AP Chemistry
Chapter 19 Notes
CHEMICAL
THERMODYNAMICS
REVIEW
1st Law of
Thermodynamics
Energy in the universe
is conserved
REVIEW
Path Function
value depends on
how process takes
place (i.e. q, w)
heat
q = mCpT
REVIEW
State Function
value independent of
path - is a defined
reference or zero
point (i.e. H)
Enthalpy - H
zero point
0
Hf of elements in
natural form at
0
25 C
Enthalpy - H
types:
Hrxn = nHf
0
nHf (react)
0(prod)
Enthalpy - H
types:
Hfus , Hvap
ENTROPY “S”
a measure of
randomness or
disorder of a
system
Entropy is NOT
conserved. The
universe seeks
maximum disorder.
2nd Law of
Thermodynamics
for a spontaneous
process, entropy
increases
Entropy - state function
zero reference: S = 0
for a perfect crystal at
absolute zero
Ssolid < Sliquid << Sgas
0
S
= standard entropy
of elements & cmpds
0
at P = 1 atm; T = 25 C
units = J/K mol
[Appendix L, text]
Calculate S using
Hess’ Law
S(rxn) =
0
nS (prod)
0
nS (react)
Example:
Ca(s) + C(gr) + 3/2 O2  CaCO3(s)
Suniv = Ssys + Ssurr
if Suniv > 0
process spontaneous
Suniv = Ssys + Ssurr
if Suniv < 0
process ?
Suniv = Ssys + Ssurr
if Suniv = 0
process ?
to relate S and H
consider
H2O(s)  H2O(l)
where the water is the
system & everything
else is the surroundings
Temperature
Dependence
S = J/mol (1/T)
Ssurr = -H/T
Ssys Ssurr Suniv
+
+
spon
Ssys Ssurr Suniv
+
+
-
-
+
spon
Y
Ssys Ssurr Suniv
spon
+
+
+
Y
-
-
-
N
+
-
Ssys Ssurr Suniv
spon
+
+
+
Y
-
-
-
N
+
-
?
D
-
+
Ssys Ssurr Suniv
spon
+
+
+
Y
-
-
-
N
+
-
?
D
-
+
?
D
For a phase change:
S
0
phase change

Hphase change
Tphase change
GIBB’S FREE
ENERGY
G
most abstract of
thermodynamic state
functions
w = reversible
w1
work
P
w2
V
Definition:
G = w + PV
w: reversible work
PV: pressure/volume work
isothermal, reversible path
G = w + PV + VP
at constant P
P = 0 so VP = 0
G = w + PV
G = w + PV + VP
at constant P
G = w + PV
at constant V
V = 0 so PV = 0
G = w = useful work
G cannot be
measured
must measure G
over a process
ZERO REFERENCE
G = 0 for elements in
stable form under
Standard
Thermodynamic
Conditions
o
T = 25 C ; P = 1 atm
Gf = standard Free
Energy of Formation
from the elements
0
Appendix L, text
G follows Hess’ Law:
0
G (rxn)
=
0
0
nGf (p) - nGf (r)
Summary of Laws of
Thermodynamics
Zeroth Law:
Heat Gain = Heat Loss
Summary of Laws of
Thermodynamics
First Law:
Law of Conservation
of Energy
Summary of Laws of
Thermodynamics
Second Law:
Defines Entropy
Summary of Laws of
Thermodynamics
Third Law:
Defines Absolute Zero
GIBB’S
HELMHOLTZ
EQUATION
combine
G
T
H
G = -aT + H
G, H are state
functions, thus “a”
must be a state function
G = -S(T) + H
Gibb’s Helmholtz
Equation
G = H - TS
Units on the State Functions
kJ
H 
mol
J
S 
K  mol
kJ
G 
mol
G  H  TS
 G  H

 S
T
T
but 
H

 S surr
T
G

 Ssurr  S
T
but 
S surr  S  S univ
G

 S univ
T
thus, a process is
spontaneous
if and only if
G is negative
Spontaneity
controlled by
enthalpy
(minimum energy)
Sponaneity
controlled by
enthalpy
entropy
(maximum disorder)
Sponaneity
controlled by
enthalpy
entropy
both
Predict Spontaneity
IF H(-) and S(+)
G = -H - T(+S)
G < 0, => spontaneous
Predict Spontaneity
IF H(+) and S(-)
G = +H - T(-S)
G > 0, => NOT spontan
Summary of Spontaneity
H
+
+
-
S
+
+
-
G Spont.
yes
+
no
+ or ?
+ or ?
Uses of the
Gibb’s Helmholtz
Equation
1. Find the molar
entropy of formation
for ammonia.
2. Elemental boron, in
thin fibers, can be
made from a boron
halide:
BCl3(g) + 3/2 H2(g) ->B(s) + 3HCl(g)
Calculate:
0
0
0
H , S and G .
Spontaneous?
Driving force?
3. Using thermodynamic
information, determine
the boiling point of
bromine.
Thermodynamic
Definition of
Equilibrium
Geq = 0
by definition
G = H - TS
&
H = E + PV
thus,
G = E + PV - TS
take derivative of both sides
dG = dE + PdV + VdP - TdS - SdT
for a reversible process
TdS = q
derivative used for
state function
while
partial derivative used
for path function
if the only work is PV
work of expansion
PdV = w
First Law of
Thermodynamics
E=q-w
dE = q - w = 0
thus
q = w
or
TdS = PdV
by substitution
dG = 0 + PdV + VdP
- PdV - SdT
or
dG = VdP - SdT
let’s assume we have
a gaseous system at
equilibrium, therefore,
examine Kp at that
constant temperature
at constant T
SdT = 0
thus
dG = VdP
G2
P2
G1
P1
dG

V
dP


but
nRT
V
P
G2
P2
dP
dG

nRT
G
P P
1
1
G2
P2
dP
dG

nRT
G
P P
1
1
P2
G2  G1  nRT  ln
P1
 P1 
G1  G2  RT  ln 
 P2 
n
if “condition 2” is
at standard
thermodynamic
conditions, then
0
G2 = G and P2 = 1 atm
thus
G  G  RT  ln P
0
n
determine G for
aA + bB  cC + dD
where all are gases
G = Gprod - Greact
= cGC + dGD - aGA - bGB
but
cG C  cG  RT  ln P
0
C
and likewise
for the others
c
and
cG  dG  aG  bG
0
C
0
D
 G
0
rxn
0
A
0
B
G  G
 (PC ) (PD ) 
 RT  ln
a
b
 (PA ) (PB ) 
c
0
rxn
but
 (PC ) (PD ) 

?
 (P )a (P )b 
 A

B
c
d
d
Thus,
G =
0
G
+ (RT) ln Q
But
aA + bB  cC + dD
G = 0
thus
G
0
rxn
  RT  ln K P
or in general
G
0
rxn
  RT  ln K eq
THERMODYNAMICS
&
EQUILIBRIUM
3. Using thermodynamic
information, determine
the boiling point of
bromine.
Thermodynamics and Keq

FACT: Product-favored
systems have Keq > 1.
Thermodynamics and Keq

Therefore, both
∆G˚rxn and Keq are
related to reaction
favorability.
Thermodynamics and Keq
Keq is related to reaction
favorability and thus to ∆Gorxn.
The larger the value of K the
more negative the value of
∆Gorxn
Thermodynamics and Keq
o
∆G rxn=
- RT lnK
where R = 8.314 J/K•mol
Thermodynamics and Keq
∆Gorxn = - RT lnK
Calculate K for the reaction
N2O4 2 NO2
∆Gorxn = +4.8 kJ
K = 0.14
When ∆G0rxn > 0, then K < 1
∆G, ∆G˚, and Keq
 ∆G
is change in free energy at
non-standard conditions.
 ∆G
is related to ∆G˚
 ∆G = ∆G˚ + RT ln Q
where Q = reaction quotient
∆G, ∆G˚, and Keq
 When
Q < K or Q > K, reaction is
spontaneous.
 When
Q = K reaction is at equilibrium
 When ∆G = 0 reaction is at equilibrium
 Therefore, ∆G˚ = - RT ln K
∆G, ∆G˚, and Keq
Product favored
reaction
–∆Go and K > 1
In this case
∆Grxn is < ∆Gorxn ,
so state with
both reactants
and products
present is
MORE STABLE
than complete
conversion.
∆G, ∆G˚, and Keq
Product-favored reaction.
2 NO2  N2O4
∆Gorxn = – 4.8 kJ
Here ∆Grxn is less than ∆Gorxn , so the
state with both reactants and
products present is more stable
than complete conversion.
Thermodynamics and Keq
Overview
o
 ∆G rxn
is the change in free
energy when reactants
convert COMPLETELY to
products.
Keq is related to reaction
favorability.
When ∆Gorxn < 0, reaction
moves energetically
“downhill”
4. For the following
reaction, calculate the
temperature at which the
reactants are favored.
1
3
N2 (g)  H2 (g)  NH3 (g)
2
2
THERMODYNAMICS
OF
CHEMICAL
REACTIONS
5. How much useful
work can be obtained
from an engine fueled
with 75.0 L of hydrogen
at 10 C at 25 atm?
6. The reaction to split water
into hydrogen and oxygen
can be promoted by first
reacting silver with water.
2 Ag(s) + H2O(g) Ag2O(s) + H2(g)
Ag2O(s)  2 Ag(s) + 1/2 O2(g)
0
H ,
0
S
Calculate
0
and G for each
reaction.
Combine the reactions
0
and calculate H and
0
G for the combination.
Is the combination
spontaneous?
At what temperature
does the second reaction
become spontaneous?
7. The conversion of
coal into hydrogen for
fuel is:
C(s) + H2O(g)  CO(g) + H2(g)
Calculate
0
G
and Kp at
0
25 C.
Is the reaction spontaneous?
At what temperature
does the reaction
become spontaneous?
Calculate the
temperature at which
-4
K = 1.0 x 10 .
8. The generation of
nitric acid in the upper
atmosphere might
destroy the ozone
layer by:
NO(g) + O3(g)  NO2(g) + O2(g)
0
G
Calculate
(reaction)
0
and K at 25 C.