) H2O (l) Δ Hro = -40.66 kJ/molEXOTHERMIC
SPONTANEOUS at T < 100°C
H2O (l) H2O (g) Δ Hro = +40.66 kJ/molENDOTHERMIC
SPONTANEOUS at T > 100°C
Spontaneity
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Week 14 CHEM 1310 - Sections L and M 7
S = kB ln ΩkB = Boltzmann constantΩ = # microstates available
RN0
= 8.314472 J K-1 mol-1
6.022142 x 1023 mol-1kB =
kB = 1.308630 x 10-23 J/K
- a measure of the disorder of a system
Entropy
Week 14 CHEM 1310 - Sections L and M 8
The entropy of a substance increases withincreasing temperature in part because the
number of degrees of freedom increases withtemperature.
Translation RotationVibration
Degrees of Freedom
Week 14 CHEM 1310 - Sections L and M 9
Entropy & Spontaneous Change
S = kB ln Ω Why do increases in entropy, S,correspond to spontaneous changes?Probability
2 Molecules: 1 in 4 chance
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Week 14 CHEM 1310 - Sections L and M 10
Entropy & Spontaneous Change
S = kB ln Ω Why do increases in entropy, S,correspond to spontaneous changes?Probability
4 Molecules
2 possibilities
8 possibilities
6 possibilities
1 in 16chance
Week 14 CHEM 1310 - Sections L and M 11
Entropy & Spontaneous Change
Week 14 CHEM 1310 - Sections L and M 12
Spontaneous change in an isolated systemalways goes to increase the # of microstates inthat system entropy increases.
Spontaneous change in a non-isolated systemoften leads to a decrease in entropy.
Determining the number of microstates is not practical!
Entropy & Spontaneous Change
S = kB ln ΩHow can we describe entropy in more useful terms?
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Week 14 CHEM 1310 - Sections L and M 13
Entropy in Terms of Volume
S = kB ln Ω ∆ S = S2 - S1
∆ S = kB ln Ω2 - kB ln Ω1 = kB ln (Ω2/Ω1)
Consider one gas molecule at V1 and 2V1…
∆ S = kB ln 2Ω1 - kB ln Ω1 = kB ln 2
Week 14 CHEM 1310 - Sections L and M 14
Entropy in Terms of Volume
S = kB ln Ω ∆ S = S2 - S1
∆ S = kB ln Ω2 - kB ln Ω1 = kB ln (Ω2/Ω1)
Consider a mole of gas molecules at V1 and 2V1…
∆ S = kB ln N0Ω1 - kB ln Ω1 = kB ln 2 N0
∆ S = N0 kB ln (V2/V1)…
Week 14 CHEM 1310 - Sections L and M 15
Entropy in Terms of Volume
Consider a mole of gas molecules at V1 and V2…
∆ S = kB ln N0Ω1 - kB ln Ω1 = kB ln 2 N0
∆ S = N0 kB ln (V2/V1)…
∆ S = N0 (R/N0) ln (V2/V1)∆ S = R ln (V2/V1)∆ S = n R ln (V2/V1)
Recall:kB = R/N0
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Week 14 CHEM 1310 - Sections L and M 16
Entropy in Terms of Work
2
1
2
1
ln
ext
V
extV
rev rev
w P V
Work P dV
Vw nRT q
V
= - D
=
Ê ˆ= - = -Á ˜
Ë ¯
Ú Pexternal = nRTV
PV = nRT
This expression is very similar to that for ∆S…
∆ S = n R ln (V2/V1) ∆ S = nRT ln (V2/V1)T∆ S = qrev / T
Week 14 CHEM 1310 - Sections L and M 17
Entropy Expressions
∆ S = qrev / TS = kB ln Ω
kB = 1.308630 x 10-23 J/K
Notice that the units for S are J/K!
As T increases, S increases.So, why the inverse relationship?
T increasing
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