Chemistry

Chemical equilibrium-II

Session Objectives

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 Chatellier’s principle.

Types: Equilibrium

Homogeneous equilibria All the reactants and products of an equilibrium reaction are present in the same phase For example, 4 RCOOR ( ) Ester    RCOOH Acid    R ' OH Alcohol  

Types: Equilibrium

Heterogeneous equilibria: All the reactants and products are present in more than one phases For example

CaO (s)

 The concentration of pure solids and pure liquids is taken as 1.

Types: Equilibrium

Heterogeneous equilibria: Thus, K C   CaO (s)    and K = p CO 2 p

Prediction of direction of reaction – reaction quotient

aA + bB Q c  cC + dD, d b When, Q c > K c — (Backward reaction will occur, the reaction will proceed in direction of reactants) When, Q c < K c — (Forward reaction will occur, the reaction will proceed in direction of products) When, Q c = K c — The reaction will be in equilibrium

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.

Calculation of Equilibrium concentration

(i) When Δ n = 0 Considering the equilibrium 2 2HI(g) Let the reaction starts with ‘a’ moles of H moles of I 2 ‘x’ moles of H 2 is attained, then ‘x’ moles of I 2 2 and ‘b’ taken in a container of volume V litres. have reacted when the equilibrium will also react and 2x moles of HI are produced at T K temperature.

Calculation of Equilibrium concentration

Initial moles 1 Moles at eqm a – x (a  x) Conc. at eqm or K 1    V     V 2x v       2   V    2 1 b – x (b  x) V 2HI(g) 0 2x (2x) V 4x 2

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 (a- x) (a+b) (b- x) (a+b) , Partial pressure of H 2  and HI  2x  P  P, I 2  HI= (2x) (a+b)  P,

Calculation of Equilibrium concentration

K P      2x .P

    P  2 .P

   4x 2 1. K P = K C .

2. No volume or pressure term is involved in the expression for K P or K C .

3. Pressure has no effect on the state of equilibrium.

Reactions in which 2 Suppose initially one mole of PCl 5 is present in a vessel of V litres and x moles of PCl 5 dissociates at equilibrium at T K temperature. As per stoichiometry of the reaction, ‘x’ moles of PCl 5 dissociates to give ‘x’ moles of PCl 3 and ‘x’ moles of Cl 2 at equilibrium, Initial moles

Conc. at eqm.

v

1 0

x v

2 0

x v

Reactions in which K C  x x v  1 x v    x 2 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   1- x 1+x   P   x 1+x   P PCl =   x 1+x   P

Reactions in which x 1+x  P × 1- x  1+x x 1+x P  P =   x 2 (1- x) (1+x)   P =   x 2   P If the degree of dissociation is very small then Hence K P = x 2 ·P 1 – x 2  1 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 PCl 5 favours the forward while the increase in the concentration of PCl 3 or Cl 2 favours the backward reaction.

Characteristics of Equilibrium constant (K) For example in the reaction 2 K = [H O] 2 2 If the same reaction is written as 1 2

   

1 2 i.e, for this reaction which can be balanced in two ways K   K

Degree of dissociation and vapour Density

PCl 5 Initial conc.

1 Conc. at eqm. (1 – a ) PCl 3 + Cl 2 0 a 0 a Where a = degree of dissociation Total number of moles at equilibrium = 1 + a From ideal gas equation, PV = nRT = W RT M M   RT P

Degree of dissociation and vapour Density

\ Vapour density =  RT 2P [M = 2 × V.D.]  RT Again, Vapour density = \ Vapour density  2nRT 1  V   V 2n Number of moles Let initial and equilibrium vapour density are D and d respectively, then D d  1  a 1 \ a  D d This is valid for those equilibrium where K p exists D  d d

Question

Illustrative example 1

The vapour density of PCl temperature.

5 at 250° C at equilibrium is found to be 57.9. Calculate percentage dissociation at this

Solution:

Degree of dissociation is related with vapour density as a   D  d  D  \ a  M PCl 5 2 104.25

   208.5

2  57.9

 57.9

 104.25, d 0.8

 80%  57.9, n  2 dissociated.

Free energy change and equilibrium constant



G ° = – RT ln K

Where K is equilibrium constant, R is gas constant and T is absolutely temperature.

 G 0  s tandard free energy change of the reaction

Question

Free energy change and equilibrium constant



G ° = – RT ln K

Where K is equilibrium constant, R is gas constant and T is absolute temperature.

 G 0  s tandard free energy change of the reaction

Le Chatelier’s 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.

PCl 5 PCl 3  Cl 2 Let a, b and c moles of PCl eqm. Pressure.

5 , PCl 3 , Cl 2 respectively are present at equilibrium and P is

Le Chatelier’s principle K p  p PCl 3  p PCl 5 p Cl 2    a b c   a  P     a a c c  P   c  P   K p  bc a  a P c

Le Chatelier’s principle P 



a V (Ideal gas equation.) \ K p  bc a  RT V   Moles of gaseous reac tan ts

Effect of concentration

 Increase in concentration of reactants and decrease in product the reaction moves to forward direction and vice-versa.

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

Effect of inert gas

Inert gas at constant volume there will be no effect on the equilibrium Inert gas at constant pressure For reactions  n > 0, the equilibrium shifts to greater number of moles and vice versa.

If  n = 0, there will be no effect.

Effect of temperature

l og K 2 K 1   H   1 1  1 T 2    , T 2  T 1 For an endothermic reaction, (  H is +ve) K 2 > K 1 , i.e. the reaction will move towards forward direction. Favoured by high temperature While for an exothermic reaction (  H is –ve), then K 2 < K 1 and the reaction moves toward backward direction.

Favoured by low temperature

Class exercise

Class exercise 1

In the following gaseous equilibrium p 1 , p 2 and p 3 are partial pressures 2

2XY

2 p 1 p 2 p 3 The value of Kp: (a) 2p 3 (b) 1 2 p 3 (c)   2 2 2 (d)

Solution:

K P  p 1

 

p 2

   

2 Hence, the answer is (c).     2 2 2

Class exercise 2

The equilibrium constant for the reaction W + X Y + Z is 9. If one mole of each of W and X are mixed and there is no change in volume, number of moles of Y formed is (a) 0.10

(b) 0.50

(c) 0.75

(d) 0.90

Solution

W + X 1 Y + Z 1 0 0 1 – x 1 – x x x then x 2  1- x  2 =9 x = 3 – 3x x = 0.75

Hence, the answer is (c).

Class exercise 3

The equilibrium constant at 323°C is 1000. What would be the value in the presence of a catalyst for the following reaction

A + B C + D + 38 Kcal

(a) 1000 (Concentration of catalyst) (b) 1000 (c) 1000/Catalyst (d) Cannot be determined

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 K C .

Hence, the answer is (b).

Class exercise 4

The K eq for the dissociation of iodine 2I(g) If the equilibrium concentration of atomic iodine is -2 What is the concentration of molecular iodine?

(a) 0.8M

(c) 0.3 M (b) 0.4 M (d) 0.2 M

Solution

    I 2 2   4×10 -2 4×10 -3   = 4×10 -3 2 = 0.4 M Hence, the answer is (b).

Class exercise 5

NO 2 associates (or dimerises) as 2NO 2 N 2 O 4 the apparent molecular weight of a sample of NO 2 calculated from the vapour density under certain conditions was 60, the mole fraction of dimer is (a) 14/46 (b) 16/46 (c) 28/46 (d) 46/60

Solution

2NO 2 N 2 O 4 Initial 1 0 At eqm. 1–2x x (moles) n Total M  M eqm.

 60 Now, V.D

i V.D

e  M i M e  n e n i 14 60   V.D

 1 no. of moles  

Solution

Mole fraction of the dimer  x   14 / 60 46 / 60

14 46

Hence, answer is (a).

Class exercise 6

At 700 K, hydrogen and bromine react to form hydrogen bromide. The value of equilibrium constant for this reaction is 5 10 8 . Calculate the amount of H 2 , Br 2 , and HBr at equilibrium if a mixture of 0.6 mol of H 2 and 0.2 mol of Br 2 is heated to 700 K.

Solution

2 2HBr Initial 0.6 0.2 0 At eqm. 0.6–x 0.2–x 2x given      2    8 Such a large value indicates near completion reaction H 2 = 0.6 – 0.2 = 0.4 moles Br 2 = 0 moles HBr = 0.4 moles

Class exercise 7

The K p for the reaction 2NO 2 is 640 mm of Hg at 775 K. Calculate the percentage dissociation of N be 50% 2 O 4 at equilibrium pressure of 160 mm of Hg . At what pressure, the dissociation will

Solution:

Initial At eqm. 1 0 2 1 – a 2a

solution

   1 2 a  a    1 1  a  a .160

      2   640  4 a   a 2  a 2 a  1 2  0.707

\ % dissociation of N2O4 = 0.707 × 100 = 70.7% Let P be the required pressure 4α 2  1- α 2  . P = 640 a = 0.5

 P 0.75

= 640  P = 480 mm of Hg

Class exercise 8

The degree of dissociation of N 2 O 4 into NO 2 at one atmosphere and 40°C is 0.310. Calculate its K 40°C. Also report the degree of dissociation at 10 atm pressure and same temperature.

p at

Solution:

2NO 2 Initial 1 Final 1–0.31 0 0.62

Total number of moles = 1.31 moles

Solution

K  0.62

1.31

0.69

1.31

 2  0.425 atm At 10 atm 2 1 a 1 1  a  a  a . 10  2 .10

0.425

 94.12

a 2 =1 a 2 a = 0.1025=10.25%  4 a 2 1  a 2  0.0425

Class exercise 9

In an equilibrium A + B C + D A and B are mixed in a vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the equilibrium has reached, concentration of C was thrice the equilibrium concentration of B. Calculate K C .

Solution

A + B C + D Initial 2x x 0 0 Final 2x – y x – y y y x y  y  3 K C      3x 5x 4 4  2  y  3x 4  9 5  1.8

Class exercise 10

       What will be the effect on the reaction equilibrium of (i) increased temperature (ii) decreased pressure (iii) presence of a catalyst (iv) lower concentration of N 2

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 N of reactants and reaction will move towards backward direction.

2 indicates lower concentration

Thank you