Chemical Equilibria
Chapter 16
1
Chemical reactions
• Can reverse (most of the time)
• Though might require a good deal of
energy
•  or double arrows = reversible rxn
• It “swings both ways”
• Once fwd/rev rxns occur at equal rates
• = equilibrium (no net change seen)
2
Equilibrium constant
• aA + bB  cC + dD
• Thus,
K eq =
c
[C ] [D ]
d
a
b
[A ] [B ]
3
Equilibrium constant in action
• Give the equilibrium
constant for:
H2(g) + I2(s)  2HI(g)
K eq 
[H I]
2
[H 2 ][I 2 ]
• Initially, 0.0175 M of
•
•
reactants
Decrease to 0.0037 M
reactants
What is the
concentration of HI
formed?
4
Solution
0.0175M - 0.0037M =0.0138M
0.0138m ol H 2 
2 m ol H I
 0.0276 m ol H I = 0.0276M H I
1m ol H 2
5
ICE Table
• Initial Change Equilibrium Table
• Great way to explain and show
concentration changes
• Let’s make an ICE table for the previous
reaction on the board
6
Calculate Keq for the reaction
K eq =
[H I]
2
[H 2 ][I 2 ]
=
[0.0276]
2
= 56
[0.0037][0.0037]
U nitless, since concentrations are ratio s
to reference concentration of exactly 1 M .
7
Problems
• Express Keq for:
CH3OH(g)  CO(g) + 2H2(g)
• Express Keq for:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
• Solve Keq for the following:
A(g)  2B(g)
– Given [A]i = 1.00M, [B]i = 0.00M, and
[A]eq =0.75M
8
K eq =
[C O ][H 2 ]
2
[C H 3 O H ]
3
K eq =
[C O 2 ] [H 2 O ]
[C 3 H 8 ][O 2 ]
4
5
A (g)  2B (g)
I 1.00M
0.00M
C -x (0.25M )
E 0.75M
K eq =
[B ]
+ 2x (0.50M )
0.50M
2
[A ]

[.50]
2
 0.33
[.75]
9
Caveats to Keq: solids
• Solids in reversible rxns are excluded from
expression since concentration derived
from constant densities:
• S(s) + O2(g)  SO2(g)
• Keq = ?
10
Caveats to Keq: aqueous solns
• Same rule for pure liquids
• Ex: NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
• Keq = ?
11
Caveats to Keq: gases & Kp
• For gases we can use
•
•
partial pressures
– Why?
• Hint: Relationship
between pressure
and concentration
– See ideal gas
law
Given:
aA + bB  cC + dD
Kp not necessarily equal
to Keq
Kp 
c
(P C ) (P D )
d
a
b
(P A ) (P B )
12
The relationship between Keq and
Kp
nRT
PV  nRT; P 
n
;P 
V
 RT;
V
n

V
mol
 [ X];  P  [X]RT & [X] 
L
P
RT
Given aA  bB  cC  dD
(
K eq 
(
PC
RT
PA
) (
c
) (
a
RT
 K eq  K p  (
PD
)
RT
PB

)
1
)
d
1
b
RT
1
PC  P D  (
a
PA  PB  (
b
RT
)
cd
c

)
ab
PC  P D
a
PA  PB
d
b
(
1
RT
)
c  d - (a  b)
 Kp(
1
)
c  d - (a  b)
RT
RT
c  d - (a  b)
RT
And, K p  K eq ( RT)
Finally,
c
d
c  d - (a  b)
K p  K eq ( RT)
n
13
Problem
• Given
2NO(g) + O2(g)  2NO2(g)
• Kp = 2.2 x 1012 @ 25°C
• Find Keq
14
Solution
K p  K eq (RT)
2 . 2  10
12
n
 K eq (
K eq  5 . 4  10
0 . 08206 L  atm
mol  K
 298 K)
(2 - 3)
13
15
Determining equilibrium constant
• Given 2SO2(g) + O2(g)  2SO3(g)
• And: [SO2]i = 1.00 M, [O2]i = 1.00 M &
[SO3]f = 0.925 M
• What is Keq?
16
Solution
2S O 2(g) + O 2(g)  2S O 3(g)
I
C
E
1.00
1.00M
+ 2x
-x
-2x
0
(1.00-2x) (1.00-x)
0.925
S ince 2x= 0.925, x= 0.463
K eq 
[S O 3 ]
2
2
[S O 2 ] [O 2 ]

[0.925]
2
[1.00  0.925] [1.00  0.463]
2
 300
17
More problem solving
• Sulfuryl chloride (SO2Cl2) dissociates into sulfur dioxide
and chlorine in the gas phase:
SO2Cl2(g)  SO2(g) + Cl2(g)
• In an experiment, 3.174 g of SO2Cl2 (MW =
134.96g/mol) is placed in a 1.000 L flask and is at
100.0°C. At equilibrium, the total pressure in the flask is
1.30 atm.
Calculate:
• a) The partial pressures of each gas at equilibrium.
• b) The Kp at 100.0°C for the reaction.
18
Solution
PV  nRT
mol
P  (1.000L)  (3.174g 
)  ( 0 . 08206
134.96g
atm  L
K  mol
)  ( 373 . 2 K )
PSO 2 Cl 2  0.7202 atm
i
SO 2 Cl 2 (g)  SO
I
C
E
2(g)
 Cl 2(g)
0.7202 atm
0
0
x
x
x
x
-x
(0.7202 atm - x)
P total  1 . 30 atm  (0.7202atm
- x)  x  x  0.7202atm
x
x  0.58atm
Thus, (SO 2 ) eq  (Cl 2 ) eq  0.58 atm & (SO 2 Cl 2 )  0 . 7202 atm  0.58  0.14atm
And, K p 
(SO 2 )  (Cl 2 )
( SO 2 Cl 2 )

(0.58)
2
( 0 . 14 )
 2 .4
19
Yet another
• Isopropanol can dissociate into acetone
and hydrogen:
(CH3)2CHOH(g)  (CH3)2CO(g) + H2(g)
• At 179°C, the equilibrium constant is
0.444. Calculate the equilibrium partial
pressures of all three gases if 10.00 g
(MW = 60.10g/mol) of isopropanol are
initially placed in a 10.00 L vessel.
20
Solution
(10.00g 
mol
)  ( 0 . 08206
60.10g

Pisopropano
li
I
0.617
0
C
-x
x
E
0.617 - x
x
KP 
L  atm
mol  K
)  ( 452 C)

 0 . 617 atm
10.00L
(x)(x)
0
x
x
 0 . 444
(0.617 - x)
x  0 . 274  0 . 444 x
2
x  0 . 444 x  0 . 274  0
2
x 
-b
b  4 ac
2
2a
x  0.346
x  0.346atm
 (acetone)
eq
 ( H 2 ) eq
0 . 617  x  0 . 271 atm  (isopropan ol)
eq
21
Another but with a twist
• At 25°C, the equilibrium constant for the
•
•
•
reaction below is 5.9 x 10-13.
2NO2(g)  2NO(g) + O2(g)
Suppose a container is filled with 0.89 atm of
NO2
Calculate the equilibrium partial pressures of
each gas
OK to approximate if “x  [A]0 < 5%“
– “5% rule”
22
Solution
2 NO
2(g)
 2NO
 O 2(g)
(g)
I 0.89
0
0
C - 2x
 2x
x
E (0.89 - 2x)
2x
x
2
Kp 
(NO) ( O 2 )
( NO 2 )
2
 5 . 9  10
2
5 . 9  10
 13

x  4.9  10
4.9  10
2
(2x) ( x )
( 0 . 89  2 x)
 13
2

(2x) ( x )
( 0 . 89 )
2
-5
-5
 100 %  0 . 0055 % (  5%)  VALID
0 . 89
Thus, x  4.9  10 atm  (O 2 ) eq
-5
2 x  9 . 8  10 atm  (NO)
-5
eq
( 0 . 89  2 x)  0 . 89 atm  (NO 2 ) eq
23
More on Keq
• Is the rxn product or reactant
favored?
– I.e., Will it form more
product or reactant?
• If K  1, then prod
concentration higher than
reactant concentration 
2
2
[H
I]
[0.0276]
prod favored
K eq =
=
= 56
[H 2 ][I 2 ] [0.0037][0.0037]
– Makes mathematical
sense
• See right
• If K < 1, then reactant
concentration higher than
prod concentration 
reactant favored
• If K = 1, neither favored;
both equal concentration
24
Reaction quotient, Q
• When rxn not at
•
equilibrium, use Q
Where
aA + bB  cC + dD
– Then 
• Remember, Q used
Q =
c
[C ] [D ]
d
a
b
[A ] [B ]
for system when
system NOT in
equilibrium
25
Q – its benefits
• Answers the question:
Is the system at equilibrium (i.e., does Q = K)?
• If not, we can predict which way the reaction will continue to
proceed
• If Q < K, rxn still needs to go to prod side to achieve equilibrium
(i.e., where Q = K)
– In other words, insufficient product formed for equilibrium conditions
• If Q > K, rxn has “overshot” K and needs to go to reactant side to
achieve equilibrium (i.e., where Q = K)
– In other words, an excess of product is formed for equilibrium
conditions
26
Work on this…
• N2(g) + 3H2(g)  2NH3(g)
• The equilibrium constant at 400°C is
Keq=0.5.
• Suppose we make a mixture with the
following concentrations:
• [NH3] = 5.0M, [N2] = 3.5M, [H2] = 1.9M
• In which direction will the reaction go?
• a) products b) reactants
27
Q =
[NH 3 ]
2
[N 2 ][H 2 ]
3

[5.0]
2
[3.5][1.9]
3
 1.0
28
Manipulating Keq
• C(s) + ½ O2(g)  CO(g)
• Kp’ = ?
• Changing it: 2C(s) + O2(g)  2CO(g)
• Kp” = ?
• How are Kp’ and Kp” related mathematically?
29
More on Keq
• Given 2NO2(g)  N2O4(g)
• Keq = ?
• What is Keq when rxn is reversed?
• Therefore, what can we say about Keq fwd
and Keq rev?
30
Even more on Keq
•
•
•
•
•
•
Remember Hess’s Law?
1) A+B  C
2) B+C  D
What is the net rxn and it’s Knet using Hess’s
Law?
Can one obtain the same values as above not
using Hess’s Law?
What can we say about Knet using each rxn’s Keq?
31
Disturbing chemical equilibria
• Le Châtelier’s Principle
• Change one component of the rxn & the
rxn will attempt to rectify it
• Think of it this way:
• If something is changed, how can it be
undone or controlled so that equilibrium is
achieved once again?
32
Temperature variation on
equilibrium
• 2NO2(g)  N2O4(g) + heat
• Kp = ?
• H° = -57.1 kJ
• @ 273 K, Kp = 1300, and @ 298 K, Kp =
170
• Hence, if one raises the temp to 298 K,
which way will it swing?
33
Pressure & volume change on
equilibrium
• If volume decreased (pressure increased)
 favors smaller # of molecules
• If volume increased (pressure decreased)
 favors larger # of molecules
• If reversible rxn has = # of molecules on
each side, a volume/pressure change will
do nothing
34
Question
• What change in equilibrium will be seen
when one adds a solid and why?
35
How about this?
• 2H2S(g) + O2(g)  2S(s) + 2H2O(g); ΔH = -221.19 kJ/mol
• If O2(g) is added to the reaction vessel, what happens to the amount
of S(s)?
– a) It increases b) It decreases c) Nothing
• If the volume of the vessel is cut in half, what happens to the ratio
of PH O/PH S?
2
2
– a) It increases b) It decreases c) Nothing
• If the temperature is increased, what happens to the equilibrium
constant K?
– a) It increases b) It decreases c) Nothing
• If S(s) is added to the reaction, what happens to PH2O?
– a) It increases b) It decreases c) Nothing
36