Transcript Review - University of Illinois at Urbana–Champaign

Review
Ssystem + Ssurroundings = Suniverse > 0
for spontaneous processes
Ssurroundings = - Hsys
energetic disorder
T
Ssystem
=
qr
T
positional disorder
-
Ssystem + Ssurroundings = Suniverse > 0
for spontaneous processes
at 298 K
H2O(s)  H2O(l)
Ssurroundings = - H
T
Ssurr < 0
spontaneous
H
a) > 0
b) <
Ssurr does not contribute to spontaneity
-
+
Ssystem + Ssurroundings = Suniverse
for spontaneous processes
at 298 K
H2O(s)  H2O(l)
solid  liquid
spontaneous
increasing disorder
S > 0
S
positional disorder
S does contribute to spontaneity
-
+
Ssystem + Ssurroundings = Suniverse
H2O(s)  H2O(l)
S
solid  liquid
positive
- H / T
endothermic
negative
At high T a) spontaneous
Suniv > 0
b) non-spontaneous
At low T non-spontaneous Suniv < 0
+
Suniverse= Ssystem+ Ssurroundings > 0
2NO2 (g)  N2O4 (g)
NO2 - brown, toxic gas
N2O4 - colorless gas
Ssurr = - H / T
H = Hof N2O4 - 2 Hof NO2
= (9.66) - 2 (33.5) = -58 kJ mol-1
Ssurr > 0
favors spontaneity
-
+
Suniverse= Ssystem+ Ssurroundings > 0
2NO2 (g)  N2O4 (g)
S = Soproducts - Soreactants
N 2O 4
NO2
So (J mol-1 K-1)
304.18
239.95
SoN O a) > SoNO2
2 4
b) <
S = [304.18 - 2 (239.95)]= -175.7 J/mol K
2 mol gas  1 mol gas non-spontaneous
-
+
Suniverse= Ssystem+ Ssurroundings > 0
2NO2 (g)  N2O4 (g)
S
-175.7 J mol-1 K-1
-H / T
+58 kJ mol-1
T(K)
At high T a) spontaneous
b) non-spontaneous
At low T
spontaneous
Free Energy G
Ssystem+ Ssurroundings = Suniverse > 0
-T
S - H
T
H - TS
= Suniverse > 0
= - TSuniverse< 0
G  H - TS
G = H - TS
G = H - TS
G < 0 spontaneous reaction
G > 0 non-spontaneous reaction
G = 0 equilibrium
G = wmax maximum useful work
G is an extensive State function
Gof = 0
elements in standard states
Gorxn = Gof products - Gof reactants
G = H - TS
calculate Go for:
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
Gorxn = [Gof CO2 (g)+ 2 (Gof H2O (l)) ]
- [Gof CH4 (g) + 2(Gof O2 (g))]
= - 819 kJ
a)
>
o
S rxn
0
o
o
o
b)
<
G rxn = H rxn - TS rxn
= [-892 kJ]-[(298K) (-242 J/K) ] = -819 kJ
Rubber band Thermodynamics
State 1 = relaxed
State 2 = stretched
go from State 1 to State 2
What is sign of Go
+
What is sign of Ho a) +
o
What is sign of S
b) Go = Ho - TSo
+
Ho
Go = Ho - TSo
So
Go
+
-
always positive
-
+
+
+
-
-
always negative
negative a) high T
b)
low
T
positive
low T
a) high T
negative b) low T
positive
high T
Equilibrium
Go = Ho - TS
Go = 0
phase changes
chemical reactions
Ho- TSo =H
0 o
Ho = So
T
T=
Ho
So
2NO2  N2O4
o prod - Ho react
H
___________________________
f
f
Soprod - Soreact
-58 kJ
-175.7 J/K
=T
= 331 K
Napoleon - 1812
tin buttons
ΔHof (kJ/mol) So (J/mol K)
white tin
0.0
51.55
grey tin
-2.1
44.14
ΔHo = -2.1 - 0.0 = -2.1 kJ
Snwhite

SngreyΔSo = 44.14 - 51.55
= -7.4 J/mol K
T = -2100 J = 283 K = 10oC
-7.4 J/K
∆G298 = .105 kJ ∆G233 = -.376 kJ