Chapter 19: Chemical Thermodynamics Spontaneous processes… …happen without outside help …are...
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Transcript of Chapter 19: Chemical Thermodynamics Spontaneous processes… …happen without outside help …are...
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Spontaneous processes…
…happen without outside help
…are “product favored”
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Spontaneous processes
at 25oC
H2O (s) → H2O (l)
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Spontaneous processes
CO2 (s) → CO2 (g)
at 25oC
‘dry ice’
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Spontaneous processes
4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)
at 25oC
rust
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Spontaneous processes
…occur in a definite direction: towards the formation of product
H2O (s) → H2O (l) CO2 (s) → CO2 (g)
4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)
at 25oC
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
The direction of a spontaneous processes may depend on temperature
at -10oC
H2O (l) → H2O (s)
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
H2O (l) → H2O (s)
CO2 (s) → CO2 (g)
2 Fe2O3 (s) → 4 Fe (s) + 3 O2 (g)
at 25oC
• At a given temperature and pressure, processes arespontaneous only in one direction
• If a processes is spontaneous in one direction it is non-spontaneous in the other direction
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Which reactions are spontaneous?
• many spontaneous reactions are exothermic (H < 0)
… but not all!
• Some reactions are endothermic (H > 0) and still spontaneous
NH4NO3 (s) → NH4+ (aq) + NO3
- (aq)
H > 0
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Which reactions are spontaneous?
Reactions proceed towards a more probable state
In general, the more probable state is associated
with more disorder
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Entropy can be thought of as a measure of disorder
Ludwig Boltzmann (1844-1906)
S = k log W
k = 1.38 x 10-23 J / K
W = Wahrscheinlichkeit (probability)
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
The change in entropy for any process is:
S = Sfinal - Sinitial
What is the sign of S for the following processes at 25oC ?
H2O (s) → H2O (l)
CO2 (g) → CO2 (s)
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Second Law of Thermodynamics
For any spontaneous process, the entropy of the universe increases
Souniverse = So
system + Sosurroundings > 0
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Ssolid < Sliquid < Sgas
gasliquidsolid
Of all phase states, gases have the highest entropy
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Larger Molecules generally have a larger entropy
Ssmall < Smedium < Slarge
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Often, dissolving a solid or liquid will increase the entropy
dissolves
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Dissolving a gas in a liquid decreases the entropy
dissolves
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
The entropy of a substance increases with temperature
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
S = k ln W
• W is a measure for how many different ways there areof arranging a molecule or an ensemble of molecules (the system)
• W reflects the number of microstates of a system
• The larger the possible number of microstates the higher the entropy
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Ssolid < Sgas
gassolid
Of all phase states, gases have the highest entropy
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Ssolid < Sgas
gassolid
Of all phase states, gases have the highest entropy
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Larger Molecules generally have a larger entropy
Ssmall < Slarge
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Often, dissolving a solid or liquid will increase the entropy
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Often, dissolving a solid or liquid will increase the entropy
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Dissolving a gas in a liquid decreases the entropy
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
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Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
What is the sign of S for the following reactions?
FeCl2 (s) + H2 (g) → Fe (s) + 2 HCl (g)
Ba(OH)2 (s) → BaO (s) + H2O (g)
2 SO2 (g) + O2 (g) → 2 SO3 (g)
Ag+ (aq) + Cl- (aq) → AgCl (s)
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
For each of the following pairs, which substance has a highermolar entropy at 25oC ?
HCl (l) HCl (s)
C2H2 (g) C2H6 (g)
Li (s) Cs (s)
Pb2+ (aq) Pb (s)
O2 (g) O2 (aq) HCl (l) HBr (l)
CH3OH (l) CH3OH (aq)N2 (l) N2 (g)
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Sorxn = Σ n So(products) – Σ m So(reactants)
If you know the standard molar entropies of reactants andproducts, you can calculate
S for a reaction:
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
substance So (J/K-mol)
H2 130.6
C2H4 (g) 219.4
C2H6 (g) 229.5
What is So for the following reaction?
C2H4 (g) + H2 (g) → C2H6 (g)
Sorxn =
Do you expect S to be positive or negative?
So for elements are NOT zero
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Which reactions are spontaneous?
We need to know the magnitudes of both S and H !
Go = Gibbs free energy…
… is a measure of the amount of “useful work” a system can perform
Go = Ho - TSo
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Go = Ho - TSo
A reaction is spontaneous if Go is negative
J. Willard Gibbs(1839 – 1903)
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g)
The reaction of sodium metal with water:
Is the reaction spontaneous?
What is the sign of Go?
What is the sign of o?
What is the sign of So?
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Go < 0 => reaction is spontaneous(“product favored”)
Go > 0 => reaction is non-spontaneous
Go = 0 => reaction is at equilibrium
“exergonic”
“endergonic”
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Go = Ho - TSo
Ho So
+ -
Go
- +
- -
+ +
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Go = Ho - TSo
H2O (s) → H2O (l) Go < 0
at 25oC (298K):
spontaneous:
Ho > 0
So > 0
So > 0 > Ho => Go < 0
but: if T becomes very small:
at 298K:
So > 0 < Ho => Go > 0
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
A diamond left behind in a burning house reacts according to
2 C (s) + O2 (g) → 2 CO (g)
What is the value of Go for the reaction at 298K ?
substance Hfo (kJ/mol) Gf
o (kJ/mol) So (J/K-mol)
O2 0 0 205.0
C (diamond, s) 1.88 2.84 2.43
C (graphite, s) 0 0 5.69
CO2 (g) -393.5 -394.4 213.6
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
There are two possible ways to calculate Go:
I) Go = Ho - TSo
II) Go = Σ n Gfo (products) – Σ m Gf
o (reactants)
calculate Go from Ho and So :
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
I) calculate Go from Ho and So : Go = Ho - TSo
substance Hfo (kJ/mol) Gf
o (kJ/mol) So (J/K-mol)
O2 0 0 205.0
C (diamond, s) 1.88 2.84 2.43
C (graphite, s) 0 0 5.69
CO (g) -110.5 -137.2 197.9
2 C (s) + O2 (g) → 2 CO (g)
Go = - 280.2 kJ
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
substance Hfo (kJ/mol) Gf
o (kJ/mol) So (J/K-mol)
O2 0 0 205.0
C (diamond, s) 1.88 2.84 2.43
C (graphite, s) 0 0 5.69
CO (g) -110.5 -137.2 197.9
2 C (s) + O2 (g) → 2 CO (g)
II) Go = Σ n Gfo (products) – Σ m Gf
o (reactants)
Go = -280.1 kJ
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Consider the following reaction:
2 H2 (g) + O2 (g) → 2 H2O (l)
S = -326.3 J/K, H = -571.7 kJ , and G = -475.3 kJ at 25oC. At what temperature does the reaction become spontaneous in the
opposite direction?
G = H - TS
The reaction “switches” direction if H = TS , i.e. if G = 0
H < 0 S < 0 G < 0 at 298 K
G > 0 at high T
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
0 = H - TS
TS = H
T =ΔSΔH
x 1 kJ / 1000JT =
T = 1752.1 K
-571.7 kJ
-326.3 J/K
At temperatures greater than 1752.1 K liquid water will spontaneously decompose into H2 and O2 !!
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
At the normal melting point, the Gibbs free energies ofthe solid and liquid phase of any substance are equal:
H2O (s) → H2O (l)at 0oC
Go = 0
H2O (l) → H2O (g)at 100oC
Go = 0
At the normal boiling point, the Gibbs free energies ofthe liquid and gas phase of any substance are equal:
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
At a phase change: Go = 0
Go = Ho - TSo
0 = Ho - TSo
if we know Hofus (or Ho
vap) , we can
calculate
So at the melting (or boiling point):
Ho = TSo
So = Ho
T
Chapter 19: Chemical ThermodynamicsChapter 19: Chemical Thermodynamics
Sofus
= Hofus
T
The entropy of melting for H2O
= 6.02 kJ/mol
273.15 K
= 22.0 J/mol-K
Sovap
= Hovap
T
The entropy of vaporization for H2O
= 40.7 kJ/mol
373.15 K
= 109 J/mol-K
units!melting temp
boiling temp