Degree of dissociation
(x)
In the
study of dissociation equilibrium, it is easier to derive the equilibrium
constant expression in terms of degree
of dissociation (x). It is considered as the fraction of total molecules
that actually, dissociate into the simpler molecules x has no units. If x is
the degree of dissociation then for completely dissociating molecules x = 1.0.
For all dissociations involving equilibrium state, x is a fractional value. If
x is known, K_{c} or K_{p} can be calculated and vice-versa.
Equilibrium constants in terms of degree of
dissociation
Formation of HI from H_{2} and I_{2}
The formation of HI from H_{2} and I_{2}
is an example of gaseous homogeneous equilibrium reaction. It can be
represented as
H_{2(g)} + I_{2(g)-- >} _{<
-- }2HI_{(g)} ∆H=-10.4 kJ
This
equilibrium is an exothermic one.
Let us consider that one mole of H_{2}
and one mole of I_{2} are present initially in a vessel of volume V dm^{3}.
At equilibrium let us assume that x mole of H2 combines with x mole of I2 to
give 2x moles of HI. The concentrations of H2, I2 and HI remaining at
equilibrium can be calculated as follows :
H2(g) I2(g) HI(g)
Initial
number of moles I I 0
Number of
moles reacted x x -
Number of
moles remaining at equilibrium 1-x 1-x 2x
Equilibrium
concentration 1-x/V 1-x/1-x 2x/V
According
to the law of mass action,
K_{c} = [HI]^{2} / [H2] [I2]
Substituting
the values of equilibrium concentrations in the above equation, we get
K_{c }= 1/ (1-x)^{2}
If the initial concentration of H_{2} and I_{2} are
equal to a and b moles dm^{-3} respectively, then it can be shown that
_{Kc} _{= }4x^{2 } / (a-x)(b-x)
Derivation of Kp in
terms of x
Let us
consider one mole of H2 and one mole of I2 are present initially. At
equilibrium, let us assume that x mole of H2 combines with x mole of I2 to give
2x moles of HI. Let the total pressure at equilibrium be P atmosphere. The
number of moles of H2, I2 and HI present at equilibrium can be calculated as
follows :
H_{2(g)} I_{2(g)} HI_{(g)}
Initial
number of moles I I O
Number of
moles reacted x x -
Number of
moles I-x I-x 2x
remaining
at equilibrium
The total
number of = 1 - x + 1 - x + 2x = 2
moles at
equilibrium
We know that partial pressure is the product of
mole fraction and the total pressure. Mole fraction is the number of moles of
that individual component divided by the total number of moles in the mixture.
Therefore,
P_{H2}
= (1-x / 2) P
P_{I2}
= (1-x / 2) P
P_{HI}
= (2x / 2) P
We know that K_{p} = P^{2}_{HI} / P_{H2}.P_{I2}
Substituting
the values of partial pressures in the above equation, we get
K_{p}=
(2xp/2)^{2 } / ((1-x)/2)P((1-x)/2)P
K_{p }= 4x^{2} / (1-x)^{2}
we see that K_{p} and K_{e} are equal in terms of x
values. The influence of various factors on the chemical equilibrium can be
explained as below:
Influence of pressure : The
expressions for the equilibrium constants
K_{c} and K_{p} involve neither the pressure nor volume term.
So the equilibrium constants are independent of pressure and volume. Pressure
has therefore no effect on the equilibrium.
Influence of concentration : The
addition of either H_{2} or
I_{2} to the equilibrium mixture well increase the
value of the denominator in the equation Ke = [HI]^{2}/[H2][I2] and
hence tends to decrease the value of Ke. In order to maintain the constancy of
Kc, the increase in the denominator value will be compensated by the
corresponding increase in the numerator value. In other words, the forward
reaction will be favoured and there will be corresponding increase in the
concentration of HI.
Influence of catalyst : A catalyst
affects both the forward and reverse
reactions to the same extent. So it does not change the relative amounts of
reactants and products at equilibrium. The values of Ke and Kp are not
affected. However the equilibrium is attained quickly in the presence of a
catalyst.