you will find it help ful

2. Equilibrium is a state in which there are no
observable changes as time goes by.
When a chemical reaction has reached the
equillibrium stae,the concentrations of reactants
and products remain contant over time and there
are no visible changes in the system.

3.  In reversible reaction like-
R + R’ P + P’
Initially only reactants are present
R and R’ conbine to form P and P’.As soon as
P and P’ are formed they start the backward
reaction.As concentrations of R and R’ decrease
rate of forward reaction decreasesand rate of
backward reaction increases.Ultimately a stage is
reached when both the rates become equal.Such a
state is known as ‘CHEMICAL
EQUILIBRIUM’.

4. Attainment of chemical equilibrium

5. 1. Equilibrium is Dynamic in nature.
2. Equilibrium can be observed by constancy of some
observable properties like
colour,pressure,concentration,density,temperature etc.which
may be suitable in a given reaction.
3. Equilibrium state can be affected by altering factors like
pressure,volume,concentration and temperature.

6. At a given temperature,the product of concentrations of
the reaction products raised to the rerespective
stoichiometric coefficient in the balanced chemical
equation divided by the product of concentrations of the
reactants raised to their individual coefficients has a
constant value.This is known as the equilibrium law or
law of chemical equilibrium.

7. For a general reaction :
aA + bB <--> cC + dD
the equillibrium constant can be defined as:
is a constant and is called the equilibrium constant in terms of
concentration, where all the concentrations are at equilibrium and are
expressed in moles per litre.

8. EXAMPLE
Q.1 The following concentrations were
obtained for the formation of NH3 from N2
and H2 at equilibrium at 500K.
[N2 ] = 1.5 X 10-2 M
[H2] = 3.0 X 10-2 M
[NH3] = 1.2 X 10-2 M
Calculate equilibrium constant.

9. SOLUTION
The equilibrium constant for the reaction
3 H2 + N2 → 2 NH3 can be written as:
Kc = [NH3]2 / [H2]3 [N2]
(1.2 X 10-2)2 / (1.5 X 10-2)(3.0 X 10-2)3
=
0.106 x 104 = 0.106 x 103

10. Kp in homogeneous gaseous equilibria
A homogeneous equilibrium is one in which everything in the equilibrium mixture is
present in the same phase. In this case, to use Kp, everything must be a gas.
A good example of a gaseous homogeneous equilibrium is the conversion of sulphur
dioxide to sulphur trioxide at the heart of the Contact Process:
Writing an expression for Kp
If you allow this reaction to reach equilibrium and then measure (or work out) the
equilibrium partial pressures of everything, you can combine these into the equilibrium
constant, Kp.

11. RELATION BETWEEN KP AND KC
 PV = nRT
P=CRT where C= n/v = (moles per litre)
Pc=[C] RT;
Pd=[D] RT;
PA=[A] RT;
PB=[B] RT;
KP =[C]c(RT)c [D]d(RT)d / [A ]a (RT)a[B]b(RT)b
=[C]c [D]d (RT)(c+d) - (a+b)
[A]a [B]b

12. Kp = Kc (RT) n
where ng = (c+d) – (a+b), calculation of n
involves only gaseous components.
n = sum of the no. of moles of gaseous products- sum of the no.
of moles of gaseous reactants

13. 1. Equilibrium constant is applicable only when
concentrations of the reactants and products have attained
their equilibrium state.
2. The value of equilibrium constant is independent of initial
concentrations of the reactants and products.
3. Equilibrium constant is temperature dependent.
4. The equilibrium constant for the reverse reaction is equal to
the inverse of the equilibrium constant for the forward
reaction.

14. 5. The equilibrium constant K for a reaction is related to the equilibrium
constant of the corresponding reaction,whose equation is obtained by
multiplying or dividing the equation for the original reaction by a small
integer.
LET US CONSIDER APPLICATIONS OF EQUILIBRIUM
CONSTANT TO:
1.Predict the extent of a reaction on the basis of its magnitude.
2.Predict the direction of the reaction.
3.Calculate equilibrium concentrations.

15.  K = [Product]eq / [Reactant]Eq
 CASE 1
If K is large (K>103) then the product concentration is
very very larger than the reactant {[Product] >>
[Reactant]}.Hence concentration of reactant can be
neglected with respect to the product. In this case, the
reaction is product favourable and equilibrium will be
more in forward direction than in backward direction.
 CASE 2
If K is small (K<10-3)

16.  Hence the concentration of product can be neglected as
compared to the reactant.
In this case ,the reaction is reactant is favourable.

17.  Reaction Quotient (Q)
 At each point in the reaction , we can write a ratio of
concentration terms having the same form as the
equilibrium constant expression.This ratio is called the
reaction qoutient denoted by symbol Q.
It helps in predicting the direction of a reaction.
 The expression Q = [C]c [D]d / [A]a [B]b at any time
during the reaction is called reaction quotient.The
concentrations [C] , [D] , [A] , [B] are not necessarily at
equilibrium.

18.  If Q > Kc reaction will proceed in backward direction
until equilibrium is reached.
 If Q < Kc reaction will proceed in forward direction
until equilibrium is established.
 If Q = Kc reaction is at equilibrium.

19. Q. The value of Kc for a reaction
2A B + C is 2 x 10 -3. At a given time,the
composition of reaction mixture is [A]=[B]=[C] =
3x10-4 M. In which direction the reaaction will
proceed?
ANS : For the reaction the reaction quotient Q is given
by,
Qc = [B][C]/[A]2
As [B] = [C] = [A] = 3x10-4 M
Qc = (3x10-4)(3x10-4) / (3x10-4)2 = 1
As Qc > Kc so the reaction will proceed in the reverse
direction

20.  The concentration of various reactants and products can
be calculated using the equilibrium constant and the
initial concentrations.
In case of a problem in which we know the initial
concentrations but do not know any of the equilibrium
concentrations,the following three steps shall be followed :
1.Write the balanced chemical equation for the reaction.
2.Under the balanced equation,make a table that lists for
each substance involved in the reaction:
(d)The initial concentration
(e) the change in concentration on going to equilibrium
(f) the equlibrium concentration

21. 3. Substitute the equlibrium concentrations into the
equilibrium equation for the reaction and solve for x.If you
are to solve a quadratic equation choose the mathematical
solution that makes chemical sense.
4. Calculate the equilibrium concentrations from the
calculated value of x.
5. Check your results by substituting then into the
equilibrium equation.

22. Q.13.8g of N2 O4 was placed in a 1L reaction vessel at 400K and
allowed to attain equilibrium.
N2 O 4 2NO2
The total pressure at equilibrium was found to be 9.15 bar.Calculate KC , Kp and
Partial pressure at equilibrium.
We know pV = nRT
Total volume = 1L
Molecular mass of N2O4 = 92g
Number of moles = 13.8 / 92 = 0.15
Gas constant = 0.083 bar L mol-1 K-1
Temperature = 400K
pV=nRT
P = 4.98

23. N2O4 2NO2
Initial pressure 4.98 0
At equilibrium 4.98 – x 2x
Hence,
Ptotal at equilibrium = P (N2O4) + P(NO2)
9.15 = 4.98-x + 2x
X = 9.15 – 4.98 = 4.17.
Partial pressures at equilibrium are,
P(N2O4) = 4.98 – 4.17 = 0.81 bar
P(NO2) = 2x = 2 x 4.17 = 8.34 bar
Kp = [P(NO2)]2 / P(N2O4)
= (8.34)2 / 0.81 = 85.87
KP = Kc (RT) n
85.87 = Kc (0.083 X 400)
Kc = 2.586 = 2.6

24.  Effect of concentration change
 If the concentration of a component is
increased,reaction shifts in a direction which
tends to decrease its concentration.
Eg : N2 + 3H2 2NH3
[Reactant] Forward shift
[Product] Backward shift

25.  If concentration of reactant is increased at
equilibrium then reaction shifts in the forward
direction.
 If concentration of product is increased then
reaction shifts in the backward direction.

26.  On increasing pressure,equilibrium will shift in the direction in
which pressure decreases i.e no. of moles in the reaction decreases
and vice versa.
 For n = 0 (No effect)
 For n>0
 If P ; QP (Forward shift)
 If P ; Qp (Backward shift)

27.  For n<0
 P QP (Forward shift)
 P QP (Backward shift)

28. EFFECT OF INERT GAS
ADDITION
 At constant volume:
Inert gas addition has no effect at constant volume.
 At constant pressure:
If inert gas is added then to maintain the pressure
constant,volume is increased.Hence equilibrium
will shift in the direction in which larger no. of
moles of gas is formed.
f ng > 0 reaction will shift in forward direction
i ng < 0 reaction will shift in backward direction
l ng = 0 no effect

29. EFFECT OF TEMPERATURE
CHANGE
 Whenever an equilibrium is disturbed by a
change in concentration , pressure or volume,
the composition of the equilibrium mixture
changes because the reaction quotient Qc no
longer equals the equilibrium constant Kc.
However,when a change in temperature occurs
the value of equilibrium constant is changed.
 In general the temperature dependence of the
equilibrium constant depends on the sign of
H for the reaction.
 The equilibrium constant for an exothermic
reaction decreases as the temperature
increases.

30.  The equilibrium constant for an endothermic reaction
increases as temperature increases.

31. EFFECT OF CATALYST
 Due to catalyst,the state of equilibrium is not
affected i.e no shift will occur as catalyst lowers
the activation energy of both the forward and
reverse reaction by same amount,thus altering
the forward and reverse rate equally and
hence,the equlibrium will be altered faster.

32. LE CHATELIER’S PRINCIPLE
 It states that a change in any of the factors that
determine the eqiulibrium conditions of a system will
cause the system to change in such a manner so as to
reduce or to counteract the effect of the change.