Chemical Equilibrium
Chemistry 100
The concept


A condition of balance between
opposing physical forces
A state in which the influences or
processes to which a thing is
subject cancel one another and
produce no overall change
Oxford English Dictionary
Static and Dynamic

A book sitting on a desk is in static
equilibrium;


The book remains at rest; its position is
constant.
The moon circles the earth.

There is movement but the (average)
distance between the two is unaltered.
This is dynamic equilibrium.
Equilibrium


The molecules of A are able to turn
into molecules of B
The rate at which this happens is
proportional to [A].
Ratefor = kfor[A]

Likewise, if B can turn into A, then
Raterev = krev[B]
The Equilibrium Condition

Start with pure A.


[A] decreases and [B] increases as A
turns into B
What happens to the rate at which A
turns into B, and the rate at which B
turns into A?

The rate of A B decreases, while BA
increases
Ratefor = Raterev  kfor[A] = krev[B]

What eventually happens?

Rate of A  B = Rate B  A
The Equilibrium Condition #2
Concentration
A
B
Time
A
Rate
B
Time
And then what?

We have as an equilibrium
condition
kfor[A] = krev[B]
B  k for
A k rev
 K eq
Keq  the thermodynamic
equilibrium constant
The Meaning of PB/PA = K



K is a constant number such as 2.3,
0.65, etc
What the equilibrium expression
means is:
No matter how much A or B we start
with, when the system reaches
equilibrium
PB
K eq 
PA
Reversible reactions


If these two reactions are possible
A  B and B  A,


we have a reversible reaction A ⇌ B
Here is a real reversible reaction
N2(g) +3H2(g) ⇌ 2NH3(g)
Equilibrium can be reached from either
side
At start
PH2 = 3; PN2 = 1; PNH3 = 0
At start
PH2 = 0; PN2 = 0; PNH3 =2
Law of Mass Action
Expression for K
For the reaction
aA (g) + bB (g) ⇌ pP (g) + qQ (g)
K eq 
p
PP PQ
q
a
b
PA PB
where PP , PQ , ..
are the partial pressures at equilibrium
Examples of Keq

P 

P P 
2
N2(g) +3H2(g) ⇌ 2NH3(g)
K eq
NH3
3
N2
H2

PBrC l

2
Br2(g) +Cl2(g) ⇌ 2BrCl(g)
SO2(g) +½O2(g) ⇌ SO3(g)
K eq
K eq 
PBr2 PC l2
PSO3
 
PSO2 PO2
1
2
Magnitude of Keq


2 HI(g) ⇌ H2(g) + I2(g)
= 0.016
The magnitude (size) of Keq provides
information



Keq
K >> 1 the products are favoured
K << 1 the reactants are favoured
CO(g) + Cl2(g) ⇌ COCl2(g) Keq=
4.57109

Equilibrium lies far to the right - there is
very little CO and Cl2 in the equilibrium
mixture.
Heterogeneous Equilibrium



When the substances in the reaction
are in the same phase (e.g., all
gases), reactions are termed
homogeneous equilibria.
When different phases are present,
we speak of heterogeneous
equilibrium.
We will look at reactions involving
gases and solids, and gases and
liquids
Solids do not appear in Keq

Examine the reaction
CaCO3(s) ⇌ CaO(s) + CO2(g) K eq 
[CaO]PC O2
[CaCO3 ]
For a pure solid X (or liquid X)
[X] = density/molar mass.
Note  molar mass and density are intensive
properties!!  [X] = constant
[constant 1][CO2 ]
constant 2
K eq 
or K eq
 K 'eq  PC O2
[constant 2]
constant 1
Heterogeneous Equilibrium


At a given temperature, the
equilibrium between CaCO3(s),
CaO(s), and CO2(g) yields the same
concentration (same partial pressure)
of CO2(g).
True as long as all three components
are present. K  P
eq
C O2
Note that it does not matter how much of the
two solids are present; we just need some.
More heterogeneous equilibria
CO2(g) + H2(g) ⇄CO(g) + H2O(l)
K eq
PC O

PC O2 PH2
SnO2(s) + 2CO(g) ⇄ Sn(s) + 2CO2(g)
K eq 
2
C O2
2
CO
P
P
Keq values for forward and reverse
reactions

For the reaction
2 HI(g)
⇌ H2(g) + I2(g), Keq =
0.016
for H2(g) + I2(g)
What is Keq
⇌
2HI(g) ?
2
 Call the
first
reaction
(F)
and
the
PH2 PI2
PHI
F
R
(R)
Ksecond
and K eq 
eq 
2

PHI 
 
PH2 PI2
Forward and Reverse (II)
R
eq
1
 F
K eq
K
F
eq
 1 or K
so K
R
eq
1

 62.5
0.016
K
R
eq
An aside

For the equilibrium H2O(l) ⇌ H2O(g), write
down the expression for Keq.
K eq  PH2O
When liquid water and water vapour are in
equilibrium the vapour has a fixed pressure at a
given temperature!
Applications


Obtaining the equilibrium constant
from the measured equilibrium
concentrations
Calculating the composition of the
equilibrium system


have the concentration of all but one
component at equilibrium and the value
of Keq
given initial amounts of reactants and
the equilibrium constant
Applications (II)

K eq
For a given reaction, Keq has a set
value for a given temperature
 PP pPQ q 
 a b
P P 
 A B eq
Q
p
PP PQ
q
a
b
PA PB


Q depends on the
experimental conditions
Q = Keq at equilibrium
Summary of the Q and Keq story

When Q > Keq


When Q = Keq


reaction shifts left 
equilibrium
When Q < Keq

reaction shifts right 
Applications (III)

Equilibrium is
approachable
from either
side of the
reaction.
Le Châtelier’s Principle

Perturb a system at equilibrium



Change in temperature, pressure, or the
concentration of a component
The system will shift its equilibrium
position so as to counteract the
disturbance.
The effect of the last two
disturbances can also be be predicted
by the Law of Mass Action
Changing concentration (I)


Examine the system
2 NO2Cl(g) ⇌ 2 NO2(g) + Cl2(g)
Introduce a small amount of
substance X that reacts with Cl2 to
make XCl.


The value of PCl2 has been decreased
Le Châtelier’s Principle predicts that
more NO2Cl will react to increase
PCl2
Changing concentration (II)
K eq 
2
NO2 C l2
2
NO2 C l
P P
P
2
(0.216) (0.108)

 11.2
2
(0.0212)
Remove Cl2 - theAt
newequilibrium
value of PCl2 is 0.05 atm
Q
2
PNO
P
2 C l2
P
2
NO2 C l
(0.216)2 (0.05)

 5.2
2
(0.0212)
Q is now smaller than Keq. The reaction moves
to the right to increase Qeq. Same prediction!
Changing concentration (III)
CaCO3(s) ⇌ CaO(s) + CO2(g)
If we have this system in
equilibrium and add either CaCO3(s)
or CaO(s), there will be no effect on
the equilibrium.
Changing the concentration
Changing pressure (I)


2 NO2Cl(g) ⇌ 2 NO2(g) + Cl2(g)
Increase the pressure in the system
by making the vessel smaller.
Note that there are a total of 3
moles on the right of the reaction
and 2 on the left.


The left “takes up less space”
Le Châtelier’s Principle predicts that
the species on the left will react to
form more NO2Cl.
Changing pressure (II)
2 NO2Cl (g)  2 NO2 (g) + Cl2 (g)
 System is initially at equilibrium.
 Increase the pressure by making
the vessel smaller.
 We could use the Keq expression to
predict what happens but Le
Châtelier’s Principle is much easier
to use!
Cautionary Note


Le Châtelier’s Principle predicts
what occurs when we change the
partial pressure of one or more of
the species in the reaction
Change the total pressure by
adding/removing an inert gas (not
involved in the reaction)
NO EFFECT ON THE
EQUILIBRIUM
More pressure changes
Predict what happens
 N2(g) + 3H2(g) ⇌ 2NH3(g); total
pressure is decreased
Reaction shifts to left; more moles
on left

H2(g) + I2(g) ⇌ 2HI(g); total
pressure is increased
No effect; 2 moles on each side
What is Heat - not a substance!

Some textbooks


Heat is treated as a chemical reagent
when applying Le Châtelier’s Principle
to change of temperature problems.
Treating heat as a substance can
lead to confusion. There is a better
way!
Changing the temperature



An exothermic reaction causes an
increase in temperature. The reverse
causes cooling.
Warm up an exothermic reaction
Le Châtelier’s Principle predicts the
system will move in the direction that
will bring the temperature back down

The direction that cools. So the reverse
reaction () occurs
Changing Temperature

N2O4(g)  2 NO2(g) H = 58.0 kJ
The forward reaction is endothermic


Decrease the temperature - reaction
shifts to the left


Absorbs heat.
Brings T back up.
If we increase the temperature,
opposite effect

Reaction takes in heat and lower the
temperature
Temperature and Keq


Endothermic reactions – increasing
temperature increases the value of
the equilibrium constant!
Exothermic reactions – increasing
temperature decreases the value of
the equilibrium constant!
Temperature changes are the only stresses
on the systems that change the numerical
values of Keq
Temperature and Keq (II)
Co(H2O)62+(aq)+ 4 Cl- (aq) ⇌ CoCl42- (aq) + 6 H2O (l)
H> 0
Catalyst does NOT change K


A catalyst speeds up a reaction by
providing and alternate reaction
pathway with a lower Ea.
Reversible reaction


the forward and backward reactions
have their Ea’s changed by the same
amount.
Keq is not altered.
A catalyst cannot alter K!! Otherwise we
would be able to build a perpetual
motion machine!!