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

All the reactants and products of an equilibrium reaction are present in the same phase

2 4 2N O (g) 2 NO (g)

2Ester Acid Alcohol

RCOOR ( ) H O RCOOH R 'OH

Homogeneous equilibria

For example,

Types: Equilibrium

All the reactants and products are present in more than one phases

3 2CaCO (s) CaO (s) CO (g)

The concentration of pure solids and pure liquidsis taken as 1.

Heterogeneous equilibria:

For example

Types: Equilibrium

2C 2

3

CaO (s) CO (g)Thus, K [CO (g)]

CaCO (s)

Heterogeneous equilibria:

2P COK = pand

Where is partial pressure of carbon dioxide.2COp

Prediction of direction of reaction – reaction quotient

aA + bB cC + dD,

c d

c a b

[C] [D]Q

[A] [B]

When, Qc > Kc — (Backward reaction will occur, the reaction will proceed in direction of reactants)

When, Qc < Kc — (Forward reaction will occur, the reaction will proceed in direction of products)

When, Qc = Kc — 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

(ii) When Δ n 0

Reactions in which n = 0

Considering the equilibrium

2 2H (g) + I (g) 2HI(g)

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.


2 2H (g) + I (g) 2HI(g)

Initial moles 1 1 0

Moles at eqm a – x b – x 2x (a x) (b x) (2x)

Conc. at eqmV V V

2

2

1

2x4xv

or Ka x b x (a x) (b x)

V V

Calculation of Equilibrium concentrationSuppose the total pressure at equilibrium is ‘P’ atm. and total moles at equilibrium

= a – x + b – x + 2x = a + b

2 2(a- x) (b- x) (2x)

Mole fraction of H = , I = , HI =(a+b) (a+b) (a+b)

2 2(a x) (b x)

Partial pressure of H P, I P,(a b) (a b)

2xand HI P

(a b)


2

2

P

2xP

(a b) 4xK

(a x) (b x)(a x) (a x).P .P

(a b) (a b)

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.

Note that for reactions having n= 0

Reactions in which n 0

5 3 2 PCl (g) PCl (g) + Cl (g)

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,

5 3 2 PCl (g) PCl (g) + Cl (g)

Initial moles 1 0 01 x x x

Conc. at eqm.v v v

Reactions in which

2

C

x xxv v

K1 x v(1 x)

v

If the total equilibrium pressure is ‘p’ and total moles at equilibrium are 1 – x + x + x = 1+ x,

51- x

PCl = P1+x

n 0

The partial pressure of each gas is

3x

PCl = P1+x

xPCl = P

1+x

Reactions in which

2 2

P 2

x xP× P

x x1+x 1+xor K = = P = P1- x (1- x) (1+x) (1- x )P1+x

n 0

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.

If the degree of dissociation is very small then Hence KP = x2·P

21 – x 1

Characteristics of Equilibrium constant (K)

For example in the reaction

2 2 22H +O 2H O

222

2 2

[H O]K =

[H ] . [O ]

If the same reaction is written as

2 2 21

H + O H O2

2

122 2

[H O]K =

H . O

i.e, for this reaction which can be balanced in two waysK K

Degree of dissociation and vapour Density

PV = nRT =W

RTM

RTM

P

PCl5 PCl3 + Cl2

Initial conc. 1 0 0Conc. at eqm. (1 –)

Where = degree of dissociation

Total number of moles at equilibrium = 1 + From ideal gas equation,

Degree of dissociation and vapour Density

Vapour density = RT

2P[M = 2 × V.D.]

Again, Vapour density =

RT V

V2nRT 2n

1Number of moles

Vapour density

Let initial and equilibrium vapour density are D and d respectively, then

D 1

d 1

D D d1

d d

This is valid for those equilibrium where Kp exists

Question

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.

Degree of dissociation is related with vapour density as

D d

n 1 d

5PClM 208.5D 104.25, d 57.9, n 2

2 2

104.25 57.90.8 80%

2 1 57.9dissociated.

Solution:

Free energy change and equilibrium constant

G° = – RT ln K

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

0G 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.

0G s tandard free energy change of the reaction

Le Chatelier’s principle

Considering the following example.

5 3 2PCl g PCl g Cl g

Let a, b and c moles of PCl5, PCl3, Cl2 respectively are present at equilibrium and P is eqm. Pressure.

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.


3 2

5

PCl Clp

PCl

p pK

p

b cP P

a b c a b c

aP

a b c

p

bc PK

a a b c


n 1

pbc RT

Ka V

a b c RTP

V(Ideal gas equation.)

n Moles of gaseous products

Moles of gaseous reac tants

the reaction moves to forward direction and vice-versa.

Effect of concentrationIncrease in concentration of reactants and decrease in product

Effect of pressure/volume

If the pressure decreases (increasing the volume)

With increased pressure

Effect will be opposite

The reaction moves to more no. of moles

of gas

Effect of inert gas

Inert gas at constant volume

For reactionsn > 0, the equilibrium shifts to greater number of moles and vice versa.

there will be no effect on the equilibrium

Inert gas at constant pressure

If n = 0, there will be no effect.

Effect of temperature

2

2 11 1 2

K H 1 1log , T T

K 2.303 R T T

For an endothermic reaction, (H is +ve) K2 > K1, i.e. the reaction will move towards forward direction.

Favoured by high temperature

While for an exothermic reaction (H is –ve), thenK2 < K1 and the reaction moves toward backward direction.

Favoured by low temperature

Class exercise

Class exercise 1

In the following gaseous equilibrium p1, p2 and p3 are partial pressures

2 2 2X + 2Y 2XY

p1 p2 p3

The value of Kp:

3

1 2

2p(a)

p p1 2

3

p p(b)

p

23

21 2

p(c)

p p

21 2

23

p p(d)

p

23

P 21 2

pK

p p

Hence, the answer is (c).

Solution:

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 Y + Z

1 1 0 0

1 – x 1 – x x x

2

2x

then =91- x

x = 3 – 3x

x = 0.75

Hence, the answer is (c).

Class exercise 3The equilibrium constant at 323°C is 1000. What would be the value in the presence of a catalyst for the following reactionA + B C + D + 38 Kcal

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

SolutionA 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).

Class exercise 4The Keq for the dissociation of iodine

2I (g) 2I(g)

If the equilibrium concentration of atomic iodine is -2 4 10 M

What is the concentration of molecular iodine?

(a) 0.8M (b) 0.4 M(c) 0.3 M (d) 0.2 M

Solution

2-3

eq2

IK = =4×10

I

2-2

2-3

4×10= I

4×10

= 0.4 M

Hence, the answer is (b).

Class exercise 5NO2 associates (or dimerises) as

the apparent molecular weight of a sample of NO2 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

2NO2 N2O4

Solution

2NO2 N2O4

Initial 1 0

At eqm. 1–2x x(moles)

Totaln 1 2x x 1 x

2i(NO )M 14 32 46

eqm.M 60

ei i

e e i

n 1V.D MNow, V.D

V.D M n no. of moles

461 x

60 14

x60

Solutionx

Mole fraction of the dimer1 x

14 / 6046 / 60

1446

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 5 108. 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.

Solution

Initial 0.6 0.2 0At eqm. 0.6–x 0.2–x 2x

282x

given 5 100.6 x 0.2 x

Such a large value indicates near completion reaction

H2 = 0.6 – 0.2 = 0.4 moles

Br2 = 0 moles

HBr = 0.4 moles

2 2H + Br 2HBr

Class exercise 7

The Kp for the reaction

2 4 2N O 2NO

is 640 mm of Hg at 775 K. Calculate the percentage dissociation of N2O4 at equilibrium pressure of 160 mm of Hg . At what pressure, the dissociation will be 50%

Initial 1 0

At eqm. 1 – a 2a

Where, is the degree of dissociation.

Solution:

2 4 2N O 2NO

solution

22

. 1601

6401

.1601

2 24 4 1

28 = 4 1

0.7072

% dissociation of N2O4 = 0.707 × 100 = 70.7%

Let P be the required pressure

2

2

4α. P=640

1- α

P= 0.5 = 640

0.75 P = 480 mm of Hg

Class exercise 8

The degree of dissociation of N2O4 into NO2 at one atmosphere and 40°C is 0.310. Calculate its Kp at 40°C. Also report the degree of dissociation at 10 atm pressure and same temperature.

Solution:

Initial 1 0

Final 1–0.31 0.62

Total number of moles = 1.31 moles

2 4 2N O 2NO

Solution

20.621.31

K 0.425 atm0.691.31

22

. 101

At 10 atm 0.4251

.101

2

2

40.0425

1

2 294.12 =1-

= 0.1025=10.25%

Class exercise 9

In an equilibriumA + 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 KC.

Solution

A + B C + D

Initial 2x x 0 0

Final 2x – y x – y y y

y 3x

3 yx y 4

2

C

3x94

K 1.85x x 54 4

Class exercise 10

2 2N g O g 2NO g , H 179 kJ

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 N2

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.

Thank you

Chemistry

Documents

Transcript of Chemistry