Equilibrium Law, Kc
Another
way chemists may describe a state of equilibrium is using an equilibrium law. We can study this, using the Haber Process
(named after Fritz Haber, who has a despicable history):
N2(g) + 3H2(g) ↔
2NH3(g) (several 100 kPa, a
few 100°C and a catalyst)
 |
| Notice that the equilibrium [ ] of each substance is the same whether we begin with a 3:1 ratio of H2:N2 or pure NH3, thus the equilibrium condition can be reached from either direction. |
Using
the simple general reaction, A ↔ B, let’s derive its equilibrium law:
⇰ since
A → B and B →
A are elementary processes, ratef = kf [A] and rater
= kr [B]
⇰ at equilibrium, ratef = rater
⇰ thus, kf [A] = kr [B]
⇰ which rearranges to give,[B]/[A] = kf/ kr = constant
⇰ at equilibrium, ratef = rater
⇰ thus, kf [A] = kr [B]
⇰ which rearranges to give,[B]/[A] = kf/ kr = constant
A
similar ratio governs all reactions.
Equilibrium Law
The
equilibrium law (eq law) is an expression of the relationship between the [P]
and the [R] at equilibrium in any reaction.
In general,
aA + bB ↔ cC
+ dD
[D]d[C]c
Kc = ----------
[A]a[B]b
Kc = ----------
[A]a[B]b
This
is called the equilibrium expression
or equilibrium law or law of mass action (which is simply a ratio of products:reactants at eq). The equilibrium
constant (Kc) is obtained upon substitution of actual
equilibrium concentrations* into the equilibrium expression.
*hence, the subscript "c" to represent that we are subbing in concentrations.
For
the Haber Process, N2(g) + 3H2(g) ↔
2NH3(g), the eq law is:
[NH3]2
Kc
= ----------
[N2][H2]3
The
equilibrium law depends only on the temperature at which the reaction occurs, as
well as, the stoichiometry of the reaction. The equilibrium constant does not depend on
[initial] and it doesn’t matter if other non-reactive species are present.
Magnitude of Kc, Kp
For
the reaction, CO(g) + Cl2(g) ↔
COCl2(g)
[COCl2]
Kc
= ----------- = 4.57 x
109
[CO][Cl2]
or likewise,
[COCl2] 4.57 x
109
Kc = ------------- = --------------
[CO][Cl2] 1
Kc = ------------- = --------------
[CO][Cl2] 1
Since
the numerator > denominator (that is, 4.57 x 109 > 1 ), [COCl2] > [CO][Cl2]. Thus, there are a lot of products (COCl2) and very
little reactants (CO and Cl2).
- K > 1 equilibrium lies to the right and products are favoured (chemists love lots of products so this results in a happy chemist 😁)
- K < 1 equilibrium lies to the left and reactants are favoured (chemists do not love lots of reactants, so this results in a sad chemist 😦)
- K = 1 equilibrium lies in the middle and neither reactants nor products are favoured (this results in a neutral chemist 😐)
Direction of the Chemical Equation and
K
For
the reaction N2O4(g) ↔
2NO2(g); (note: Kcf ↝ eq constant in the forward direction)
[NO2]2
Kcf
= ----------
[N2O4]
For
the reaction 2NO2(g) ↔ N2O4(g); (note: Kcr ↝ eq constant in the reverse direction)
[N2O4]
Kcr
= ----------
[ NO2]2
Thus,
1
Kcf
= ------
Kcr
So, if we know the rate constant for the forward reaction and desire it for the reverse reaction, just take the inverse.
Homogeneous and Heterogeneous
Equilibria
The
Haber Process, N2(g) + 3H2(g) ↔
2NH3(g), has all substances all in one phase and it is said to be homogeneous. We have seen the eq law for the Haber
Process above.
The
reaction, CaCO3(s) ↔ CaO(s) + CO2(g), has substances
in different phases and it is said to be heterogeneous.
[CaO][CO2]
Kc
= -----------------
[CaCO3]
c1[CO2]
Kc
= ------------
c2
Kc(c2/c1)
= [CO2]
Kc
= [CO2]
Since
[s] (and [l]) do not change, we know that they remain constant and can be
replaced by c1 and c2. Gathering all the
constants together on the left side of the equation, results in a “super duper” constant,
which is still just a constant and can continue to be represented by Kc. So, include solutions and gases in eq laws;
do not include solids and liquids in eq laws.
Calculating Equilibrium Constants
Haber
conducted many experiments to determine if his process was a viable method to
produce ammonia.
ex. In an experiment, Haber introduced a
mixture of H2 and N2 into a reaction vessel and allowed
the system to go to equilibrium at 472°C.
At equilibrium, [H2] = 0.1207 M, [N2] = 0.0402 M
and [NH3] = 0.00272 M. Was
this a trial a success?
[NH3]2
Kc = ------------
[N2][H2]3
(0.00272)2
Kc = -----------------------
(0.0402)(0.1207)3
Kc = 0.105
Since Kc < 1, reactants
are favoured and this is trial was not a success.
Homework #8-16
Homework #8-16
