Chemistry. Chemical equilibrium-I Session Objectives

Chemistry

Chemical equilibrium-I

Session Objectives

Session objectives

1. Dynamic nature of equilibrium

2. Equilibrium in physical processes.

3. General characteristics of equilibrium involving physical processes.

4. Equilibrium in chemical process

5. Law of mass action

6. Equilibrium constant in gaseous systems. KP, KC

7. Relation between KP and KC

8. Application of equilibrium constant

Dynamic equilibrium

At equilibrium, the reaction does not stop. moles of reactant converted into product = moles of product converted into reactant in the same time.It is considered as dynamic equilibrium.

A B+ C + D

Equilibrium in Physical Processes Solid-Liquid Equilibrium

This state is represented as:

2 2 H O (s) H O (l) At 0°C

ice   water

At equilibrium: Rate of melting = Rate of freezing

Liquid-Gas Equilibrium

2 2H O (l) H O (g)

ice   water vapour

Rate of evaporation = Rate of condensation.

Gases in Liquids

2 2CO (gas) CO  (dissolved in solution)

At equilibrium

Henry’s law

or m kp m p

Where,

m is the mass of gas

p is the applied pressure

k is the Henry’s constant.

Equilibrium in Chemical Process

The state in which both the reactants and products co-exist considered chemical equilibrium.

Forward reaction

Backward reaction

Rate

of

react

ion

Time

Equilibrium in Chemical Process

Rate of forward reaction = Rate of backward reaction.

Rate of formation of HI = Rate of decomposition of HI.

H2(g) + I2 (g) 2HI(g)

At equilibrium:

At equilibrium:

Characteristics of chemical equilibrium

1. The concentration of each of the reactants and its products becomes constant at equilibrium.

2. At equilibrium, the rate of forward reaction =rate of backward reaction

3. A chemical equilibrium can be attained from either direction, i.e., from the direction of the reactants as well as from the direction of products.

4. At equilibrium the free energy changes of the system, i.e., G 0

Law of Mass Action

Law of mass action states that rate of a chemical reaction is proportional to the product of the active masses of the reactants raise to the power of stoichiometric coefficients of balanced chemical equation.

For example,

aA + bB cC + dD

Law of Mass Action

Rate of forward reaction [A]a [B]b

Rate of forward reaction = kf [A]a [B]b

Similarly

Rate of backward reaction = kb [C]c [D]d

kb = rate constant of backward reaction then

At equilibrium kf [A]a . [B]b = kb [C]c . [D]d. c d

fa b

b

k [C] [D]K

k [A] [B]

KC used for molar concentration

where kf = rate constant of forward reaction

Illustrative example Determine the value of the equilibrium constant for the reaction

+ 2C2BA

if 1.0 mol of A and 1.5 mol of B are placed in a 2.0-litre vessel and allowed to come to equilibrium. The equilibrium concentration of C is 0.35 mol L-1.

Solution

+ 2C2BAInitial 1 1.5 0

At equ. 1–x 1.5 –2x 2x(moles)

1-x 1.5 -2x 2xConc.

2 2 2

2x[C] 0.35

2 x 0.35

1 0.35[A] 0.325

2

1.5 2 0.35[B] 0.4

2

2 2

c 2

[C] (0.35)K

[A][B] 0.325(0.4)

12.36 mol L

Equilibria in gas-phase reactions(Kp)

Let A, B, C and D be gases in the following gaseous equilibrium:

aA  +  bB  cC  +  dD

c dC D

P a bA B

p .pK =

p .p

Where pA, pB, pC, and pD are the partial pressures of the gases A, B, C and D, respectively in the mixture, at equilibrium.

Relation between Kp and Kc

c d

fa b

b

k[C] . [D]K

k[A] . [B]

We know, partial pressure nRT

V

aA + bB cC + dD

AA

n RTp

Vpartial pressure of A ,

Suppose volume is 1 L then

A A Ap C RT, C molar conc. of A

Relation between Kp and Kc

c d

DCP a b

A B

p . pK

p . p

C DC D

p A BA B

[C RT] [C RT]K

[C RT] [C RT]

Similarly for B,C and D

c dc d (a b)C D

a bA b

C CRT

C C

gnp cK K RT

n = (No. of gaseous moles of products) – (No. of gaseous moles of reactants)

Application: Equilibrium Constant

A B C D, [C][D]

K[A][B]

K = Equilibrium constant

(i) C D A B, 1

K 'K

2K '' K

1 1 1 1

(iii) A B C D2 2 2 2

K ''' K

(ii) 2A + 2B 2C + 2D

Application : Equilibrium Constant

13

2

KK '

K

A + B C + D K3 = K1 × K2

(v) And for A – B C – D

(iv) A C K1

B D K2

Questions

Class exercise 1The equilibrium system

was found to contain [SO2] = 0.40 M, [O2] = 0.13 M, [SO3] = 0.70 M. Find Kc:

(a) 12 (b) 23.6

(c) 9 (d) 48

2SO2 (g) + O22SO3(g)

Solution2SO2 (g) + O2

2SO3(g)

At eqm. 0.4 0.13 0.7conc.

23

c 22 2

[SO ]K

[SO ] [O ]

2

2

(0.7)

0.13 (0.4)

23.6

Hence, answer is (b).

Class exercise 2If N2O4 is 25% dissociated in a 4 litre vessel at given temperature. What is the Kc for the reaction

?

(a) 1/16 (b) 1/4

(c) 1/6 (d) 1/12

N2O4 2NO2

SolutionN2O4 2NO2

Initial 1 0At eqm. 1- 2

(moles)

1- 2conc.

4 4

2

c

24

K1

4

Since 0.25

2

c4 (0.25)

K4 0.75

0.25 0.250.75

112

Hence, answer is (d).

Class exercise 3

Calculate the equilibrium constant for the reaction

H g CO g H O g CO g 2 2 2

If the equilibrium constant at 1395 K for the following are:

at 1395 K.

2 2 22H O g 2H g +O g

2 22CO g 2CO g +O g

k1 = 2.1 × 10–13... (i)

k2 = 1.4 × 10–12 ... (ii)

(a) 3.26 (b) 4.28 (c) 1.34 (d) 2.58

Solution

2 2 22H O g 2H g O g

k1 = 2.1 × 10–13... (i)

2 22CO g 2CO g O g

k2 = 1.4 × 10–12... (ii)

Subtracting equation (ii) from equation (i), we get

2 2 22H O- 2CO 2H - 2CO 1

2

KK=

K

2 2 22H O + 2CO 2CO + 2H ... (iii)

Divide equation (iii) by equation (ii), we get

2 2 2H O + CO CO + H K' K

Solution

2 2 2CO H H O CO 1

K ''K '

1K

K ''K '

122

131

K1 1.4 102.58

KK 2.1 10

Class exercise 4Equimolar quantities of HI, H2 and I2are brought to equilibrium. If the total pressure in the vesselis 1.5 atm and Kp for the reaction

2 22HI(g) H (g) + I (g)

is 49. Which of the following is the correct value for the equilibrium partial pressure of I2?

(a) 0.4 (b) 0.9 (c) 0.7(d) 0.5

2 22HI H I

Initial p p p

Final p – 2x p + x p + x

Solution:

solution

3p = 1.5 atm

p = 0.5 atm

2

2

p+x=49

p- 2x

p c[K = K ( ng = 0)]

p + x = 7p – 14x

15x 6p6p 2

x p15 5

2x 0.5 0.2 atm

5

2I = p + x = 0.7 atm

Class exercise 5In the following gaseous equilibrium P1, P2 and P3 are partial pressures of X2, Y2 and XY2 respectively.

2 2 2X + 2Y 2XY

P1 P2 P3

The value of Kp:

3

1 2

2P(a)

P P(b) (P1P2)P3

23

21 2

P(c)

P P

21 2

23

P P(d)

P

2 2 2X 2Y 2XY

23

P 21 2

PK

P P

P1 P2 P3

Solution:

Class exercise 660 ml of H2 and 42 ml of I2 are heated in a closed vessel. At equilibrium the vessel contains 28 ml of HI. Calculate percentage of dissociation of HI.

Solution:

At constant temperature and pressure, Moles volume of gas

2 2H I 2HI

Initial moles 60 42 0 (Given)

At equilibrium 60 – x 42 – x 2x 2x = 2860 – 14 42 – 14 28 x = 14 = 46 = 28

Solution2

c

2828 141

K = = =46 28 46 231 1

Again

2 22HI H + I

Initial moles 1 0 0

Where is degree of dissociation

At eqm. 1 – 2 2

Solution

2

c 2k '

4 1

2

2c

1 23k 14 4 1

On solving

= 0.719 or 71.9%

Class exercise 7The percentage of dissociation is 0.4 at 400 K and 1.0 atm for the gaseous reaction . Assuming ideal behaviour of all the gases, calculate the density of equilibrium mixture at 400 K and 1.0 atmosphere. (Given observed molecular mass at equilibrium = 148.92)

5 3 2PCl PCl Cl

5 3 2PCl PCl Cl

Solution:

Moles at (1 – 0.4) 0.4 0.4equilibrium

Total moles = 1.4

Initial moles 1 0 0

Solution

WPV = RT

M at equilibrium

PM 1×148.92d = = = 4.53 g/ litre

RT 0.0821×400

Class exercise 8

KC for the reaction, is 0.5 mol-2 litre2 at 400 K. Find Kp. Given R = 0.082 litre atm degree–1 mol–1.

2 2 3 N + 3H 2NH

P C

g

K = K (RT) nn = –2

p 2

0.5K

0.0821 400

= 4.636 × 10–4 atm–2

Solution:

Class exercise 9For a reaction,At equilibrium 7.8 g, 203.2 g and 1638.4 g of H2 . I2 and HI were found respectively at equilibrium. Calculate Kc.

2 22HI H + I

Solution:2 22HI H + I

(128) (2) (254)

1638.4 7.8 203.2At eqm.

128 2 254

= 12.8 = 3.9 = 0.8

Since, n = 0

2 2C 2 2

H I 3.9 0.8K = =

12.8HI= 0.019

Class exercise 100.5 moles of H2 and 0.5 moles of I2 react in 10 litre flask at 448°C. The equilibrium constant (KC) is 50 for

2 2H I 2HI

(a) What is the value of Kp?(b) Calculate moles of I2 at equilibrium

nP CK K RT

n = 0 KP = KC = 50

2 2H I 2HI

Initial 0.5 0.5 0

Final 0.5 – x 0.5 – x 2x

Solution:

Solution

22xGiven, 50

0.5 – x 0.5 – x

2x=7.07

0.5– x

Moles of I2 at equilibrium = 0.50 – 0.39 = 0.11 mole

x = 0.39

Thank you