Drill:
•List five factors &
explain how each
affect reaction rates
Drill: Solve Rate Law
A+B
4C+A
2K
G+K
Q+W
C+D
2G
4D + B
2Q+2W
Prod.
fast
fast
fast
fast
slow
Review Drill
&
Check HW
CHM II HW
• Review PP-19 & 20
• Complete the attached
worksheet & turn it in
tomorrow
• Lab Thursday
Chemical
Equilibria
Equilibrium
•The point at which the
rate of a forward
reaction = the rate of its
reverse reaction
Equilibrium
•The concentration of all
reactants & products
become constant at
equilibrium
Equilibrium
•Because concentrations
become constant,
equilibrium is sometimes
called steady state
Equilibrium
•Reactions do not stop at
equilibrium, forward &
reverse reaction rates
become equal
Reaction
• aA(aq)+ bB(aq)
pP(aq)+ qQ(aq)
a
b
• Ratef = kf[A] [B]
p
q
• Rater = kr[P] [Q]
• At equilibrium, Ratef = Rater
• kf[A]a[B]b = kr[P]p[Q]q
At equilibrium, Ratef = Rater
a
b
c
d
kf[A] [B] = kr[C] [D]
kf /kr =
c
d
([C] [D] )/
(
a
b
[A] [B] )
kf /kr = Kc = Keq in terms of
concentration
c
d
a
b
Kc = ([C] [D] )/ ( [A] [B] )
Reaction
aA(g)+ bB(g)<-->cC(g)+ dD(g)
a
b
Ratef = kfPA PB
c
d
Rater = krPC PD
At equilibrium, Ratef = Rater
a
b
c
d
kfPA PB = krPC PD
At equilibrium, Ratef = Rater
a
b
c
d
kfPA PB = krPC PD
kf /kr =
c
d
(PC PD )/
(
a
b
PA PB )
kf /kr = Kp = Keq in terms of
pressure
c
d
a
b
Kp = (PC PD )/ ( PA PB )
All Aqueous
aA + bB
Kc 
pP + qQ
p
q
P Q
a
b
A B
aA + bB(g)
pP + qQ
 
p
q
 PP  PQ
KP 
a
b
 PA   PB 
Equilibrium
Expression
p
( Products)
Keq=
r
(Reactants)
AP CHM HW
• Read: Chapter 12
• Work problems: 5, 7, &
12
• Page: 365
CHM II HW
• Read: Chapter 17
• Work problems: 17 &
21
• Page: 745
Equilibrium
Applications
•When K >1, [p] > [r]
•When K <1, [p] < [r]
Equilibrium
Calculations
K p = Kc
Dn
gas
(RT)
Equilibrium Expression
•Reactants or products
not in the same phase
are not included in the
equilibrium expression
Equilibrium Expression
aA(s)+ bB(aq)<--> cC(aq)+ dD(aq)
Keq=
c
d
[C] [D]
b
[B]
Reaction Mechanism
•Sequence of steps that
make up the total
reaction process
Reaction Mechanism
•1) A + B <---> CFast
•2) A + C <---> D Fast
•3) B + D <---> H Fast
•4) H + A -----> P Slow
Reaction Mechanism
•The rate determining
step is the slowest step
•H + A ----> P Slow
•Rate = k4[H][A]
Reaction Mechanism
•Rate = k4[H][A]
•Because H is not one of
the original reactants, H
cannot be used in a rate
expression
Reaction Mechanism
•3) B + D <---> H
•K3 = [H]/([B][D])
•[H] = K3[B][D]
Reaction Mechanism
•[H] = K3[B][D]
•Rate = k4[H][A]
•Rate = k4K3[B][D][A]
Reaction Mechanism
•2) A + C <---> D
•K2 = [D]/([A][C])
•[D] = K2[A][C]
Reaction Mechanism
•[D] = K2[A][C]
•Rate = k4K3[B][D][A]
• Rate = k4K3[B]K2[A][C][A]
• Rate = k4K3 K2
2
[B][A] [C]
Reaction Mechanism
•1) A + B <---> C
•K1 = [C]/([A][B])
•[C] = K1[A][B]
Reaction Mechanism
• [C] = K1[A][B]
2
• Rate = k4K3 K2[B][A] [C]
• Rate = k4K3 K2[B][A]2K1[A][B]
• Rate = k4K3 K2K1
2
3
• Rate = K [B] [A]
2
3
[B] [A]
Solve Rate Expression
•1) A + B <---> 2C
•2) A + C <---> D
•3) B + D <---> 2H
•4) 2H + A ----> P
Fast
Fast
Fast
Slow
Reaction Mechanism
• When one of the
intermediates anywhere in
a reaction mechanism is
altered, all intermediates
are affected
Reaction Mechanism
•1) A + B <---> C + D
•2) C + D <---> E + K
•3) E + K <---> H + M
•4) H + M <----> P
Lab Results
%
100
80
60
40
RT 5.21 8.42 11.9 21.7
WR 2.75 4.23 7.96 11.2
Applications of Equilibrium Constants
aA + bB(g)
Q
pP + qQ
p
q
P Q
a
b
A B
where [A], [B], [P], and [Q] are molarities at any time.
Q = K only at equilibrium.
NH3
H2 + N2
At a certain temperature at
equilibrium Pammonia = 4.0
Atm, Phydrogen = 2.0 Atm,
& Pnitrogen = 5.0 Atm.
Calculate Keq:
Equilibrium
Applications
•When K > Q, the
reaction goes forward
•When K < Q, the
reaction goes in reverse
Drill: SO2 + O2
SO3
• Determine the magnitude of
the equilibrium constant if
the partial pressure of each
gas is 0.667 Atm.
Review Drill
&
Collect HW
CHM II HW
• Review PP-20
• Complete the attached
worksheet & turn it in
Friday
• Lab Tomorrow
Equilibrium Calculations
•aA + bB <--> cC + dD
•Stoichiometry is used
to calculate the
theoretical yield in a
one directional rxn
Equilibrium Calculations
•aA + bB <--> cC + dD
•In equilibrium rxns,
no reactant gets used
up; so, calculations are
different
Equilibrium Calculations
•Set & balance rxn
•Assign amounts
•Write eq expression
•Substitute amounts
•Solve for x
Equilibrium Calculations
•CO + H2O
CO2 + H2
• Calculate the partial
pressure of each portion at
eq.when 100.0 kPa CO &
50.0 kPa H2O are combined:
•Kp = 3.4 x
-2
10
Equilibrium Calculations
• CO
H2O
CO2
H2
• 100 -x
50 - x
x
x
•Kp = PCO2PH2 =
•
PCOPH2O
• Kp = 3.4 x
-2
10
2
x
(100-x) (50-x)
Equilibrium Calculations
2
x
2
x
(100 -x)(50 - x) = 5000 -150x +
= 3.4 x 10-2
2
x =
170 - 5.1x + 0.034x2
2
0.966x +
5.1x - 170 = 0
2
x
Equilibrium Calculations
Xe (g) + F2(g)
XeF2(g)
Calculate the partial pressure
of each portion when 50.0
kPa Xe & 100.0 kPa F2 are
combined:
-4
Kp = 4.0 x 10
Drill: Write the
equilibrium expression
& solve when PNO2 &
PN2O4 = 50 kPa each:
N2O4(g)
2 NO2(g)
Review Drill
&
Check HW
CHM II HW
•Review PP-20
•Complete the
attached HW
•Test: Tuesday
Le Chatelier’s
Principle
•If stress is applied to a
system at equilibrium,
the system will readjust
to eliminate the stress
LC Eq Effects
•A(aq) +2 B(aq) <--->
C(aq) + D(aq) + heat
•Write equilibrium exp:
•What happens if:
LC Eq Effects
•2 A(aq) + B(s) <--->
C(aq) +2 D(aq) + heat
•Write equilibrium exp:
What happens if:
LC Eq Effects
•2 A(g) + 2 B(g) <--->
3 C(g) + 2 D(l)
•What happens if:
Equilibrium
Applications
DG = DH - TDS
DG = - RTlnKeq
Equilibrium Calculations
Xe (g) + 2 F2(g)
XeF4(g)
Calculate the partial pressure
of each portion when 75 kPa
Xe & 20.0 kPa F2 are
combined:
-8
Kp = 4.0 x 10
A(aq)+ B(aq)
D(aq)
Calculate the equilibrium
concentration of each species
when equal volumes of
0.40 M A & 0.20 M B are
combined.
Keq = 0.50
AP CHM HW
•Read: Chapter 12
•Problems: 37 & 39
•Page: 367
CHM II HW
•Read: Chapter 17
•Problems: 45
•Page: 747
Drill: Solve for K
A(aq)+ 2 B(aq)
C(s)+ 2 D(aq)
Calculate Keq if:
[A] = 0.30 M
C = 5.0 g
[B] = 0.20 M
[D] = 0.30 M
Drill: A + B
C+D
Calculate the equilibrium
concentration of each species
when equal volumes of
0.60 M A & 0.80 M B are
combined.
Keq = 0.50
The Test on Rxn
Rates & Chemical
Equilibria will be
on Tuesday
Working with
Equilibrium
Constants
When adding
Reactions
Multiply Ks
A
B
A
K3 =
B K1
C K2
C K3
(K1)(K2)
Solve K for each:
A+B
C+D
A+B
C+D
P+Q
P+Q
When doubling
Reactions
Square Ks
A
2A
B
2B
K2 = (K1
K1
K2
2
)
When a rxn is
multiplied by any
factor, that factor
becomes the
exponent of K
A
1/3 A
B
1/3 B
K2 =
1/3
(K1)
K1
K2
When reversing
Reactions
Take 1/Ks
A
B
B
A
K1
K2
K2 = 1/K1
Equilibrium Calculations
Kr (g) + F2(g)
KrF2(g)
Calculate the partial pressure
of each portion when 4.0 kPa
of Kr & 8.0 kPa of F2 are
combined:
-2
Kp = 4.0 x 10
Equilibrium Calculations
Rn (g) + F2(g)
RnF2(g)
Calculate the partial pressure
of each portion when 50 kPa
Rn & 75 kPa F2 are
combined:
-2
Kp = 4.0 x 10
Drill:
A+B
P+ Q
Calculate the concentration of
each portion at equilibrium
when 100.0 mL 0.50 M A is
added to 150 mL 0.50 M B:
Kc = 4.0 x
-2
10
Equilibrium Calculations
-2
-2
I 2 + 2 S 2 O3
S4O6 + 2 I
Calculate the concentration of
each portion when 100 mL
0.25 M I2 is added to 150 mL
-2
0.50 M S2O3 :
Kc = 4.0 x
-8
10
Clausius-Claperon Eq
(T2)(T1)
k2
Ea= R(T – T ) ln k
2
1
1
Clausius-Claperon Eq
(T2)(T1) P2
Hv= R (T – T ) ln P
2
1
1
Clausius-Claperon Eq
DH =
(T2)(T1)
K2
R (T – T )ln K
2
1
1
DG  DH - TDS
o
DG = -RTlnK
All aqueous
aA + bB
pP + qQ
Kc =
p
q
[P] [Q]
a
b
[A] [B]
at equilibrium
All aqueous
aA + bB
pP + qQ
Q=
p
q
[P] [Q]
a
b
[A] [B]
at the other conditions
AP CHM HW
•Problems: 41 & 43
•Page: 368
CHM II HW
•Problems: 47
•Page: 747
Write the Eq Expression
AB(aq)
A(aq)+ B(aq)
Calculate [A], [B], &
[AB] at equilibrium when
[AB] = 0.60 M at the start
-5
Keq = 6.0 x 10
Drill:
Calculate the heat of
reaction when K =
-6
o
2.5 x 10 at 27 C, &
-4
o
K = 2.5 x 10 at 127 C.
Drill:
1A+ 1B
1Z +1Y
Calculate the concentration of
each portion at equilibrium
when 1.0 L 0.50 M A is added
to 1.5 L 0.50 M B:
Kc = 2.0 x
-2
10
Next Test
•Tuesday
Drill:
1A+ 1B
1Z
Calculate the concentration of
each portion at equilibrium
when the original solution has
0.50 M A & 0.50 M B:
Kc = 2.0
Review Drill
&
Check HW
Review
for the
Test
The data on the next slide
was obtained in lab. Use
that data to solve for the
order with respect to each
reactant, the reaction order,
the rate expression, k, & Ea.
Experimental Results
• Exp # [A]
• 127 1.0
• 227 2.0
• 327 1.0
• 477 1.0
[B]
Rate
-2
1.0 2.0 x 10
-2
1.0 4.0 x 10
-2
2.0 8.0 x 10
1.0
2.0
Reaction Mechanism
• Step 1
• Step 2
• Step 3
• Step 4
A <--> B
2 B <--> 3C
C <--> 2D
D  P
fast
fast
fast
slow
LC Eq Effects
•2 A(aq) + B(s) <--->
C(aq) +2 D(aq) + heat
•Write equilibrium exp:
What happens if:
LC Eq Effects
•3 A(g) + B(g) <--->
2 C(g) + 2 D(l)
•Write equilibrium exp:
•What happens if:
SO + O2
SO3
Calculate the equilibrium
pressures if SO at 80.0 kPa
is combined with O2 at
40.0 kPa. K = 4.00
Experimental Results
• Exp # [A]
• 1
1.0
• 2 2.0
• 3
1.0
• 4 1.0
[B]
1.0
1.0
2.0
1.0
[C] time
1.0 16
1.0
2
1.0 8
2.0
4
Write the Eq Expression
PQ(aq)
P(aq)+ Q(aq)
Calculate [P], [Q], &
[PQ] at equilibrium when
[PQ] = 0.90 M at the start
-5
Keq = 9.0 x 10
Write the Eq Expression
AB(aq)
A(aq)+ B(aq)
Calculate [A], [B], &
[AB] at equilibrium when
[AB] = 0.60 M at the start
-5
Keq = 6.0 x 10
Experimental Results
• Exp # [A]
• 1
0.1
• 2 0.1
• 3
0.1
• 4 0.2
[B]
0.1
0.3
0.1
0.1
[C] Rate
0.2 2
0.2 18
0.8 8
0.2 64
A + B <---> C + D
C + H <---> M + N
N + T <---> P + Q
•What happens all
intermediates if:
1A + 1B
1P + 1Q
[Ai] = 0.20 M Calcu[Bi] = 0.30 M late the
[Pi] = 0.20 M eq con[Qi] = 0.30 M centraKc = 0.020 tion of ea.