Chemical equilibrium-II

1
(No Transcript)
2
Chemical equilibrium-II
3
(No Transcript)
4
Session Objectives

1. Homogeneous equilibria
2. Heterogeneous equilibria
3. Prediction of the direction of a reaction
   reaction quotient.
4. Characteristics of equilibrium constant (K).
5. Calculation of equilibrium concentrations.
6. Degree of dissociation and vapour density.
7. Le Chatelliers principle.

5
Types Equilibrium
Homogeneous equilibria
All the reactants and products of an equilibrium
reaction are present in the same phase
For example,
6
Types Equilibrium
Heterogeneous equilibria
All the reactants and products are present in
more than one phases
For example
The concentration of pure solids and pure
liquidsis taken as 1.
7
Types Equilibrium
Heterogeneous equilibria
8
Prediction of direction of reaction reaction
quotient
When, Qc gt Kc (Backward reaction will occur,
the reaction will proceed in direction of
reactants) When, Qc lt Kc (Forward reaction
will occur, the reaction will proceed in
direction of products) When, Qc Kc The
reaction will be in equilibrium
9
Characteristics of Equilibrium constant (K)

1. Its value is independent of original
   concentration of reactants, pressure or
   presence of a catalyst.
2. It is independent of the direction from which
   equilibrium is attained.
3. Its value is constant at certain temperature and
   would change with temperature.
4. The larger the value of K, greater will be the
   value of products as compared to reactants.
5. Its value, however, depends on the coefficients
   of balanced reactants and products in the
   chemical equation.

10
Calculation of Equilibrium concentration
Considering the equilibrium
Let the reaction starts with a moles of H2 and
b moles of I2 taken in a container of volume V
litres. x moles of H2 have reacted when the
equilibrium is attained, then x moles of I2
will also react and 2x moles of HI are produced
at T K temperature.
11
Calculation of Equilibrium concentration
Initial moles 1 1 0 Moles at eqm a x b
x 2x
12
Calculation of Equilibrium concentration
Suppose the total pressure at equilibrium is P
atm. and total moles at equilibrium
a x b x 2x a b
13
Calculation of Equilibrium concentration
1. KP KC. 2. No volume or pressure term is
involved in the expression for KP or
KC. 3. Pressure has no effect on the state of
equilibrium.
14
Reactions in which
Suppose initially one mole of PCl5 is present in
a vessel of V litres and x moles of PCl5
dissociates at equilibrium at T K temperature. As
per stoichiometry of the reaction, x moles of
PCl5 dissociates to give x moles of PCl3 and
x moles of Cl2 at equilibrium,
Initial moles 1 0 0
15
Reactions in which
If the total equilibrium pressure is p and
total moles at equilibrium are 1 x x x 1
x,
The partial pressure of each gas is
16
Reactions in which
1. Equilibrium is affected by the pressure. An
increase in the value of P will prefer
backward reaction and the value of x
decreases. 2. Increase in the conc. of PCl5
favours the forward while the increase in the
concentration of PCl3 or Cl2 favours the backward
reaction.
17
Characteristics of Equilibrium constant (K)
For example in the reaction
If the same reaction is written as
i.e, for this reaction which can be balanced in
two ways
18
Degree of dissociation and vapour Density
Initial conc. 1 0 0Conc. at eqm. (1
a) a a
Where a degree of dissociation Total number of
moles at equilibrium 1 aFrom ideal gas
equation,
PV nRT
19
Degree of dissociation and vapour Density
\ Vapour density
M 2 V.D.
Again, Vapour density
\ Vapour density
Let initial and equilibrium vapour density are D
and d respectively, then
This is valid for those equilibrium where Kp
exists
20
(No Transcript)
21
Illustrative example 1
The vapour density of PCl5 at 250 C at
equilibrium is found to be 57.9. Calculate
percentage dissociation at this temperature.
Solution
Degree of dissociation is related with vapour
density as
22
Free energy change and equilibrium constant
DG RT ln K
Where K is equilibrium constant, R is gas
constant and T is absolutely temperature.
23
(No Transcript)
24
Free energy change and equilibrium constant
DG RT ln K
Where K is equilibrium constant, R is gas
constant and T is absolute temperature.
25
Le Chateliers principle
According to this principle, When the
equilibrium is disturbed in a chemical reaction
by changing any external factor such as,
concentration, pressure, temperature etc. then
the equilibrium will be shifted in a direction to
minimize the effect of the change.
Considering the following example.
Let a, b and c moles of PCl5, PCl3, Cl2
respectively are present at equilibrium and P is
eqm. Pressure.
26
Le Chateliers principle
27
Le Chateliers principle
(Ideal gas equation.)
Effect of concentration
Increase in concentration of reactants and
decrease in product
the reaction moves to forward direction and
vice-versa.
28
Effect of pressure/volume
If the pressure decreases (increasing the
volume)
The reaction moves to more no. of moles of gas
With increased pressure
Effect will be opposite
29
Effect of inert gas
Inert gas at constant volume
there will be no effect on the equilibrium
Inert gas at constant pressure
For reactions Dn gt 0, the equilibrium shifts to
greater number of moles and vice versa.
If Dn 0, there will be no effect.
30
Effect of temperature
For an endothermic reaction, (DH is ve) K2 gt
K1, i.e. the reaction will move towards forward
direction. Favoured by high temperature While
for an exothermic reaction (DH is ve), thenK2 lt
K1 and the reaction moves toward backward
direction.Favoured by low temperature
31
(No Transcript)
32
Class exercise 1
Solution
Hence, the answer is (c).
33
Class exercise 2
34
Solution
1 1 0 0 1 x 1 x
x x
x 3 3x x 0.75
Hence, the answer is (c).
35
Class exercise 3
36
Solution
A catalyst only speeds up or slows down the rate
of reaction in a particular direction. Its only
the temperature that can change the value of KC.
Hence, the answer is (b).
37
Class exercise 4
38
Solution
0.4 M
Hence, the answer is (b).
39
Class exercise 5
40
Solution
Initial 1 0
At eqm. 12x x (moles)
41
Solution
Hence, answer is (a).
42
Class exercise 6
At 700 K, hydrogen and bromine react to form
hydrogen bromide. The value of equilibrium
constant for this reaction is 5 108. Calculate
the amount of H2, Br2, and HBr at equilibrium if
a mixture of 0.6 mol of H2 and 0.2 mol of Br2 is
heated to 700 K.
43
Solution
Such a large value indicates near completion
reaction H2 0.6 0.2 0.4 moles Br2 0
moles HBr 0.4 moles
44
Class exercise 7
Solution
45
solution
46
Class exercise 8
The degree of dissociation of N2O4 into NO2 at
one atmosphere and 40C is 0.310. Calculate its
Kp at 40C. Also report the degree of
dissociation at 10 atm pressure and same
temperature.
Solution
Total number of moles 1.31 moles
47
Solution
48
Class exercise 9
49
Solution
Initial 2x x 0
0 Final 2x y x y y y
50
Class exercise 10
51
Solution

1. Since it is an endothermic reaction, increase in
   temperature will favour forward reaction.
2. If the pressure will be decreased, then reaction
   will move to the more number of moles, both
   sides have same number of moles. So no effect of
   pressure.
3. Presence of a catalyst does not affect the
   position of equilibrium
4. Lower concentration of N2 indicates lower
   concentration of reactants and reaction will move
   towards backward direction.

52
(No Transcript)