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 BA increases
What eventually happens?
Rate of A B = Rate B A
Ratefor = Raterev kfor[A] = krev[B]
The Equilibrium Condition #2
A
Concentration
B
Time
A
B
Rate
Time
And then what?
We have as an equilibrium
condition
kfor[A] = krev[B]
B k for K eq
A k rev
Keq the thermodynamic
equilibrium constant
The Meaning of [B]/[A] = 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
[B]
K eq
[A]
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 At start
PH2 = 3; PN2 = 1; PNH3 = 0 PH2 = 0; PN2 = 0; PNH3 =2
Law of Mass Action
Expression for K
For the reaction
aA (g) + bB (g) ⇌ pP (g) + qQ (g)
p q
PP PQ
K eq a b
PA PB
where PP , PQ , ..
are the partial pressures at equilib rium
Examples of Keq
K eq
P NH 3
2
P P
3
N2(g) +3H2(g) ⇌ 2NH3(g) N2 H2
K eq
PBrCl 2
Br2(g) +Cl2(g) ⇌ 2BrCl(g)
PBr2 PCl2
PSO 3
SO2(g) +½O2(g) ⇌ SO3(g) K eq
PSO 2 PO2
1
2
Magnitude of Keq
2 HI(g) ⇌ H2(g) + I2(g) Keq = 0.016
The magnitude (size) of Keq provides
information
K >> 1 the products are favoured
K Keq reaction shifts left
When Q = Keq equilibrium
When Q 0
H
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!!