chemical equilibrium for iit jee

ikB’kkyk Tower : Saras Dairy Road, Near ITI, Shastri Circle, Jodhpur

Ph.: 0291-3291970.

1

1. Irreversible and Reversible Reactions Those reactions which proceeds in forward direction and reaches almost to completion are called

irreversible reactions. For example

23(s) (s) 2(g)MnO

2KClO 2KCl 3O (Thermal decomposition)

(aq) (aq) (aq) 2NaOH HCl NaCl H O (Strong acid–strong base neutralisation reaction)

3 2 3 22AgNO BaCl 2AgCl Ba(NO ) (Precipitation reaction)

Whereas, those reactions which proceed in forward and backward directions both and never

reaches completion are called reversible reactions. These reactions can be initiated in any direction.

For example,

5(g) 3(g) 2(g)PCl PCl Cl (Thermal dissociation)

3 3 2

strong baseWeak acid

CH COOH NaOH CH COONa H O (Neutralisation reactions)

2(g) 2(g) 3(g)N 3H 2NH (synthesis reaction)

But when Fe(s) is heated and water vapour is passed over it in open vessel, it is converted to

Fe3O4(s) along with the evolution of hydrogen gas.

(s) 2 (g) 3 4(s) 2(g)3Fe 4H O 3Fe O 4H

and when Fe3O4 is reduced with hydrogen gas, it gives Fe(s) and H2O

3 4(s) 2(g) (s) 2 (g)Fe O 4H 3Fe 4H O

But, if the reaction is carried out in a closed container, this reaction becomes reversible.

(s) 2 (g) 3 4(s) 2(g)3Fe 4H O Fe O 4H

2. State of Equilibrium It has generally been observed that many changes (physical and chemical) do not proceed to

completion when they are carried out in a closed container. Consider for example vapourisation of

water,

Water Vapour

At any temperature, vapourisation of water takes place, initially the concentration of water is much

greater than the concentration of vapour, but with the progress of time, concentration of vapour

increases whereas that of water remains constant and after a certain interval of time, there is no

change in concentration of vapour, this state is known as state of physical equilibrium.

In a similar way, this has also been found for chemical reactions, for example, when PCl5(g) is

heated in a closed container, its dissociation starts with the formation of PCl3(g) and Cl2(g).

Initially, only PCl5(g) was taken, but with the progress of reaction, PCl3(g) and Cl2(g) are formed

due to dissociation of PCl5(g). After a certain interval of time, the concentration of PCl5(g),

PCl3(g) and Cl2(g) each becomes constant. It does not mean that at this point of time, dissociation

of PCl5(g) and its formation from PCl3(g) and Cl2(g) has been stopped. Actually the rate of

ikB'kkyk

Highest Selection in IIT-JEE since 2006

CHEMISTRY CHEMICAL EQUILIBRIUM


Ph.: 0291-3291970.

2

dissociation of PCl5 and the rate of formation of PCl5(g) becomes equal. This state is called the

state of chemical equilibrium. So, the state of chemical equilibrium is dynamic.

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

This can be shown graphically.

Equilibrium

Reactant

Time

Con

cent

rat i

on

State of equilibrium

Time

Rat

eo

fre

acti

on

Backward reaction

Forward

reaction

So, ―state of chemical equilibrium can be defined as the state when the rate of forward reaction

becomes equal to rate of reverse reaction and the concentration of all the species becomes

constant‖.

3. Law of mass action Guldberg and Waage in 1807 gave this law and according to this law, ―At constant temperature, the

rate at which a substance reacts is directly proportional to its active mass and the rate at which a

chemical reaction proceeds is directly proportional to the product of active masses of the reacting

species‖.

The term active mass of a reacting species is the effective concentration or its activity (a) which is

related to the molar concentration (C) as

a = f C

where f = activity coefficient. f 1 but f increases with dilution and as V i.e. C 0, f 1

i.e., a C. Thus at very low concentration the active mass is essentially the same as the molar

concentration. It is generally expressed by enclosing the formula of the reacting species in a square

bracket.

To illustrate the law of mass action, consider the following general reaction at constant

temperature,

(g) (g) (g) (g)aA bB cC dD

Applying law of mass action,

Rate of forward reaction,

Rf [A]a [B]b

or Rf = kf[A]a[B]b …(i)

Where kf = rate constant of forward reaction

Similarly,

Rate of reverse reaction

Rr [C]c [D]d

or Rr = kr[C]c [D]d …(ii)

where kr = rate constant of reverse reaction.

At equilibrium,

Rate of forward reaction = Rate of reverse reaction


Ph.: 0291-3291970.

3

i.e., Rf = Rr

So, from equations (i) and (ii), we get

kf[A]a [B]b = kr[C]c[D]d

or,

c df

a br

k [C] [D]

k [A] [B]

or

c df

c a br

k [C] [D]K

k [A] [B]

Where, Kc is the equilibrium constant in terms of molar concentration.

Equilibrium Constant (Kp) in terms of Partial Pressures: Consider the same general reaction

taking place at constant temperature,

aA(g)

+ bB(g)

cC

(g) + dD

(g)

From law of mass action,

c d

c a b

[C] [D]K

[A] [B]

…(1)

From ideal gas equation

PV = nRT

or,

nP RT

V

At constant temperature,

nP

V

Thus, we can say in a mixture of gases, Partial pressure of any component (say A)

PA [A]

Similarly, PB [B]

PC [C]

PD [D]

So, equation (1) can be rewritten as

c dC D

p a bA B

P PK

P P

…(2)

4. Relationship Between Kp and Kc From the above, for the same general reaction at constant temperature.

c d

c a b

[C] [D]K

[A] [B]

…(1)

and,

c dC D

p a bA B

P PK

P P

…(2)

From the ideal gas equation,

PV = nRT

nP RT

V


Ph.: 0291-3291970.

4

So, PB = [B]RT ; PD = [D]RT

Similarly, A CP [A]RT; P [C]RT

Substituting the values of PA, PB, PC and PD in equation (2), we get

c c d d

p a a b b

[C] (RT) [D] (RT)K

[A] (RT) [B] (RT)

or

c d (c d)

p a b (a b)

[C] [D] (RT)K

[A] [B] (RT)

or n

p cK K (RT)

Where, n = (c + d) – (a + b)

i.e., n = sum of no. of moles of gaseous products – sum of no. of moles of gaseous reactants.

If n 0, Kp Kc

If n = 0, Kp = Kc

and if n 0, Kp Kc

Illustration 1: Calculate the Kc and K

p for the following reactions and also deduce the relationship between K

c and K

p

(i) 2SO2(g)

+ O2(g)

2SO

3(g) (ii)

2(g) 2(g) 3(g)

1 3N + H NH

2 2

Solution:

(i) 2SO2(g)

+ O2(g)

2SO

3(g)


23

c 22 2

[SO ]K

[SO ] [O ]

…(i)

3

2 2

2SO

p 2SO O

PK

P P

…(ii)

We know that

np cK K (RT)

n = 2 – (2+1) = – 1

cp

KK

RT

(ii) 2(g) 2(g) 3(g)

1 3N H NH

2 2

3c 1/ 2 3/ 2

2 2

[NH ]K

[N ] [H ]

…(i)

3

2 2

NH

p 1/ 2 3/ 2N H

PK

P P

…(ii)

Now,

np cK K (RT)

3 1n 1 1

2 2

cp

KK

RT


Ph.: 0291-3291970.

5

Illustration 2: At 400K for the gaseous reaction

2A + 3B 3C + 4D

the value of Kp is 0.05. Calculate the value of Kc (R = 0.082 dm3 atm K–1 mol–1)

Solution:

From the given data

For the reaction 2A + 3B 3C + 4D

n = (3 + 4) – (2 +3) = 2

Since R is given in dm3 atm. K–1 mol–1, we shall take P = 1 atm.

Substituting these values in the equation

n

p c

CRTK K , we get

P

2

c

1 0.082 4000.05 K

1

c

0.05K

0.082 0.082 400 400

Kc = 4.648 × 10–5

Illustration 3: Establish a relationship between Kc and Kp for the following reactions.

(a) N2(g) + O2(g) 2NO(g) (b)

4 2 3(s) 3(g) 2(g) 2 (g)(NH ) CO 2NH + CO + H O

Solution:

(a) ng = 0 Kp = Kc

(b) ng = 4 Kp = Kc (RT)4

5. Application of law of mass action

1. Synthesis of Hydrogen Iodide: Suppose ‗a‘ moles of H2 and ‗b‘ moles of I2 are heated at 444°C in a closed container of volume

‗V‘ litre and at equilibrium, 2x moles of HI are formed.

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

Initial concentration (mol L–1)

a

V

b

V 0

Equilibrium concentration(mol L–1)

a x

V

b x

V

2x

V

2

c

2 2

[HI]K

[H ][I ]

…(i)

Substituting the equilibrium concentrations of H2, I2 and HI in equation (i), we get

2

2

c

2x

4xVK

a x b x (a x)(b x)

V V

…(ii)

2 2

2HI

p

H I

PK

P P

…(iii)


Ph.: 0291-3291970.

6

Suppose the total pressure of the equilibrium mixture at 444°C is P, then

PHI = mole fraction of HI × Total pressure

PHI =

2xP

a b

Similarly,

2H

(a x)P P

(a b)

and 2I

b xP P

(a b)

Substituting these values in equation (iii) we get

2

2

p

2xP

a bK

a x b xP P

a b a b

2

p

4xK

(a x)(b x)

…(iv)

So, one can see from equations (iii) and (iv), that

Kp = Kc

This is so, because n= 0 for the synthesis of HI from H2 and I2.

2. Thermal Dissociation of Phosphorus Pentachloride PCl5(g) dissociates thermally according to the reaction,

PCl5(g) PCl3(g) + Cl2(g)

Let us consider that 1 mole of PCl5 has been taken in a container of volume V litre and at

equilibrium x moles of PCl5(g) dissociates. Thus


Initial concentration (mol L–1) 1 0 0

Equilibrium concentration(mol L–1)

1 x

V

x

V

x

V

According to law of mass action, at constant temperature,

Kc =

3 2

5

[PCl ][Cl ]

[PCl ] …(1)

and Kp =

3 2

5

PCl Cl

PCl

P P

P

…(2)

Substituing the values of equilibrium concentration, in equation (1), we have

Kc =

x x

V V1 x

V

or

2

c

xK

V(1 x)

Now, total number of moles at equilibrium = 1 – x + x + x = 1 + x

Mole fraction of PCl3 = Mole fraction of Cl2 =

x

1 x

and mole fraction of PCl5 =

1 x

1 x


Ph.: 0291-3291970.

7

Suppose total pressure at equilibrium is P, then we have from equation (2),

2

p

x xP

1 x 1 xK1 x

P1 x

or

2

p 2

x PK

(1 x )

Similarly, we can apply law of mass action on any reaction at equilibrium.

Illustration 4: Derive an expression for Kc and Kp for the reaction

N2(g) + 3H2(g) 2NH3(g)

Assuming that in a container of volume V, initially 1 mole of N2 and 3 moles of H2 were taken and

at equilibrium 2x moles of NH3 is formed.

Solution:

N2 + 3H2 2NH3

t = 0 1 3 –

t = teq 1 – x 3 – 3x 2x = 4 – 2x

Kc =

2

3

(2 )

(1 ) (3 3 )

x

x x × V2 = 4

22

)x1(27

Vx

Kp =

2

3

(2 )

(1 ) (3 3 )

x

x x ×

2

2

(4 2 )x

P

= 24

22

P)x1(27

)x2(x16

6. Equilibrium Constant for Heterogeneous Equilibria The equilibrium which involves reactants and products in different physical states. The law of mass

action can also be applied on heterogeneous equilibria as it was applied for homogeneous equilbria

(involving reactants and products in same physical states).

(i) Thermal Dissociation of Solid Ammonium Chloride :

The thermal dissociation of NH4Cl(s) takes place in a closed container according to the equation:

NH4Cl(s) NH3(g) + HCl(g)

Let us consider 1 mole of NH4Cl(s) is kept in a closed container of volume ‗V‘ litre at temperature

TK and if x mole of NH4Cl dissociates at equilibrium, then

NH4Cl(s) NH3(g) + HCl(g)

Initial moles 1 0 0

Moles at equilibrium 1 – x x x


3c

4

[NH ][HCl]K

[NH Cl]

As NH4Cl is a pure solid, so there is no appreciable change in its concentration. Thus,

Kc = [NH3] [HCl]

2

c 2

x x xK

V V V

and 3p NH HClK P P


Ph.: 0291-3291970.

8

(ii) Thermal Dissociation of Ag2CO3:

Ag2CO3(s) dissociates thermally according to the equation.

Ag2CO3(s) Ag2O(s) + CO2(g)

Applying law of mass action, at constant temperature, we get,

2 2c

2 3

[Ag O][CO ]K

[Ag CO ]

Now, let us consider that 1 mol of Ag2CO3(s) is heated in a closed container of volume V and x

mol of Ag2CO3(s) dissociates at equilibrium, then

Ag2CO3(s) Ag2O(s) + CO2(g)

Initial moles 1 0 0

Equilibrium moles 1 – x x x

Now, as Ag2CO3 and Ag2O are solids, so their concentration can be assumed to be constants thus

c 2

xK [CO ]

V

2P COK P

Illustration 5: Calculate the partial pressure of HCl gas above solid a sample of NH4Cl(s) as a result of

its decomposition according to the reaction:

NH4Cl(s) NH3(g) + HCl(g) Kp = 1.04 × 10–16

Solution:

NH4Cl(s) NH3(g) + HCl(g) Kp = 1.04 ×10–16

Kp = 3NHP × PHCl P2 = Kp P = pK

P =161.04 10 = 1.02 × 10–8 atm.

7. The Le-Chatelier’s Principle This principle, which is based on the fundamentals of a stable equilibrium, states that

“When a chemical reaction at equilibrium is subjected to any stress, then the equilibrium

shifts in that direction in which the effect of the stress is reduced”.

Confused with ―stress‖. Well by stress here what we mean is any change of reaction conditions e.g.

in temperature, pressure, concentration etc.

This statement will be explained by the following example.

Let us consider the reaction: 2NH3 (g)

endo

exo

N2 (g) + 3H2 (g)

Let the moles of N2, H2 and NH3 at equilibrium be a, b and c moles respectively. Since the

reaction is at equilibrium,

Where,

2 2

3

3

N H

2

NH

P P

P

= Kp =

2 2

3

3

N T H T

2

NH T

X .P X .P

X .P

X terms denote respective mole fractions and PT is the total pressure of the system.


Ph.: 0291-3291970.

9

3

T T

p2

T

a bP P

a b c a b cK

cP

a b c

Here,

a

a b c = mole fraction of N2

b

a b c = mole fraction of H2

c

a b c = mole fraction NH3

23T

2 2

Pab

c a b c

= Kp

Since PT =

a b c RT

V

( assuming all gases to be ideal)

23

2

ab RT

Vc

= KP …(1)

Now, let us examine the effect of change in certain parameters such as number of moles, pressure,

temperature etc.

If we increase a or b, the left hand side expression becomes QP ( as it is disturbed from

equilibrium) and we can see that QP > KP

The reaction therefore moves backward to make QP = KP.

If we increase c, QP < KP and the reaction has to move forward to revert

back to equilibrium.

If we increase the volume of the container (which amounts to decreasing the pressure),

QP < KP and the reaction moves forward to attain equilibrium.

If we increase the pressure of the reaction, then equilibrium shifts towards backward direction since

in reactant side we have got 2 moles and on product side we have got 4 moles. So pressure is

reduced in backward direction.

If temperature is increased, the equilibrium will shift in forward direction since the forward

reaction is endothermic and temperature is reduced in this direction.

However from the expression if we increase the temperature of the reaction, the left hand side

increases (QP) and therefore does it mean that the reaction goes backward (since QP > KP)?. Does

this also mean that if the number of moles of reactant and product gases are equal, no change in

the reaction is observed on the changing temperature (as T would not exist on the left hand side)?.

The answer to these questions is No. This is because KP also changes with temperature. Therefore,

we need to know the effect of temperature on both QP and KP to decide the course of the reaction.

Effect of Addition of Inert Gases to A Reaction At Equilibrium 1. Addition at constant pressure

Let us take a general reaction

aA + bB cC + dD


Ph.: 0291-3291970.

10

We know,

c d

C DT T

p a b

A BT T

n nP P

n nK

n nP P

n n

Where,

nC nD, nA, nB denotes the no. of moles of respective components and PT is the total pressure and

n = total no. of moles of reactants and products.

Now, rearranging ,

nc dc D T

P a bA B

n n PK

nn n

Where n = (c + d) – (a + b)

Now, n can be = 0, 0 or 0

Let us take each case separately.

(a) n = 0 : No effect (b) n = ‗+ve‘ :

Addition of inert gas increases the n i.e.

TP

n

is decreased and so is

n

TP

n

. So products

have to increase and reactants have to decrease to maintain constancy of Kp.

So the equilibrium moves forward.

c) n = ‗–ve‘ :

In this case

TP

n

decreases but

n

TP

n

increases. So products have to decrease and reactants

have to increase to maintain constancy of Kp. So the equilibrium moves backward.

2. Addition at Constant Volume : Since at constant volume, the pressure increases with addition of inert gas and at the same time n

also increases, they almost counter balance each other. So

n

TP

n

can be safely approximated as

constant. Thus addition of inert gas has no effect at constant volume.

Dependence of Kp or Kc on Temperature

Now we will derive the dependence of KP on temperature.

Starting with Arrhenius equation of rate constant

af

E /RT

f fk A e

… (i)

Where, kf = rate constant for forward reaction, Af = Arrhenius constant of forward reaction,

faE= Energy of activation of forward reaction

ar

E / RT

r rk A e

…(ii)

Dividing (i) by (ii) we get,

a ar fE E

f f RT

r r

k Ae

k A

We know that (equilibrium constant )


Ph.: 0291-3291970.

11

K =

a ar fE E

f f RT

r r

k Ae

k A

At temperature T1

a ar f

1

1

E E

RTfT

r

AK e

A

…(iii)

At temperature T2

a ar f

2

2

E E

RTfT

r

AK e

A

… (iv)

Dividing (iv) by (iii) we get

a ar f

2 2 1

1

E E 1 1

T R T T

T

Ke

K

log

2 r f

1

T a a

T 2 1

K E E 1 1

K 2.303 R T T

The enthalpy of a reaction is defined in terms of activation energies as f ra aE E = H

2

1

T

T 2 1

K H 1 1log

K 2.303 R T T

log

2

1

T

T 1 2

K ΔH 1 1= -

K 2.303 R T T

…(v)

For an exothermic reaction, H would be negative. If we increase the temperature of the system

(T2>T1), the right hand side of the equation (V) becomes negative.

2 1T TK < K , that is, the equilibrium constant at the higher temperature would be less than that at

the lower temperature.

Now let us analyse our question. Will the reaction go forward or backward?

Before answering this, we must first encounter another problem. If temperature is increased, the

new KP would either increase or decrease or may remain same. Let us assume it increases.

Now, QP can also increase, decrease or remain unchanged. If KP increases and QP decreases, then,

T T2 2P PQ K

therefore the reaction moves forward. If KP increase and QP remains same, then

T T T2 1 2

P P PQ Q K . Again, the reaction moves forward. What, if KP increase and QP also increases?

Will 22 TT PP KQ or T T2 2

P PQ K or T T2 2

P PQ K? This can be answered by simply looking at the

dependence of QP and KP on temperature. You can see from the equation that KP depends on

temperature exponentially. While Q‘s dependence on T would be either to the power g,l,t……..

Therefore the variation in KP due to T would be more than in QP due to T.

KP would still be greater than QP and the reaction moves forward again.

Therefore, to see the temperature effect, we need to look at KP only. If it increases, the reaction

moves forward, if it decreases, reaction moves backward and if it remains fixed, then no change at

all.


Ph.: 0291-3291970.

12

Illustration 6:

Under what conditions will the following reactions go in the forward direction ?

1. N2(g)+ 3H2(g) 2NH3(g) + 23 k cal.

2. N2(g) + O2(g) 2NO(g) - 43.2 k cal.

3. C(s) + H2O(g) CO2(g) + H2(g) + X k cal.

4. N2O4(g) 2NO2(g) - 14 k cal.

Solution:

1. Low T, High P, excess of N2 and H2.

2. High T, any P, excess of N2 and O2

3. Low T, Low P, excess of C and H2O

4. High T, Low P, excess of N2O4.

Illustration 7:

Under what conditions will the following reactions go in the forward direction?

1. 2SO2(g) + O2(g) 2SO3(g) + 45 k cal.

2. 2NO(g) + O2(g) 2NO2(g) + 27.8 k cal.

3. PCl5(g) PCl3(g) + Cl2(g)- X k cal.

Solution : (i) Low T, high P, excess of O2 and SO2

(ii) Low T, high P, excess of NO and O2

(ii) High T, low P, excess of PCl5

8. Le-Chatelier’s Principle and Physical Equilibria

Le Chatelier‘s principle, as already stated, is applicable to all types of equilibria involving not only

chemical but physical changes as well. A few examples of its application to physical equilbria are

discussed below.

1. Vapour pressure of a liquid: Consider the equilibrium

Liquid Vapour

It is well known that the change of a liquid into its vapour is accompanied by absorption of heat

whereas the conversion of vapour into liquid state is accompanied by evolution of heat. According

to Le Chatelier‘s principle, therefore, addition of heat to such a system will shift the equilibrium

towards the right. On raising the temperature of the system, liquid will evaporate. This will raise

the vapour pressure of the system. Thus, the vapour pressure of a liquid increases with rise in

temperature.

2. Effect of pressure on the boiling point of a liquid: The conversion of liquid into

vapour, as represented by the above equilibrium, is accompanied by increase of pressure (vapour

pressure). Therefore, if pressure on the system is increased, some of the vapours will change into

liquid so as to lower the pressure. Thus, the application of pressure on the system tends to condense

the vapour into liquid state at a given temperature. In order to counteract it, a higher temperature is

needed. This explains the rise of boiling point of a liquid on the application of pressure.

3. Effect of pressure on the freezing point of a liquid (or melting point of a solid): At the melting point, solid and liquid are in equilibrium:

Solid Liquid


Ph.: 0291-3291970.

13

Now, when a solid melts, there is usually a change, either increase or decrease, of volume. For

example, when ice melts, there is decrease in volume, or at constant volume, there is decrease in

pressure. Thus, increase of pressure on ice water system at a constant temperature will cause

the equilibrium to shift towards the right, i.e., it will cause the ice to melt. Hence, in order to retain

ice in equilibrium with water at the higher pressure it will be necessary to lower the temperature.

Thus, the application of pressure will lower the melting point of ice.

When sulphur melts, there is increase in volume or at constant volume, there is increase in

pressure. From similar considerations, it follows that if the pressure on the system, sulphur (solid)

sulphur (liquid) is increased, the melting point is raised.

4. Effect of temperature on solubility: In most cases, when a solute passes into solution,

heat is absorbed, i.e., cooling results. Therefore according to Le Chatelier‘s principle, when heat is

applied to a saturated solution in contact with solute, the change will take place in that direction

which absorbed heat (i.e., which tends to produce cooling). Therefore, some more of the solute will

dissolve. In other words, the solubility of the substance increases with rise in temperature.

Dissociation of a few salts (e.g., calcium salts of organic acids) is accompanied by evolution of

heat. In such cases, evidently, the solubility decreases with rise in temperature.

9. Free Energy and Chemical Equilibrium The Gibbs free energy function is a true measure of chemical affinity under conditions of constant

temperature and pressure. The free energy change in a chemical reaction can be

defined as

G = G(products) – G(reactants)

When G = 0, there is no net work obtainable. The system is in a state of equilibrium. When G is

positive, net work must be put into the system to effect the reaction, otherwise it cannot take place.

When G is negative, the reaction can proceed spontaneously with accomplishment of the net

work. The larger the amount of this work that can be accomplished, the farther away is the reaction

from equilibrium. For this reason –G has often been called the driving force of the reaction. From

the statement of the equilibrium law, it is evident that the driving force depends on the

concentration of the reactants and products. It also depends upon the temperature and pressure

which determine the molar free energies of the reactants and products.

The reaction conducted at constant temperature (i.e., in a thermostat)

–G = – H + TS

The driving force is made up of two parts, –H term and TS term. The –H term is the heat of

reaction at constant pressure and TS is heat involved when the process is carried out reversibly.

The difference is the amount of heat of reaction which can be converted into net work (–G), i.e.,

total heat minus unavailable heat.

If the reaction is carried out at constant volume, the decrease in Helmholtz function –G =

–E + TS would be the proper measure of affinity of the reactant or the driving force of the

reaction.

Now we can see why Berthollet and Thompson were wrong in assuming that driving force of the

reaction was the heat of reaction. They neglected the TS term. The reasons for the apparent

validity of their principle was that for many reactions, H term far outweighs the TS term. This is

especially true at low temperature, since at higher temperature, TS term increases.

The fact that driving force for a reaction is large (G is large negative quantity) does not mean that

the reaction will necessarily occur under any given conditions.

For example, the reaction


Ph.: 0291-3291970.

14

2 2 2

1H O H O; G 228.6kJ

2

does not occur at the laboratory temperature. The reaction mixture may be kept for years without

any detectable formation of water. Here H factor favours, but S factor disfavours the reaction.

Similarly, the reaction

2Mg + O2 2MgO; G = – 570.6kJ

is not favoured. However, the thermite reaction

2Al +

3

2 O2 Al2O3

with large value of – G proceeds favourably.

Standard Free Energy and Equilibrium Constant: The change in free energy for a

reaction taking place between gaseous reactants and products represented by the general equation.

aA bB cC dD According to Van‘t Hoff reaction isotherm

c d0 C D

a bA B

p pG G RTln

p p

= G0 + RTlnQp

the condition for a system to be at equilibrium is that

G = 0

Thus at equilibrium c d

0 0 0C Dpa b

A B

p p0 G RTln G RTln K

p p

Whence G0 = – RTlnK0p

Hence

00p

Gln K

RT

Note: 1. In the reaction, where all gaseous reactants and products; K represents Kp

2. In the reaction, where all solution reactants and products; K represents Kc

3. a mixture of solution and gaseous reactants; Kx represents the thermodynamic equilibrium

constant and we do not make the distinction between Kp and Kc.

we may conclude that for standard reactions, i.e., at 1 M or 1 atm.

When G0 = –ve or K 1: forward reaction is feasible

G0 = +ve or K 1: reverse reaction is feasible

G0 = 0 or K = 1: reaction is at equilibrium (very rare)

Illustration 8: Kc for the reaction N2O4 2NO2 in chloroform at 291 K is 1.14. Calculate the free

energy change of the reaction when the concentration of the two gases are 0.5 mol dm–3 each at

the same temperature. (R = 0.082 lit atm K–1 mol–1.)

Solution:

From the given data

T = 291 K; R = 0.082 lit atm K–1 mol–1

Kc = 1.14; 2 2 4NO N OC C= 0.5 mol dm–3

The reaction quotient Qc for the reaction 2 4 2N O 2NO ,


Ph.: 0291-3291970.

15

22

c

2 4

[NO ] 0.5 0.5Q 0.5

[N O ] 0.5

Since Qp = Qc(RT)n and n = 2 – 1 = 1 in this case

Qc = 0.5 (0.082 × 291) = 11.93

Kp = Kc(RT)n = 1.14 (0.082 × 291) = 27.1

Substituting these values in the equation

G = G0 + RTlnQp = – RT ln Kp + RT ln Qp, we get

= – 2.303 RT (log Kp – log Qp)

G = – (0.082 × 291 × 2.303) [log 27.2 – log 11.93]

= – 54.95 (1.4346 – 1.0766) = – 19.67 lit atm

Illustration 9: Calculate the pressure of CO2 gas at 700K in the heterogeneous equilibrium reaction

CaCO3(s) CaO(s) + CO2(g)

if G0 for this reaction is 130.2 kJ mol–1.

Solution:

Here Kp = 2COP

Also, 0

pG RTln K

0 3 1

p 1 1

G 130.2 10 Jmolln K

RT (8.314JK mol )(700K)

2

10CO pp K 1.94 10 atm

Illustration 10:

For the equilibrium NiO(s) + CO(g) Ni(s) + CO2(g), G0 (J mol–1)

= – 20,700 – 11.97 T. Calculate the temperature at which the product gases at equilibrium at 1

atm will contain 400 ppm (parts per million) of carbon monoxide.

Solution: For the given reaction 2p CO COK p / p

Since 2CO COp p , hence

p 6

CO

1 1K 2,500

p 400 10

0

pG RTln K

0

p

G 20,700 11.97Tln K

RT RT

The equation when solved for T using R = 8.314 K–1 mol–1, gives T = 399K.

Illustration 11: Calculate the equilibrium constant of a reaction at 300 K if G0 at this temperature for

the reaction is 29.29 kJ mol–1.

Solution

(ii) G0 = –2.303 RT log Kp

29.29 ×103 = –2.303 × 8.31 × 300 log Kp

Kp = 7.91 × 10–6


Ph.: 0291-3291970.

16

Illustration 12: For the formation of ammonia the equilibrium constant data at 673K and 773K

respectively are 1.64 × 10–4 and 1.44 × 10–5 respectively. Calculate heat of the reaction. Given R

= 8.314 JK–1 mol–1.

Solution: Substituting the values in the equation

2

1

p 2 1

p 2 1

K T THln , we get

K R T T

5

4

1.44 10 H 773 6732.303log

8.314 773 6731.64 10

H(100)2.303log(0.0878)

8.314 773 673

100H = 2.303 (–1.0565) × 673 × 773 × 8.314

whence,

673 773 2.303 1.0565 8.314H

100

= – 105216J = – 105.216kJ

Illustration 13: The equilibrium constant KP for the reaction N2(g) + 3H2(g) 2NH3(g) is 1.6 10-

4 atm at 400oC. What will be the equilibrium constant at 500oC if heat of the reaction in this

temperature range is 25.14 k cal?

Solution: N2(g) + 3H2(g) 2NH3(g)

Kp = 1.6 × 10–4 at T1 = 673 K

Kp = ? at T2 = 773 K

log

4

p

106.1

K

= 2303.2

1014.25 3

log [673 – 773]

Kp = 1.43 × 10–5

Relation between Vapour Density and Degree of Dissociation In the following reversible chemical equation

A yB

Initial mol 1 0

At equilibrium (1 –x) yx x = degree of dissociation

Number of moles of A and B at equilibrium = 1 – x + yx = 1 + x(y –1)

If initial volume of 1 mole of A is V, then volume of equilibrium mixture of A and B is

= [1 + x(y – 1)]V

Molar density before dissociation,

molecular weight mD

volume V

Molar density after dissociation

md

[1 x(y 1)]V

D[1 x(y 1)]1

d

D dx

d(y 1)


Ph.: 0291-3291970.

17

Note: y is the number of moles of products from one mole of reactant.

D

d is also called Van‘t Hoff

factor.

Thus for the equilibria

I: 5(g) 3(g) 2(g)PCl PCl Cl , y = 2

II: 2 4(g) 2(g)N O 2NO , y = 2

III: 2 2 4

12NO N O , y =

2

D dx

d

2(d D)x

d

Also D × 2 = Molecular weight (theoretical value)

d × 2 = Molecular weight (abnormal value) of the mixture

Illustration 14: Vapour density of the equilibrium mixture of NO2 and N2O4 is found to be 40 for the

equilibrium

N2O4 2NO2

Calculate (i) abnormal molecular weight

ii) degree of dissociation

iii) percentage of NO2 in the mixture

Solution: (i) N2O4 2NO2

Observed value of vapour density (d) = 40

Hence, abnormal molecular weight = 40 × 2 = 80

(ii) D × 2 = theoretical molecular weight = 2

92D 46

2

D d 46 40x 0.15

d 40

(iii) N2O4 2NO2

Initial mol 1 0

At equilibrium (1 – x) 2x

0.85 0.30

Total moles at equilibrium = (1 + x) = 1 + 0.15 = 1.15

Percentage of NO2 =

2x 0.30100 100

1 x 1.15

= 26.08%

Illustration 15: N2O4(g) 2NO2(g). In this reaction, NO2 is 20% of the total volume at

equilibrium. Calculate

(i) Vapour density (ii) abnormal molecular weight

(iii) percentage dissociation of N2O4

Solution : N2O4 2NO2


Ph.: 0291-3291970.

18

t = 0 1 –

t = teq 1–x 2x = 1 + x

2

NO =

2

1

x

x = 0.2 x =

1

9

(i) 1

9 =

46 d

d

10d = 46 × 9

(ii) M =92

1 x d = 41.4

(iii) = 1

9× 100 = 11.1%

Illustration 16:The vapour density of a mixture consisting of NO2 and N2O4 is 38.3 at 26.7°C.

Calculate the number of moles of NO2 in 100 gm mixture.

Solution:

N2O4 2NO2 = D d

d

t = 0 1.087 0 = 46 38.3

38.3

= .2 (20%)

t = teq 1.087 (1 –) 1.087 × 2

2NOn = 1.087 × 0.4 = 0.435

Illustration 17: The degree of dissociation at a certain or given temperature of PCl5 at 2 atm is found to

be 0.4. At what pressure, the degree of dissociation of PCl5 will be 0.6 at the same temperature?

Also calculate the equilibrium constant for the reverse reaction.

Solution:

The dissociation of PCl5 takes place according to the equation,

PCl5 PCl3(g) + Cl2(g)

Let 1 mole of PCl5 is taken in a closed container. Then


Mole(s) before dissociation 1 0 0

Mole(s) at equilibrium 1 –

As degree of dissociation() is given at 2 atm

1 – = 1.0 – 0.4 = 0.6

Total moles at equilibrium = 1 – + + = 1 – 0.4 + 0.4 + 0.4 = 1.4

Now,

3 2

5

PCl Cl

p

PCl

P PK

P

=

0.4 0.42 2

1.4 1.40.6

21.4

p

0.4 0.4 4 1.4 0.32K

0.6 2 1.4 1.4 0.84

Now, as temperature remains the same, Kp will also remain the same. So for

= 0.6


Mole before dissociation 1 0 0

Moles at equilibrium 1 – 0.6 0.6 0.6


Ph.: 0291-3291970.

19

Again,

2

p

0.6 0.6P

1.6 1.6K0.4

P1.6

p

0.6 0.6 1.6K P

0.4 1.6 1.6

32 9

84 16

P P =

32 4 128

9 21 189

= 0.677 atm

Illustration 18: (i) Calculate the percentage dissociation of H2S(g), if 0.1 mole of H2S is kept in 0.4 litre

vessel at 1000K. For the reaction, 2H2S(g) 2H2(g) + S2(g), the value of Kc = 1.6 × 10–6

(ii) A sample of HI was found to be 22% dissociated when equilibrium was reached. What will be

the degree of dissociation if hydrogen is added in the proportion of 1 mole for every mole of HI

originally present, the temperature and volume of the system being kept constant?

Solution:

(i) 2H2S(g) 2H2 + S2(g) Kc = 1.6 × 10–6

t = 0 0.1 – –

t = teq 0.1 – x x x/2

value of Kc suggest that (0.1 – x) =~

0.1

3

2

1

0.42(0.1)

x = 1.6 × 10–6 x = 2.34 ×10–3

= 32.34 10

0.1

× 100 = 2.34%

(ii) 2HI H2 + I2

t = 0 1 – –

t = teq 1– /2 /2 since = 0.22

Kc = 2

2

(0.11)

(0.88) =

1

64

now 2HI H2 + I2

t = 0 1 1 0

t = teq 1– /2 /2

1

64 =

2

(1 / 2).( / 2)

(1 )

=

2

2

2

4(1 )

4 + 42 – 8 = 128 + 642

602 + 136 – 4 = 0

= 0.29


Ph.: 0291-3291970.

20

Problem Set – 1

Q.1 For the given gaseous phase reaction, K1, K2, K3 are equilibrium constants respectively.

Correlate them.

2NO + O2 2NO2 4NO + 2Cl2 4NOCl

NO2 + ½ Cl2 NOCl + ½ O2

Q.2 In an equilibrium A + B C + D, the initial concentration of A was three times to that of B.

At equilibrium, the concentration of C was found to be equal to equilibrium concentration of B.

Calculate KC.

Q.3 For an equilibrium reaction A + 2B 2C + D; A and B are mixed in a reaction vessel at

300 K. The initial concentration of B was 1.5 times the initial concentration of A. After the

equilibrium, the equilibrium concentrations of A and D are same. Calculate KC.

Q.4 KC for N2O4(g) 2NO2(g) at 298°C is 5.7 ×10–9. Which species has higher

concentration at equilibrium ?

Q.5 From the given data of equilibrium constants of the following reactions :

CoO(s) + H2(g) Co(s) + H2O(g); K = 670

CoO(s) + CO(g) Co(s) + CO2(g); K = 490

Calculate KC for the reaction: CO2(g) + H2(g) CO(g) + H2O(g)

Q.6 The compound AB dissociates at 300°C of AB(g) A(g) + B(g); KC of reaction is 5 × 10–

2 and equilibrium concentration of A is 0.03 mol litre–1

(a) What was initial concentration of AB ?

(b) Calculate partial pressure of each gas and hence calculate Kp.

(c) Verify the value of Kp derived as above for its agreement with Kp = KC(RT)n

Q.7 At 25°C for the reaction A2(g) 2A(g), KC is 0.165. If one mole of A2 is placed in 0.1

litre flask at 25°C, calculate how many moles of A are formed at equilibrium ?

Q.8 A mixture of SO2 and O2 under atmospheric pressure in the ratio of 2 : 1 is passed over a

catalyst at 1170°C. After equilibrium has reached, the gas coming out has been found to contain

87% SO3 by volume. Calculate Kp for SO2 + ½ O2 SO3.

Q.9 KC for PCl5 PCl3 + Cl2 is 13.7 mol litre–1 at 546 K. Calculate the pressure developed

in 10 litre box in equilibrium if reactant concentration is 1.0 mole.

Q.10 n mole of PCl3 and n mole Cl2 are allowed to react at constant temperature T under a total

pressure P, as PCl3(g) + Cl2(g) PCl5(g)

If Y mole of PCl5 are formed at equilibrium, find Kp.

Q.11 n mole each of H2O, H2 and O2 are mixed at a suitable high temperature to attain the

equilibrium 2H2O 2H2 + O2. If y mole of H2O are dissociated and the total pressure

maintained is P, calculate the Kp.

Q.12 How many mole of PCl5 must be added to one litre vessel at 250°C in order to obtain a

concentration of 0.1 mol litre–1 of Cl2 at equilibrium. Kp for PCl5 PCl3 + Cl2 is 1.78 atm?

Q.13 The equilibrium constant of the reaction A2(g) + B2(g) 2AB(g)

at 100°C is 50. If a one litre flask containing one mole of A2 is connected to a two litre flask

containing two mole of B2, how many moles of AB will be formed at 373 K ?

Q.14 The equilibrium constant for the reaction.

CH3COOH(l)+ C2H5OH(l) CH3COOC2H5(l) + H2O(l)


Ph.: 0291-3291970.

21

is 4. What will be the composition of the equilibrium mixture when one mole of acetic acid is

taken along with 4 moles of ethyl alcohol ?

Q.15 At 450°C, the equilibrium constant, Kp for the reaction, N2(g) + 3H2(g) 2NH3(g)

was found to be 1.6 × 10–5 at a pressure of 200 atm. If N2 and H2 are taken in 1 : 3 ratio what is

% of NH3 formed at this temperature ?

Problem Set – 2

Q.1 When limestone is heated, quick lime is formed according to the equation,

CaCO3(s) CaO(s) + CO2(g)

The experiment was carried out in the temperature range 800 – 900°C. Equilibrium constant Kp

follows the relation,

log10Kp = 7.282 – 8500/T

where T is temperature in Kelvin. At what temperature the decomposition will give CO2(g) at 1

atm?

Q.2 Consider the equilibrium :

LiCl.3NH3(s) LiCl.NH3(s) + 2NH3(g)

with Kp = 9 atm2 at 40°C. A 5 litre flask contains 0.1 mol of LiCl.NH3. How many moles of

NH3 should be added to the flask at this temperature to drive the backward reaction practically

to completion?

Q.3 A sample of CaCO3(s) is introduced into a sealed container of volume 0.654 litre and heated to

1000 K until equilibrium is reached. The equilibrium constant for the reaction.

CaCO3(s) CaO(s) + CO2(g) is 3.9 × 10–2 atm at this temperature. Calculate the mass

of CaO present at equilibrium.

Q.4 The decomposition of ammonium carbamate at 30°C is represented as

NH2COONH4(s) 2NH3(g) + CO2(g)

The equilibrium constant Kp is 2.9 × 10–5 atm3. What is the total pressure of gases in

equilibrium with NH2COONH4(s) at 300°C.

Q.5 Solid ammonium carbamate dissociates to give ammonia and carbodioxide as follows :

NH2COONH4(s) 2NH3(g) + CO2(g)

At equilibrium, ammonia is added such that partial pressure of NH3, now equals to the original

total pressure. Calculate the ratio of total pressure now to the original total pressure.

Q.6 Show that for the reaction AB(g) A(g) + B(g), the total pressure at which AB is

50% dissociated is numerically equal to three times of Kp.

Q.7 At 55°C and one atmosphere, N2O4 was found to decompose 50.3%. At what pressure and

same temperature the equilibrium, N2O4 2NO2

will give the ratio of N2O4 : NO2 at 8 : 1 ?

Q.8 Pure PCl5 is introduced into an evacuated chamber and comes to equilibrium at 250°C and

2 atmosphere. The equilibrium mixture contains 40.7% Cl2 by volume.

(a) What are the partial pressure of each constituent at equilibrium ?

(b) What are Kp and KC ?

(c) If the gas mixture is expanded to 0.200 atm at 250°C, calculate

(i) the % of PCl5 dissociated at this equilibrium.

(ii) The partial pressure of each at equilibrium


Ph.: 0291-3291970.

22

Q.9 5 mole of PCl5 and 4 mole of neon are introduced in a vessel of 110 litre and allowed to attain

equilibrium at 250°C. At equilibrium, the total pressure of reaction mixture was 4.678 atm.

Calculate degree of dissociation of PCl5 and equilibrium constant for the reaction.

Q.10 5g of PCl5 were completely vaporised at 250°C in a vessel of 1.9 litre capacity. The mixture at

equilibrium exerted a pressure of one atmosphere. Calculate , KC and Kp for the reaction PCl5

PCl3 + Cl2.

Q.11 1.0 mole of nitrogen and 3.0 mole of PCl5 are placed in 100 litre vessel heated to 227°C. The

equilibrium pressure is 2.05 atm. Assuming ideal behaviour calculate the degree of dissociation

for PCl5 and Kp for the reaction : PCl5(g) PCl3(g) + Cl2(g)

Q.12 The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 26.7°C. Calculate the

number of moles of NO2 in 100 grams of the mixture.

Q.13 For the reaction,

NH3(g) N2(g) + H2(g)

show that degree of dissociation of NH3 is given as

=

1/2

3 31

4 p

p

K

, where ‗p‘ is equilibrium pressure. If Kp of the above reaction is 78.1 atm

at 400°C, calculate KC.

Q.14 At 25°C and 1 atm, N2O4 dissociates by the reaction,

N2O4(g) 2NO2(g)

If it is 35% dissociated at given condition, find :

(i) The percent dissociation at same temperature if total pressure is 0.2 atm.

(ii) The percent dissociation (at same temperature) in a 80 g sample of N2O4 confined to a 7

litre vessel.

(iii) What volume of above mixture will diffuse if 20 mL pure O2 diffuses in 10 minutes at same

temperature and pressure ?

Q.15 The value of KC for 2HF(g) H2(g) + F2(g) is 1.0 ×10–13 at a particular temperature. At

a certain time, the concentration of HF, H2 and F2 were found to be 0.5, 1 ×10–3 and 4 ×10–3

mol litre–1 respectively. Predict whether the reaction is in equilibrium ? If not, what is the

direction of the reaction to attain equilibrium?

Problem Set – 3

1. If for A + 2B 3C, KC = 10 and a 1 litre container contains 5 moles of C,

10 moles of B and 1.25 moles of A, then predict the direction in which the reaction

is moving.

2. A + B C + D, KC = 20. A 10 litre container contains 5 moles each of A and

B and 20 moles each of C and D.

(i) Predict direction of reaction

(ii) Calculate concentration of A, B, C, D at equilibrium.

3. A 2 litre container contains 1 moles of PCl5 2 moles of PCl3 and 2 moles of

Cl2. If KC for PCl5 PCl3 + Cl2 is 2 then, calculate concentration of PCl5,

PCl3 and Cl2 at equilibrium.


Ph.: 0291-3291970.

23

4. For the reaction , A + B 3C at 25ºC, a 3 litre vessel contains 1,2,4 mole of A, B

and C respectively . Predict the direction of reaction if :

(a) KC for the reaction is 10. (b) KC for the reaction is 15.

(c) KC for the reaction is 10.66 .

5. If Gº for the reaction given below is 1.7 kJ; the equilibrium constant of the

reaction, 2HI (g) H2(g) + I2(g) at 25 ºC is :

(A) 24.0 (B) 3.9 (C) 2.0 (D) 0.5

6. A large positive value of Gº corresponds to which of these?

(A) small positive K (B) small negative K

(C) large positive (D) large negative K

7. The equilibrium constant for a reaction is 1 × 10 20 at 300 K. The standard free

energy change for this reaction is :

(A) – 115 kJ (B) +115 kJ (C) +166 kJ (D) –166 kJ

8. At equilibrium, If Kp = 1, then:

(A) Gº = 0 (B) Gº > 1 (C) Gº < 1 (D) None

9. Gº for the reaction , X + Y Z is –4.606 kcal. The equilibrium constant

for the reaction at 227 ºC is :

(A) 100 (B) 10 (C) 2 (D) 0.01

10. For the given equilibrium A + B C + D. The reaction is at equilibrium

with 0.2 M A and 0.6M B, 3M C and 4M D. Then calculate the value of Gº at

27ºC (in kcal).

11. Gº for () N2 + (3/2) H2 NH3, is –16.5 kJ mol–1. Find out Kp for the

reaction. Also report Kp and Gº for N2 + 3H2 2NH3, at 25ºC.

12. For the given equilibrium , N2O4 2NO2 total pressure at equilibrium is 5

atm and the mole ratio of N2O4 and NO2 is 3 : 2. Find the value of Gº at 227ºC

(in cal)

13. The value of KC for 2HF(g) H2(g) + F2(g) is 1.0 ×10–13 at a particular temperature. At

a certain time, the concentration of HF, H2 and F2 were found to be 0.5, 1 ×10–3 and 4 ×10–3

mol litre–1 respectively. Predict whether the reaction is in equilibrium ? If not, what is the

direction of the reaction to attain equilibrium ?

14. The equilibrium constant, Kp for the reaction

N2(g) + 3H2(g) 2NH3(g) is 1.6 × 10–4 atm–2 at 400°C.

what will be the equilibrium constant at 500°C if heat of reaction in this range is -25.14 Kcal.

15. For a reaction X Y, heat of reaction is +83.68 kJ, energy of reactant X is 167.36 kJ and

energy of activation is 209.20 kJ. Calculate (i) threshold energy (ii) energy of product Y and (iii)

energy of activation for the reverse reaction (Y X).

Problem Set – 4

Q.1 The initial concentrations or pressure of reactants and products are given for each of the

following systems. Calculate the reaction quotient and determine the directions in which each

system will shift to reach equilibrium.

(a)2NH3(g) N2(g) + 3H2(g) K = 17

[NH3] = 0.20 M ; [N2] = 1.00 M ; [H2] = 1.00 M

(b) 2NH3(g) N2(g) + 3H2(g) Kp = 6.8 × 104 atm2


Ph.: 0291-3291970.

24

Initial pressure; NH3 = 3.0 atm; N2 = 2.0 atm ; H2 = 1.0 atm

(c) 2SO3(g) 2SO2(g) + O2(g) K = 0.230 atm

[SO3] = 0.00 M ; ]SO2] = 1.00 M; [O2] = 1.00 M

(d) 2SO3(g) 2SO2(g) + O2(g) Kp = 16.5 atm

Initial pressure : SO3= 1.0 atm ; SO2 = 1.0 atm; O2= 1.0 atm

(e)2NO(g) + Cl2(g) 2NOCl (g) K = 4.6 × 104

[NO] = 1.00 M ; [Cl2] = 1.00 M ; [NOCl] = 0M

(f) N2(g) + O2(g) 2NO(g) Kp=0.050

Initial pressure ; NO = 10.0 atm N2 = O2 = 5 atm

Q.2 Among the solubility rules is the statement that all chlorides are soluble except Hg2Cl2,, AgCl,

PbCl2 and CuCl.

(a) Write the expression for the equilibrium constant for the reaction represented by the equation.

AgCl (s) Ag+ (aq) + CI– (aq)

Is K greater than 1, less than 1, or about equal to 1? Explain your answer.

(b) Write the expression for the equilibrium constant for the reaction represented by the equation

Pb2+ (aq) + 2CI– (aq) PbCl2 (s)

Is K greater than 1, less than 1, or about equal to I? Explain your answer.

Q.3 Among the solubility rules is the statement that carbonates, phosphates, borates, arsenates, and

arsenites, except those of the ammonium ion and the alkali metals are insoluble.

(a) Write the expression for the equilibrium constant for the reaction represented by the equation

CaCO3 (s) Ca2+ (aq) + CO32– (aq)


(b) Write the expression for the equilibrium constant for the reaction represented by the

equation. 3Ba2+ (aq) + 2PO43– (aq) Ba3(PO4)2 (s)


Q.4 Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is

manufactured by the catalytic conversion of acetylene to benzene.

3C2H2 C6H6

Would this reaction be most useful commercially if K were about 0.01, about 1, or about 10?

Explain your answer.

Q.5 Show the complete chemical equation and the net ionic equation for the reaction represented by

the equation

KI (aq) + I2 (aq) KI3 (aq)

give the same expression for the reaction quotient. KI3 is composed of the ions K+ and I3–.

Q.6 Which of the following reactions goes almost all the way to completion, and which proceeds

hardly at all?

(a) N2(g) + O2(g) 2NO(g) Kc = 2.7 × 10–18

(b) 2NO(g) + O2(g) 2NO2(g) ; Kc = 6.0 × 1013

Q.7 For which of the following reactions will the equilibrium mixture contain an appreciable

concentration of both reactants and products?

(A) Cl2(g) 2Cl(g); Kc = 6.4 × 10–39

(B) Cl2(g) + 2NO(g) 2NOCl(g); Kc = 3.7 × 108

(C) Cl2(g) + 2NO2(g) 2NO2Cl(g); Kc = 1.8


Ph.: 0291-3291970.

25

Q.8 The value of Kc for the reaction 3O2(g) 2O3 (g) is 1.7 × 10–5 6 at 25ºC. Do you expect

pure air at 25ºC to contain much O3(ozone) when O2 and O3 are in equilibrium ? If the

equilibrium concentration of O2 in air at 25ºC is 8 × 10–3 M, what is the equilibrium

concentration of O3?

Q.9 At 1400 K, Kc = 2.5 × 10–3 for the reaction CH4(g) + 2H2S CS2(g) + 4H2(g) . A 10.0 L

reaction vessel at 1400 K contains 2.0 mol of CH4, 3.0 mol of CS2, 3.0 mol of H2 an 4.0 mol of

H2S. Is the reaction mixture at equilibrium? If not, in which direction does the reaction proceed

to reach equilibrium?

Q.10 The first step in the industrial synthesis of hydrogen is the reaction of steam and methane to give

water gas, a mixture of carbon monoxide and hydrogen.

H2O(g) + CH4(g) CO(g) + 3H2(g) Kc = 4.7 at 1400 K

A mixture of reactants and product at 1400 K contains 0.035 M H2O, 0.050 M CH4, 0.15 M

CO, and 0.20 M H2. In which direction does the reaction proceed to reach equilibrium?

Q.11 An equilibrium mixture of N2, H2, and NH3 at 700 K contains 0.036 M N2 and 0.15 M H2. At

this temperature, Kc for the reaction N2(g) + 3H2(g) 2NH3(g) is 0.29. What is the

concentration of NH3.

Q.12 The air pollutant NO is produced in automobile engines from the high temperature reaction

N2(g) + O2(g) 2NO(g); Kc = 1.7 × 10–3 at 2300 K. If the initial concentrations of N2 and

O2 at 2300K are both 1.40 M, what are the concentration of NO, N2, and O2 when the reaction

mixture reaches equilibrium?

Q.13 At a certain temperature, the reaction PCl5(g) PCl3(g) + Cl2(g) has an equilibrium constan

Kc = 5.8 × 10–2. Calculate the equilibrium concentrations of PCl5, PCl3 and Cl2 if only PCl5 is

present initially, at a concentration of 0.160 M.

Q.14 At 700 K Kp = 0.140 for the reaction ClF3(g) ClF(g) + F2(g). Calculate the equilibrium

partial pressure of ClF3, ClF, and F2 if only ClF3 is present initially, at a partial pressure of 1.47

atm.

Q.15 The degree of dissociation of N2O4 into NO2 at 1.5 atmosphere and 40ºC is 0.25. Calculate its

Kp at 40ºC. Also report degree of dissociation at 10 atmospheric pressure at same temperature.

Q.16 At 46ºC, Kp for the reaction N2O4(g) 2NO2(g) is 0.667 atm. Compute the percent

dissociation of N2O4 at 46ºC at a total pressure of 380 Torr.

Q.17 When 36.8 g N2O4(g) is introduced into a 1.0-litre flask at 27ºC. The following equilibrium

reaction occurs:

N2O4 (g) 2NO2(g) ; Kp = 0.1642 atm.

(a) Calculate Kc of the equilibrium reaction.

(b) What are the number of moles of N2O4 and NO2 at equilibrium?

(c) What is the total gas pressure in the flask at equilibrium?

(d) What is the percent dissociation of N2O4?

Q.18 At some temperature and under a pressure of 4 atm, PCIs is 10% dissociated. Calculate the

pressure at which PCl5 will be 20% dissociated, temperature remaining same.

Q.19 In a mixture of N2 and H2 in the ratio of 1 : 3 at 64 atmospheric pressure and 300°C, the

percentage of ammonia under equilibrium is 33.33 by volume. Calculate the equilibrium

constant of the reaction using the equation.

N2 (g) + 3H(g) 2NH3 (g).


Ph.: 0291-3291970.

26

Q.20 The system N2O 4 2 NO2 maintained in a closed vessel at 60ºC & a pressure of 5 atm has

an average (i.e. observed) molecular weight of 69, calculate Kp. At what pressure at the same

temperature would the observed molecular weight be (230/3)?

Q.21 The vapour density of N204 at a certain temperature is 30. Calculate the percentage dissociation

of N2O4 at this temperature. N2O4(g) 2NO2(g).

Q.22 In the esterfication C2H5OH (l) + CH3COOH (l) CH3COOC2H5 (l) + H2O (l) an

equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with

mole fraction = 0.333. Calculate the equilibrium constant.

Q.23 Solid Ammonium carbonate dissociates as: NH2COONH4(s) 2NH3(g) + CO2(g). In a

closed vessel solid ammonium carbonate is in equilibrium with its dissociation product. At

equilibrium, ammonia is added such that the partial pressure of NH3 at new equilibrium now

equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that

of original total. pressure.

Q.24 A sample of CaCO3(s) is introduced into a sealed container of volume 0.821 litre & heated to

1000K until equilibrium is reached. The equilibrium constant for the reaction

CaCO3(s) CaO(s) + CO2(g) is 4 × 10–2 atm at this temperature.

Calculate the mass of CaO present at equilibrium.

Q.25 Anhydrous calcium chloride is often used as a dessicant. In the presence of excess of CaCl2, the

amount of the water taken up is governed by KP = 6.4 × 1085 for the following reaction at room

temperature, CaCl2(s) + 6H2O(g) CaCl2 .6H2O(s). What is the equilibrium vapour

pressure of water in a closed vessel that contains CaCl2(s)?

Q.26 20.0 grams of CaCO3(s) were placed in a closed vessel, heated & maintained at 727°C under

equilibrium CaCO3(s) CaO(s) + CO(g) and it is found that 75 % of CaCO3 was

decomposed. What is the value of Kp ? The volume of the container was 15 litres.

Q.27 Suggest four ways in which the concentration of hydrazine. N2H2. could be increased in an

equilibrium described by the equation

N2(g) + 2H2 (g) N2H4 (g) H = 95 kJ

Q.28 How will an increase in temperature affect each of the following equilibria? An increase in

pressure?

(a) 2NH3(g) N2(g) + 3H2(g) H = 92 kJ

(b)N2(g) + O2(g) 2NO(g) H = 181 kJ

(c)2O3(g) 3O2(g) H = –285 kJ

(d) CaO(s) + CO2(g) CaCO3(s) H = –176 kJ

Q.29 (a) Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas

and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst.

Write the expression for the equilibrium constant for the reversible reaction.

2H2(g) + CO(g) CH3OH(g) H = –90.2 kJ

(b) Assume that equilibrium has been established and predict how the concentration of H2, CO

and CH3OH will differ at a new equilibrium if (1) more H2 is added. (2) CO is removed (3)

CH3OH is added. (4) the pressure on the system is increases. (5) the temperature of the system

is increased. (6) more catalyst is added.


Ph.: 0291-3291970.

27

Q.30 (a) Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction

of steam with red-hot coke, essentially pure carbon. Write the expression for the equilibrium

constant for the reversible react on.

C(s) + H2O(g) CO(g) + H2(g) H = 131.30 kJ

(b) Assume that equilibrium has been established and predict how the concentration of each

reactant and product will differ at a new equilibrium if (1) more C is added. (2) H2O is

removed. (3) CO is added. (4) the pressure on the system is increased. (5) the temperature of the

system is increased.

Q.31 Ammonia is a weak base that reacts with water according to the equation

NH3(aq) + H2O (l) NH4+ + OH–(aq)

Will any of the following increase the percent of ammonia that is converted to the ammonium

ion in water? (a) addition of NaOH. (b) Addition of HCl. (c) Addition of NH4Cl.

Q.32 Suggest two ways in which the equilibrium concentration of Ag+ can be reduced in a solution of

Na+, Cl–, Ag+ and NO3–, in contact with solid AgCl.

Na+(aq) + Cl–(aq) + Ag+ (aq) + NO3–(aq) AgCl(s) + Na+ (aq) + NO3

– (aq) H = –65.9 kJ

Q.33 Addition solid silver sulfate, a slightly soluble solid, is added to a solution of silver ion and

sulfate ion in equilibrium with solid silver sulfate. Which of the following will occur? (a) The

Ag+ and SO42– concentration will not change. (b) the added silver sulfate will dissolve. (c)

Additional silver sulfate will form and precipitate from solution as Ag+ ions and SO42– ions

combine. (d) The Ag+ ion concentration will increase and the SO42– ion concentration will

decreases.

Q.34 Consider a general, single-step reaction of the type A + B C. Show that the equilibrium

constant is equal to the ratio of the rate constant for the forward and reverse reaction Kc = kf/kr.

Q.35 Which of the following relative values of kf and kr results in an equilibrium mixture that

contains large amounts of reactants and small amounts of product?

(a) kf > kr (b) kf = kr (c) kf < kr

Q.36 Consider the gas-phase hydration of hexafluoroacetone, (CF3)2CO:

(CF3)2CO(g) + H2O (g) f

r

k

k (CF3)2C(OH)2(g)

At 76ºC, the forward and reverse rate constants are kf = 0.13 M–1s–1 and kr = 6.02 × 10–4s–1.

What is the value of the equilibrium constant Kc?

Q.37 Consider the reaction of chlorometahane with OH– in aqueous solution

CH3Cl(aq) + OH–(aq) f

r

k

k CH3OH (aq) + Cl–(aq)

At 25ºC, the rate constant for the forward reaction is 6 × 10–6 M–1s–1, and the equilibrium

constant Kc is 1 × 10+16. Calculate the rate constant for the reverse reaction at 25ºC.

Q.38 The progress of the reaction

A nB with time, is presented in figure. Determine

(i) the value of n.

(ii) the equilibrium constant k.

(iii) the initial rate of conversion of A.

1 3 5 7

0.1

0.3

0.5

Time/Hour


Ph.: 0291-3291970.

28

Q.39 Listed in the table are forward and reverse rate constants for the reaction 2NO(g) N2(g) +

O2(g)

Temperature (K) kf(M–1s–1) kr(M–1s–1)

1400 0.29 1.1 × 10–6

1500 1.3 1.4 × 10–5

Is the reaction endothermic or exothermic? Explain in terms of kinetics.

Q.40 Forward and reverse rate constant for the reaction CO2(g) + N2(g) CO(g) + N2O(g)

exhibit the following temperature dependence.

Temperature (K) kf(M–1s–1) kr(M–1s–1)

1200 9.1 ×10–11 1.5 × 105

1500 2.7 × 10–9 2.6 × 105

Is the reaction endothermic or exothermic? Explain in terms of kinetics.

Q.41 The equilibrium constant Kp for the reaction PCl5(g) PCl3(g) + Cl2(g) is 3.81 × 102 at

600 K and 2.69 × 103 at 700K. Calculate rH.

Q.42 As shown in figure a catalyst lowers the activation energy for the forward and reverse reactions

by the same amount, Ea.

(a) Apply the Arrhenius equation, K = aE /RTAe

to the forward and reverse reactions, and

show that a catalyst increases the rates of both reaction s by the same factor.

E (forward)

without catalysta

E (reverse)

without catalysta

E (reverse)

with catalystaE (forward)

with catalysta

Po

ten

tial

energ

y

Reaction progress (b)Use the relation between the equilibrium constant and the forward and reverse rate constants,

Kc = kf/kr, to show that a catalyst does not affect the value of the equilibrium. constant.

Q.43 Variation of equilibrium constant ‗K‘ with temperature ‗T‘ is given by equation

log K = log A – Hº

2.303RT

A graph between log K and 1/T was a straight line with –ve slope of 0.5 and intercept 10.

Calculate

(a)Hº.

(b)Pre exponential factor

(c)Equilibrium constant at 298 K assuming Hº to be independent of temperature.

(d)Equilibrium constant at 798 K assuming Hº to be independent of temperature.

Q.44 Rate of disappearance of the reactant A at two different temperature is given by A B

d[A]

dt

= (2 × 10–2S–1) [A] – 4 × 10–3S–1[B] ; 300 K

d[A]

dt

= ( 4 × 10–2S–1) [A] –16 × 10–4 [B] ; 400 K

Calculate heat of reaction in the given temperature range. When equilibrium is set up.


Ph.: 0291-3291970.

29

Q.45 The KP for reaction A + B C + D is 1.34 at 60ºC and 6.64 at 100ºC. Determine the free

energy change of this reaction at each temperature and Hº for the reaction over this range of

temperature?

Q.46 If Kc = 7.5 × 10–9 at 1000 K for the reaction N2(g) + O2(g) 2NO(g), what is Kc at 1000

K for the reaction 2NO(g) N2(g) + O2(g)?

Q.47 An equilibrium mixture of PCl5, PCl3 and Cl2 at a certain temperature contains 8.3 × 10–3 M

PCl5, 1.5 × 10–2 M PCl3, and 3.2 × 10–2 M Cl2. Calculate the equilibrium constant Kc for the

reaction PCl5 (g) PCl3(g) + Cl2(g).

Q.48 A sample of HI2(9.30 × 10–3 mol) was placed in an empty 2.00 L container at 1000K. After

equilibrium was reached, the concentration of I2 was 6.29 × 10–4 M. Calculate the value of Kc

at 1000K for the reaction H2(g) + I2 2HI(g).

Q.49 The vapour pressure of water at 25ºC is 0.0313 atm. Calculate the values of Kp and Kc at 25ºC

for the equilibrium H2O(l) H2O(g).

Q.50 For each of the following equilibria, write the equilibrium constant expression for Kc. Where

appropriate, also write the equilibrium constant expression for Kp.

(a) Fe2O3(s) + 3CO(g) 2Fe(l) +3CO2(g) (b) 4Fe(s) + 3O2(g) 2Fe2O3(s)

(c) BaSO4(s) BaO(s) + SO3(g) (d) BaSO4(s) Ba2+ (aq) + SO42– (aq)

Q.51 When 0.5 mol of N2O4 is placed in a 4.00 l reaction vessel and heated at 400 K, 79.3% of the

N2O4 decomposes to NO2.

Calculate Kc and Kp at 400 K for the reaction N2O4(g) 2NO2(g)

Q.52 What concentration of NH3 is in equilibrium with 1.0 × 10–3 M N2 and 2.0 × 10–3 MH2 at 700

K? At this temperature Kc = 0.291 for the reaction N2(g) + 3H2(g) 2NH3(g).

Q.53 At 100 K, the value of Kc for the reaction C(s) + H2O(g) CO(g) + H2 (g) is 3.0 × 102.

Calculate the equilibrium concentration of H2O, CO2, and H2 in the reaction mixture obtained

by heating 6.0 mol of steam and an excess of solid carbon in a 5.0 L container. What is the

molar composition of the equilibrium mixture?

Q.54 When 1.0 mol of PCl5 is introduced into a 5.0 L container at 500 K, 78.5% of the PCl5

dissociates to give an equilibrium mixture of PCl5, PCl3, and Cl2.

PCl5 (g) PCl3(g) + Cl2(g)

(a) Calculate the values of Kc and Kp.

(b) If the initial concentration s in a particular mixture of reactants and products are [PCl5] = 0.5

M. What are the concentrations when the mixture reaches equilibrium?

Q.55 The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to

propene is 1.0 × 105at 500 K.

CH2

H C2 CH2 CH3–CH=CH2 Kc = 1.0 × 105

cyclopropane Propene

(a) What is the value of Kp at 500 K?

(b) What is the equilibrium partial pressure of cyclopropane at 500 K when the partial pressure

of propene is 5.0 atm?


Ph.: 0291-3291970.

30

(c) Can you alter the ratio of the two concentrations at equilibrium by adding cyclopropane or

by decreasing the volume of the container? Explain

(d) Which has the larger rate constant, the forward reaction or the reverse reaction?

(e) Why is cyclopropane so reactive?

Q.56 -D-Glucose undergoes mutarotation to -D-Glucose in aqueous solution. If at 298 K there is

60% conversion. Calculate Gº of the reaction.

-D-Glucose -D-Glucose

Q.57 For the reaction at 298 K

A(g) + B(g) C(g) + D(g)

Hº = –29.8 kcal; Sº = – 0.1 kcal / K

Calculate Gº and K.

Q.58 The equilibrium constant of the reaction 2C3H6(g) C2H4(g) + C4H8(g) is found to fit the

expression ln K = –1.04 – 1088K

T

Calculate the standard reaction enthalpy and entropy at 400 K.

Problem Set – 5

Q.1 For an equilibrium reaction involving gases, the forward reaction is Ist order while

the reverse reaction is IInd order. The units of Kp for the forward equilibrium is:

(A) atm (B) atm2 (C) atm–1 (D) atm–2

Q.2 A chemical reaction A B is said to be at equilibrium when :

(A) Complete conversion of A to B has taken place

(B) Conversion of A to B is only 50% complete

(C) Only 10% conversion of A to B has taken place

(D) The rate of transformation of A to B is just equal to the ra te of transformation

of B to A in the system.

Q.3 In the chemical reaction , N2 + 3 H2 2 NH3 at equilibrium :

(A) equal volumes of N2 & H2 are reacting

(B)equal masses of N2 & H2 are reacting

(C) the reaction has stopped

(D) the amount of ammonia formed is equal to amount of NH3 decomposed into N2 & H2

Q.4 For N2 + 3 H2 2 NH3 ; H = – ve , then :

(A) Kp = Kc (B) Kp = KcRT (C) Kp = Kc(RT)–2 (D) Kp = Kc(RT)–1

Q.5 For the reaction , H2 (g) + I2 (g) 2 HI (g) equilibrium constant , Kp changes with :

(A) Total pressure (B) Catalyst

(C) Amount of H2 and I2 present (D) Temperature

Q.6 For the reaction H2 + I2 2HI :

(A) Kc = 2Kp (B) Kc > Kp (C) Kc = Kp (D) Kc < Kp

Q.7 Which oxide of nitrogen is the most stable:

(A) 2NO2(g) N2(g) + 2O2(g); K = 6.7 × 1016 mol litre–1

(B) 2NO(g) N2(g) + O2(g); K = 2.2 × 1030 mol litre–1


Ph.: 0291-3291970.

31

(C) 2N2O5(g) 2N2(g) + 5O2(g); K = 1.2 × 1034 mol–5 litre–5

(D) 2N2O(g) 2N2(g) + O2(g); K = 3.5 × 1033 mol litre–1

Q.8 The equilibrium constant for equilibria,

SO2(g) + 12

O2(g) SO3(g) and

2SO3(g) 2SO2(g) + O2(g) are K1 and K2 respectively. Then

(A) K2 = K1 (B) K2 = K12 (C) K2 = 1/K1 (D) K2 = 1/K1

2

Q.9 A higher value for equilibrium constant, K shows that :

(A) The reaction has gone to near completion towards right.

(B) The reaction has not yet started

(C) The reaction has gone to near completion towards left

(D) None of these

Q.10 If K1 and K2 are equilibrium constants for reactions (I) and (II) respectively for,

N2 + O2 2NO

12

N2 + 12

O2 NO

Then:

(A) K2 = K1 (B) K2 = 1K (C) K1 = 2K2 (D) K1 = (1/2)K2

Q.11 A reversible chemical reaction having two reactants is in equilibrium. If the concentrations of

the reactants are doubled then the equilibrium constant will:

(A) Also be doubled (B) Be halved

(C) Become one fourth (D) Remain the same

Q.12 For which reaction does the equilibrium constant depend on the units of concentration :

(A) NO(g) 12

N2(g) + 12

O2(g)

(B) Zn(s) + Cu2+ (aq.) Cu(s) + Zn2+ (aq.)

(C) C2H5OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l)

(D) COCl2(g) CO(g) + Cl2(g)

Q.13 Hydrogen and oxygen were heated together in a closed vessel. The equilibrium constant is

found to decrease after 200ºC. Which is responsible for this:

(A) Backward reaction predominates (B) Forward reaction predominates

(C) Both forward and backward reaction have same rate

(D) It is a property of the system, hence no reason for lower value.

Q.14 If different quantities of ethanol and acetic acid are used in the following reversible reaction,

CH3COOH + C2H5OH CH3COOC2H5 + H2O

the equilibrium constant will have values which will be:

(A) same in all cases (B) Different in all cases

(C) Higher in cases where higher concentration of ethanol is used

(D) Higher in cases where concentration of acetic acid is used.

Q.15 Which reaction has n = 2:

(A) CaCO3(s) CaO(s) + CO2(g)

(B) 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)

(C) NH4Cl(g) NH3(g) + HCl (g)

(D) CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(v)


Ph.: 0291-3291970.

32

Q.16 When a system is at equilibrium:

(A) The mass of the product is equal to the mass of the reactant

(B) The ratio of the product of the mass of the products and that of the reactants is constant

(C) The ratio of the velocity of the forward reaction and the backward reaction is one

(D) Number of moles of the product and that of the reactant are the same

Q.17 For the reaction,

CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O() which one is correct representation:

(A) Kp = 2

2( )H Op (B) Kc = [H2O]2 (C) Kp = Kc(RT)2 (D) All

Q.18 For a hypothetical equilibrium:

4A + 5B 4X + 6Y ; The equilibrium constant Kc has the unit:

(A) mol2 litre–1 (B) litre mol–1 (C) litre2 mol–2 (D) mol litre–1

Q.19 For the gaseous phase reaction,

2NO N2 + O2, Hº = + 43.5 kcal mol–1, Which statement is correct for,

N2(g) + O2(g) 2NO(g);

(A) K is independent of temperature (B) K increases as temperature decreases

(C) K decreases as temperature decreases (D) K varies with addition of NO

Q.20 In which of the following cases, does the reaction go farthest to completion:

(A) K = 103 (B) K = 10–2 (C) K = 10 (D) K = 1

Q.21 For the homogenous gaseous phase reaction,

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) the equilibrium constant, Kc, has the units of :

(A) Unit of concentration10 (B) Unit of concentration1

(C) Unit of concentration–1 (D) It is dimensionless

Q.22 If K1 and K2 are the respective equilibrium constants for the two reactions,

XeF6(g) + H2O(g) XeOF4(g) + 2HF(g)

XeO4(g) + XeF6(g) XeOF4(g) + XeO3F2(g)

The equilibrium constant for the reaction,

XeO4(g) + 2HF(g) XeO3F2(g) + H2O(g) is:

(A) K1K2 (B) K1/K22 (C) K2/K1 (D) K1/K2

Q.23 If 1.0 mole of I2 is introduced into 1.0 litre flask at 1000 K, at equilibrium (Kc = 10–6), which

one is correct:

(A) [I2(g)] > [I–(g)] (B) [I2(g)] < [I–(g)] (C) [I2(g)] = [I–(g)] (D) [I2(g)] = [I–(g)]

Q.24 The active mass of 7.0 g of nitrogen in a 2.0 L container would be -

(A) 0.25 (B) 0.125 (C) 0.5 (D) 14.0

Q.25 At 2800 K, a 1.0 mole sample of CO2 in a one litre container is 50% decomposed to carbon

monoxide and oxygen at equilibrium

2CO2(g) 2CO(g) + O2(g)

The value of Kc for the reaction is -

(A) 0.25 (B) 0.50 (C) 50 (C) 25


Ph.: 0291-3291970.

33

Problem Set – 6

Q.1 If x is the degree of dissociation of PCl5

PCl5(g) PCl3(g) + Cl2 (g)

at a given temperature and if 2 moles of PCl5 are taken in a vessel, then at equilibrium the total

number of moles of various species would be -

(A) 4 (B) 2 + x (C) 2 (1 – x) (D) 2(1 + x)

Q.2 AB dissociates as, 2AB (g) 2A(g) + B2(g)

when the initial pressure of AB is 500 mm, the pressure becomes 625 mm when the equilibrium

is attained. Calculate Kp for the reaction assuming volume remains constant -

(A) 500 (B) 125 (C) 750 (D) 375

Q.3 Starting with pure COCl2, if the initial concentration of COCl2 is 0.03 M and the equilibrium

concentration of Cl2 is 0.02 M, the equilibrium constant, KC for the reaction -

COCl2(g) CO(g) + Cl2(g) would be

(A) 4 × 10–4 (B) 4 × 10–3 (C) 4.0 (D) 4 × 10–2

Q.4 For the reaction, N2 + 3H2 2NH3

Starting with one mole of nitrogen and 3 moles of hydrogen, at equilibrium 50% of each had

reacted. If the equilibrium pressure is P, the partial pressure of hydrogen at equilibrium would

be -

(A) P/2 (B) P/3 (C) P/4 (D) P/6

Q.5 For a hypothetical reaction, A(g) + B(g) X(g) + Y(g)

occurring in a single step, the specific rate constant are 2.0 × 10–2 and 5.0 × 103 respectively

for the forward and the backward reactions. The equilibrium constant is -

(A) 4.0 ×10–4 (B) 2.5 × 10–6 (C) 2.5 × 105 (D) 4.0 × 10–6

Q.6 For the equilibrium

NH4Cl(s) NH3(g) + HCl (g)

Kp = 4.9 × 10–9 atm2 at a certain temperature. The vapour pressure of NH4Cl(s) at this

temperature would be -

(A) 4.9 × 10–9 atm (B) 7 × 10–5 atm (C) 2.45 × 10–9 atm (D) 1.4 × 10–4 atm

Q.7 The equilibrium constant for the reaction,

H2 + I2 2HI, is 49 at a certain temperature. The equilibrium constant for the reaction

HI 1

2H2 +

1

2 I2 will be -

(A) 49 (B) 7 (C) 1/7 (D) 1/49

Q.8 For an equilibrium, N2O4(g) 2NO2 (g)

the total pressure at equilibrium is P and degree of dissociation of N2O4 is x. Which one of the

following is the partial pressure of NO2 ?

(A) 2

(1 )

xP

x (B) 2 × P (C)

2

(1 )P

x (D)

2

3× P

Q.9 For a gaseous reaction, pA + qB qC + pD

Which of the following relationship is true ?

(A) KP = KC (B) KP = KC (RT)p+ q

(C) KP = KC (RT)p–q (D) KP = KC = (RT)1/p+q

Q.10 The units of equilibrium constant for the reaction,

N2 + 3H2 2NH3 + heat, are

(A) mol–2 L2 (B) mol L–1 (C) mol2L–2 (D) L mol–1


Ph.: 0291-3291970.

34

Q.11 One mole of SO3 was placed in a litre vessel at a certain temperature. When equilibrium was

established in the reaction

2SO3(g) 2SO2 (g) + O2(g)

The vessel was found to contain 0.4 moles of SO3. The value of equilibrium constant is

(A) 0.13 (B) 0.36 (C) 0.68 (D) 0.45

Q.12 When 1 mole of N2 and 1 mole of H2 is enclosed in 5 L vessel and the reaction is allowed to

attain equilibrium, it is found that at equilibrium there is x mole of H2. The number of moles of

NH3 formed would be

(A) 2

3

x (B)

2(1 x)

3

(C)

2(1 )

3

x (D)

(1 )

2

x

Q.13 A one litre vessel initially contains 2.0, 0.5 and 0.0 moles of N2, H2 and NH3 respectively. The

system after attaining equilibrium has 0.2 mole of NH3. The number of moles of N2 in the

vessel at equilibrium is

(A) 1.9 (B) 0.4 (C) 0.2 (D) 1.8

Q.14 At certain temperature, 50% of HI is dissociation into H2 and I2, the equilibrium constant of the

reaction is

(A) 1.0 (B) 3.0 (C) 0.5 (D) 0.25

Q.15 1.1 mole of A is mixed with 1.2 mole of B and the mixture is kept in a one-litre flask till the

equilibrium A + 2B 2C + D is reached. At equilibrium 0.1 mole of D is formed. The

equilibrium constant of the above reaction is

(A) 0.002 (B) 0.004 (C) 0.001 (D) 0.003

Q.16 Favourable conditions for manufacture of ammonia by the reaction

N2 + 3H2 2NH3; H = – 21.9 kcal are :

(A) Low temperature, low pressure and catalyst

(B) Low temperature, high pressure and catalyst

(C) High temperature, low pressure and catalyst

(D) High temperature, high pressure and catalyst

Q.17 For the reaction , 2X(g) + Y(g) 2Z(g) ; H = – 80 kcal the highest yield of

Z at equilibrium occurs at :

(A) 1000 atm and 500ºC (B) 500 atm and 500ºC

(C) 1000 atm and 100ºC (D) 500 atm and 100ºC

Q.18 The Haber‘s process for the manufacture of ammonia is usually carried out at about

500ºC. If a temperature of about 250ºC was used instead of 500ºC:

(A) No ammonia would be formed at all

(B) The percentage of ammonia in the equi librium mixture would be too low

(C) A catalyst would be of no use at all at this temperature.

(D) The rate of formation of ammonia would be too slow.

Q.19 At equilibrium, If Kp = 1, then:

(A) Gº = 0 (B) Gº > 1 (C) Gº < 1 (D) None

Q.20 Does Le Chatelier‘s principle predict a change of equilibrium concentration for the

following reaction if the gas mixture is compressed :

N2O4 (g) 2 NO2 (g)

(A) Yes , backward reaction is favoured (B) Yes , forward reaction is favoured

(C) No change (D) No information

Q.21 Oxidation of SO2 by O2 to SO3 is an exothermic reaction. The yield of SO3 will be maximum if:

(A) Temperature is increased and pressure is kept constant

(B) Temperature is reduced and pressure is increased

(C) Both temperature and pressure are increased

(D) Both temperature and pressure are reduced


Ph.: 0291-3291970.

35

Q.22 For the equilibrium , 2NO2(g) N2O4(g) + 14.6 kcal

An increase of temperature will:

(A) Favour the formation of N2O4 (B) Favour the decomposition of N2O4

(C) Not affect the equilibrium (D) Stop the reaction

Q.23 Which equilibrium in gaseous phase would be unaffected by an increase in pressure:

(A) N2O4 2NO2 (B) N2 + O2 2NO

(C) N2 + 3H2 2NH3 (D) CO + ½O2 CO2

Q.24 For which reaction high pressure and high temperature is helpful in obtaining a high equilibrium

yield:

(A) 2NF3(g) N2(g) + 3F2(g) – 54.40 kcal

(B) N2(g) + 3H2(g) 2NH3(g) + 22.08 kcal

(C) Cl2(g) + 2O2(g) 2ClO2(g)– 49.40 kcal

(D) 2Cl2O7(g) 2Cl2(g) + 7O2(g) + 126.8 kcal

Q.25 In the gaseous equilibrium , A + 2B C + heat, the forward reaction is favoured by:

(A) Low temperature (B) Low pressure

(C) High pressure and low temperature (D) High pressure and high temperature

Problem Set – 7

Q.1 An increase in the temperature of an equilibrium system:

(A) Favours the exothermic reaction

(B) Favours the endothermic reaction

(C) Favours both the exothermic and endothermic reactions

(D) Favours neither the exothermic nor endothermic reactions.

Q.2 The degree of dissociation of PCl5 () obeying the equilibrium,

PCl5 PCl3 + Cl2, is approximately related to the pressure at equilibrium by:

(A) P (B) 1

P (C)

2

1

p (D)

4

1

p

Q.3 A reaction in which an increase in pressure will increase the yield of products:

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

(B) H2O(g) + CO(g) CO2(g) + H2(g)

(C) H2O(g) + C(s) CO(g) + H2(g)

(D) CO(g) + 3H2(g) CH4(g) + H2O(g)

Q.4 In the thermal decomposition of potassium chlorate in a closed container given as,

2KClO3(g) 2KCl (s) + 3O2(g), law of mass action:

(A) Cannot be applied

(B) Can be applied at hig temperature

(C) can be applied at low temperature

(D) Can be applied at high temperature and pressure.

Q.5 What is the direction of a reversible reaction when one of the products of the reaction is

removed:

(A) Forward direction (B) Backward direction

(C) The reaction stops (D) All the above

Q.6 The endothermic reaction (M + N P) is allowed to attain an equilibrium at 25ºC.

Formation of P can be increased by:

(A) Raising temperature (B) Lowering temperature


Ph.: 0291-3291970.

36

(C) Keeping temperature constant (D) Decreasing the concentration of M and N

Q.7 Consider the reaction, PCl5(g) PCl3(g) + Cl2(g) in a closed container at equilibrium. At

a fixed temperature addition of more PCl5 at equilibrium will show.......in equilibrium

concentration of Cl2(g):

(A) Decrease (B) Increase (C) No change (D) None

Q.8 Vapour density of PCl5 is 104.16 but when heated to 230ºC its vapour density is reduced to 62.

The degree of dissociation of PCl5 at this temperature will be -

(A) 6.8% (B) 68% (C) 46% (D) 64%

Q.9 When pressure is applied to the equilibrium system

Ice Water

which of the following phenomenon will happen ?

(A) more ice will be formed (B) water will evaporate

(C) more water will be formed (D) equilibrium will not be disturbed

Q.10 Which of the following will not change the concentration of ammonia at the equilibrium ?

(A) increase of pressure (B) increase of volume

(C) addition of catalyst (D) decrease of temperature

Q.11 Pure ammonia is placed in a vessel at a temperature when its dissociation is appreciable. At

equilibrium

(A) does not change with pressure

(B) concentration of ammonia does not change with pressure

(C) concentration of hydrogen is less than that of nitrogen

(D) Kp does not change significantly with pressure

Q.12 Which of the following is not favourable for the formation of SO3 ?

2SO2 (g) + O2(g) 2SO3 (g) ; H = –188 kJ

(A) high pressure (B) high temperature (C) decreasing [SO3] (D) increasing [SO2]

Q.13 For which of the following gaseous equilibrium at constant temperature, doubling the volume

would cause a shift of equilibrium to the right ?

(A) 2CO + O2 2CO2 (B) N2 + 3H2 2NH3

(C) N2 + O2 2NO (D) PCl5 PCl3 + Cl2

Q.14 The yield of the product in the reaction , A2 (g) + 2B(g) C(g) + Q kJ

would be higher at

(A) high temperature and high pressure (B) high temperature and low pressure

(C) low temperature and high pressure (D) low temperature and low pressure

Q.15 For a reaction, the value of K increases with increase in temperature. The H for the reaction

would be

(A) +ve (B) – ve (C) zero (D) cannot be predicted

Q.16 For the reaction, PCl5(g) PCl3(g) Cl2(g) at equilibrium the introduction of inert gas at

constant pressure will shift the equilibrium

(A) in forward direction (B) in backward direction

(C) neither forward nor backward direction (D) cannot be predicted

Q.17 A and B are gaseous substance which react reversibly to given two gaseous substances C and D,

accompanied by the liberation of heat. When the reaction reaches equilibrium, it is observed that

Kp = Kc. The equilibrium cannot be disturbed by

(A) adding A (B) adding D

(C) raising the temperature (D) increasing the pressure

Q.18 In a vessel containing SO3, SO2 and O2 at equilibrium some helium gas is introduced so that

the total pressure increases while temperature and volume remains constant. According to Le-

chatelier‘s principle, the dissociation of SO3


Ph.: 0291-3291970.

37

(A) increases (B) decreases

(C) remain unaltered (D) change unpredictably

Q.19 For the reaction, PCl5(g) PCl3(g) + Cl(g), the forward reaction at constant temperature is

favaoured by

(A) introducing an inert gas at constant volume(B) introducing Cl2 gas at constant volume

(C) introducing inert gas at constant pressure(D) decreasing the volume of the container

Q.20 The reaction, SO2 + Cl2 SO2Cl2 is exothermic and reversible. A mixture of SO2(g),

Cl2(g) and SO2Cl2(l) is at equilibrium in a closed container. Now a certain quantity of extra SO2

is introduced into the container, the volume remaining the same. Which of the following are true ?

(A) the pressure inside the container will not change

(B) the temperature will not change

(C) the temperature will increase

(D) the temperature will decrease

Q.21 For the reaction, CaCO3(s) CaO(s) + CO2(g) the amount of CaO after attainment of

equilibrium can be increased by –

(A) adding some CaCO3(s) (B) removing some CaCO3(s)

(B) adding some lime water (D) decreasing the temperature

Q.22 To the system A(s) + 2X(g) + heat 2B(s) + 3Y(g), at equilibrium more X(g) is added with

temperature & volume constant. If the new steady pressure of X(g) becomes 2.828 times of the

previous steady pressure, the number of moles of Y(g) are increased by a factor of –

(A) 31/3 (B) 3 (C) 2 (D) 2

Q.23 If a chemical equilibrium Reactants

F.R.

B.R. Products is shifted to the right i.e. forming the

products both by an increasing of pressure & rise in temperature then :

(A) Forward reaction is exothermic with n < 0

(B) Forward reaction is endothermic with n > 0

(C) Backward reaction is exothermic with n < 0

(D) Backward reaction is exothermic with n > 0

Q.24 The equilibrium concentrations of (A) and (B) at a certain temperature for the reaction

2A(g) + B(g) 2C(g) + D(s)

are 15 M and 30 M respectively. When the volume was made four fold and the reaction was

allowed to reach equilibrium the concentration of B was found to be 8 M.

(i) The concentration of C at first equilibrium is -

(A) 2.89 (B) 11.55 (C) 5.77 (D) 23.1

(ii) What will be the concentration of C at Second equilibrium.

(A) 1.89 M (B) 1.50 M (C) 2.89 M (D) 2.50 M

(iii) Concentration of A at second equilibrium will be -

(A) 2.75 M (B) 3.75 M (C) 4.75 M (D) 15 M

Q.25 The equilibrium concentration of the reactants and products for the given equilibrium in a two

litre container are shown below :

PCl3(g) + Cl2(g) PCl5(g)

2M 1M 4M

(i) If 2 mole of Cl2 are added in the container, then concentration PCl3 , Cl2 & PCl5 at

newnequilibrium will be

(A) 2, 1, 4 (B) 1.5, 1.5, 4.5 (C) 2, 1.5, 4.5 (D) 2, 2, 5

(ii) If the equilibrium mixture reported initially is transferred into 4 litre vessel, the what

woulnbe the new concentration of PCl3, Cl2 & PCl5 at equilibrium

(A) 1.225, 0.725, 1.775 (B) 1.5, 0.5, 2

(C) 1.775, 0.725, 1.225 (D) none of these


Ph.: 0291-3291970.

38

Answer sheet of problem sets

Problem Set – 1

1. 22

31

KK =

K 2. 0.2 3. KC = 4.0 4. N2O4 5. 1.37

6. (a) 0.048 mol lit–1 (b) PAB = 0.8467 atm, PA= PB = 1.4112 atm (c) KP = 2.35 atm

7. 0.1244 mole 8. 48.22 atm–1/2 9. 8.93 atm 10. 2

(2n - y)y

(n - y) P

11.2

2

P(n + y / 2)(n + y)

(3n + y / 2)(n - y) 12. 0.341 mole 13. 1.868

14. CH3COOH = 0.07 mol, C2H5OH = 3.07 mol, CH3COOC2H5 = 0.93 mole, H2O = 0.93 mole

15. 17.64

Problem Set – 2 1. T = 894.26ºC 2. 0.7837 3. 0.0174 gm

4. 0.582 atm 5. 31/27 7. 97.55 atm

8. (a) PCl2 = PPCl3 = 0.814, PPCl5 = 0.372 atm (b) 1.78 atm, 0.04 mol lit–1

(c) (i) 94.8% (ii) 0.0053, 0.097 atm 9. 0.6, 1.75 atm

10. 84.6%, 5.87× 10–2 mol lit–1, 2.53 atm 11. x = 0.333, Kp = 0.204 atm

12. 0.43 mole NO2 14. (i) 64% (ii) 21.42 % (iii) V = 12.998 ml

15. Backward reaction

Problem Set – 3 1. Forward 2. Forward 3. 0.5 , 1 , 1

4. Backward , Forward , Equilibrium 5. D 6. A

7. A 8. A 9. A 10. 2.7636 kCal

11. Kp = 779.41 atm–1 G = –32.998 kJ/mole, Kp = 6.07 × 105 atm–2

12. 285.23

13. (a) [A = 0.52 M, B = C = 0.48 M] (b) A = 0.07 M, B = C = 0.18 M

14. Kp = 1.429 × 10–5 atm–2 15. (i) 376.56 kJ (ii) 251.04 kJ (iii) 125.52 kJ

Problem Set – 4

1. (a) 25, shifts left, (b) 0.22, shifts right (c) , shifts left

(d) 1 shifts right (e) 0, shift right (f) 4, shift left

2. (a) K = [Ag+] [Cl–] is less than 1. AgCl is insoluble thus the concentration of ions are much

less than 1M

(b) K = 1/[Pb2+][Cl–]2 is greater than one because PbCl2 is insoluble and formation of the

solid will reduce the concentration of ions to a low level.


Ph.: 0291-3291970.

39

4. K about 10 6. (a) incomplete (b) almost complete

7. c 8. ~ 9 × 10–32 mol/L

9. The reaction is not an equilibrium because Qc > Kc. The reaction will proceed from right to left

to reach equilibrium

11. 5.9 × 10–3 M 12. [NO] = 0.056 M, [N2] = [O2] = 13.7 M

13. [PCl3] = [Cl2] = 0.71 M, [PCl5] = 0.089

14. PClF = 2FP = 0.389 atm,

3ClFP = 1.08 atm 15. [KP = 0.4, a 0.1]

16. 50%

17. (a)6.667 × 10–3 mol L–1; (b) n(N2O4) = 0.374 mol; n (NO2) = 0.052 mol;

(c) 10.49 atm (d) 6.44%

18. 0.97 atm 19. KP = 1.3 × 10–3 atm–2 20. KP = 2.5 atm, P = 15 atm

21. 53.33% 22. K = 4 23. 31/27

24. 22.4 mg 25. 2H OP = 5 × 10–15 atm 26. 0.821 atm

27. add N2 add H2, increase the pressure, heat the reaction

28. (a) shift right, shift left, (b) shift right, no effect, (c) shift left, shift left,

(d) shift left, shift right

29. (a) K = [CH3OH]/[H2]2[CO],

(b) 1. [H2] increase, [CO] decrease, [CH3OH] increase;

2. [H2] increase, [CO] decrease, [CH3OH] decrease;

3. [H2] increase, [CO] increase, [CH3OH] increase;

4. [H2] increase, [CO] increase, [CH3OH] increase;

5. [H2] increase, [CO] increase, [CH3OH] decrease;

6. no change]

30. (a) K = [CO[H2]/[H2O];

(b) in each of the following cases the mass of carbon will change, but its concentration

(activity) will not change

1. [H2O] no change, [CO] increase,

2. [H2] increase, [CO] decrease, [CH3OH] decrease

3. [H2] increase, [CO] increase [CH3OH] increase

4. [H2O] increase, [CO] increase [H2] increase

5. [H2O] decrease, [CO] increase [H2] increase

31. b

32. Add NaCl or some other salt that produces Cl– in the solution. Cool the solution

33. a 34. kf [A][B] = kr[C] ; f

r

k

k =

[ ]

[ ][ ]

C

A B = kc 36. 216

38. (i) 2, (ii) 1.2 mol/L; (iii) 0.1 moles/hr

39. kr increase more than kf , this means that Ea (reverse) is greater than Ea(forward). The

reaction is exothermic when Ea ( reverse) > Ea (forward)

43. (a) +9.574 J/mol (b) A = 1010 (c) 9.96 × 109 (d) 9.98 × 109

44. +14.06 kJ 45. –810 J/mol; – 5872 J/mol and 41.3 kJ/mol

46. 1.3 × 108 47. 0.058 48. 29.0


Ph.: 0291-3291970.

40

49. KP = 0.0313 atm, KC = 1.28 × 10–3

50. (a) KC = 3

2

3

[ ]

[ ]

CO

CO, KP = 2

3

3

( )

( )

CO

CO

P

P ; (b) KC =

3

2

1

[ ]O , KP =

2

3

1

( )OP ,

(c) KC = [SO3], KP =3SOP , KC = [Ba2+][SO4

2–]

51. KC = 1.51 k, KP = 49.6 52. 1.5 × 10–6 M

53. [CO] = [H2] = 0.18 M; [H2O] = 1.02 M

54. (a) KC = 0.573 and KP = 23.5 ;

(b) to the right, [PCl5] = 0.365 M; [PCl3] = 0.285M, ; [Cl2] = 0.735M

56. –1.005 kJ/mol 57. Gº = 0 ; K = 1

58. Hº = 9.07 kJ/mol; Sº = – 8.92 J/mol–1K–1

Problem Set – 5 1. A 2. D 3. D 4. C 5. D 6. C 7. A 8. D

9. A 10. B 11. D 12. D 13. A 14. A 15. D 16. C

17. D 18. D 19. B 20. A 21. B 22. C 23. A 24. B

25. A

Problem Set – 6 1. D 2. B 3. D 4. A 5. D 6. D 7. C 8. A

9. A 10. A 11. C 12. C 13. A 14. D 15. B 16. B

17. C 18. D 19. A 20. A 21. B 22. B 23. B 24. C

25. C

Problem Set – 7 1. B 2. B 3. D 4. D 5. A 6. A 7. B 8. B

9. C 10. B 11. D 12. B 13. D 14. C 15. A 16. A

17. D 18. C 19. C 20. D 21. B 22. C 23. A

24. (i) B (ii) A (iii) A 25. (i) B (ii) A


Ph.: 0291-3291970.

41

Exercises

Exercise – 1: Single Choice Questions

Q.1 A chemical reaction, A B, is said to be in equilibrium when :

(A) rate of forward reaction is equal to rate of backward reaction

(B) conversion of A to B is only 50% complete

(C) complete conversion of A to B has taken place

(D) only 25% conversion of A to B taken place

Q.2 The equilibrium concentration of x, y and yx2 are 4, 2 and 2 respectively for the equilibrium 2x +

y yx2. The value of equilibrium constant, KC is

(A) 0.625 (B) 6.25 (C) 0.0625 (D) 62.5

Q.3 4 mole of A are mixed with 4 mole of B when 2 mole of C are formed at equilibrium, according

to the reaction, A + B C + D the equilibrium constant is :

(A) (B) 2 (C) 1 (D) 4

Q.4 HI was heated in a sealed tube at 400°C till the equilibrium was reached. HI was found to be 22%

decomposed. The equilibrium constant for dissociation is :

(A) 1.99 (B) 0.0199 (C) 0.0796 (D) 0.282

Q.5 In which of the following, the reaction proceeds towards completion ?

(A) K = 1 (B) K = 10–2 (C) K = 10 (D) K = 103

Q.6 For the following reaction at 250°C, PCl3(g) + Cl2(g) PCl5(g) the value of KC is 26 then

the value of Kp at same temperature will be

(A) 0.57 (B) 0.61 (C) 0.83 (D) 0.91

Q.7 For a reversible reaction, the rate constants for the forward and backward reactions are 2.38 ×10–

4 and 8.15 × 10–5 respectively. The equilibrium constant for the reaction is –

(A) 0.342 (B) 2.92 (B) 0.292 (D) 3.42

Q.9 If different quantities of ethanol and acetic acid was used in the following reversible reaction,

CH3COOH() + C2H5OH() CH3COOC2H5() + H2O()

the equilibrium constant will have values which will be ?

(A) different in all cases (B) same in all cases

(C) higher in cases where higher concentration of ethanol is used

(D) higher in case where higher concentration of acetic acid is used

Q.10 Which one of the following oxides is most stable? The equilibrium constants are given at the

same temperature:

(A) 2N2O5(g) 2N2(g) + 5O2(g) ; K = 1.2 × 1034

(B) 2N2O(g) 2N2(g) + O2(g) ; K = 3.5 × 1035

(C) 2NO(g) N2(g) + O2(g) ; K = 2.2 × 1030

(D) 2NO2(g) N2(g) + 2O2(g) ; K = 6.71 × 1016

Q.11 The yield of product in the reaction, A2(g) + 2B(g) C(g) + Q kJ

would be higher at :


Ph.: 0291-3291970.

42

(A) low temperature and high pressure (B) high temperature and high pressure

(B) low temperature and low pressure (D) high temperature and low pressure

Q.12 Manufacture of ammonia from the elements is represented by

N2(g) + 3H2(g) 2NH3(g) + 22.4 kcal the maximum yield of ammonia will be obtained

when the process is made to take place –

(A) at low pressure and high temperature (B) at low pressure and low temperature

(C) at high pressure and high temperature (D) at high pressure and low temperature

Q.13 In the reaction, 2SO2(g) + O2(g) 2SO3(g) + X cal, most favourable conditions of

temperature and pressure for greater yield of SO3 are :

(A) low temperature and low pressure (B) high temperature and low pressure

(C) high temperature and high pressure (D) low temperature and high pressure

Q.14 For the reaction : 2A(g) + B(g) 3C(g) + D(g)

two mole each of A and B were taken into a flask. The following must always be true when the

system attained equilibrium :

(A) [A] = [B] (B) [A] < [B] (C) [B] = [C] (D) [A] > [B]

Q.15 In a vessel containing SO2, SO3 and O2 at equilibrium, some helium gas is introduced so that

total pressure increases while temperature and volume remain the same. According to the Le

Chatelier‘s principle, the dissociation of SO3 :

(A) increases (B) decreases (C)remains unaltered (D) change unpredictably

Q.16 The equilibrium SO2Cl2(g) SO2(g) + Cl2(g) is attained at 25°C in a closed container and

an inert gas, helium, is introduced. Which of the following statements is correct ?

(A) concentrations of SO2Cl2, SO2 and Cl2 do not change

(B) more Cl2 is formed

(C) concentration of SO2 is reduced

(D) more SO2Cl2 is formed

Q.17 For the reactions,

A B, KC = 1, B C, KC = 3; C D, KC = 5

KC for the reaction A D is :

(A) 15 (B) 5 (C) 3 (D) 1

Q.18 The equilibrium constant for the reaction, N2(g) + O2(g) 2NO(g) is 4.0 × 10–4 at 2000

K. In the presence of a catalyst the equilibrium is attained ten times faster. Therefore, the

equilibrium constant in presence of the catalyst at 2000 K is –

(A) 4 × 10–4 (B) 40 × 10–4

(D) 4 × 10–2 (D) difficult to compute without more data

Q.19 The equilibrium constant for a reaction A + B C + D is 1 × 10–2 at 298 K and is 2 at 273

K. The chemical process resulting in the formation of C and D is :

(A) exothermic (B) endothermic

(C) unpredictable (D) there is no relationship between H and K

Q.20 In a flask colourless N2O4 is in equilibrium with brown coloured NO2. At equilibrium, when the

flask is heated to 100°C the brown colour deepens and on cooling, the brown colour became less

coloured. The change in enthalpy H for the system is:


Ph.: 0291-3291970.

43

(A) negative (B) positive (C) zero (D) not defined

Q.21 If 340 g of a mixture of N2 and H2 in the correct ratio gave a 20% yield of NH3, the mass

produced would be:

(A) 16 g (B) 17 g (C) 20 g (D) 68 g

Q.22 In a chemical reaction Kp is greater than Kc when

(A) the number of molecules entering into the chemical reaction is more than the number of

molecules produced.

(B) the number of molecules entering into the chemical reaction is the same as the number of

molecules produced.

(C) the number of molecules entering into the chemical reaction is less than the number of

molecules produced.

(D) the total number of moles of reactants is less than the number of moles of products.

Q.23 Which of the following statements is not correct about the equilibrium constant ?

(A) Its value does not depend upon the initial conc. of the reactants

(B) Its value does not depend upon the initial conc. of the products

(C) Its value does not depend upon nature of the reactants.

(D) Its value does not depend upon presence of catalyst.

Q.24 The equilibrium constant for the reaction, N2(g) + O2(g) 2NO(g) is 4 10–4

at 2000 K. In presence of a catalyst, equilibrium is attained ten times faster. Therefore, the

equilibrium constant, in presence of the catalyst, at 2000 K is

(A) 40 x 10–4 (B) 4 x 10–4

(C) 4 x 10–3 (D) difficult to compute without more data.

Q.25 2XY dissociates as 2(g) (g) (g)XY XY Y

. When the initial pressure of is 600 mm of Hg,

the total pressure developed is 800 mm of Hg. for the reaction is

(A) 200 (B) 50 (C) 100 (D) 150

Q.26 The reactions 5(g) 3(g) 2(g)PCl PCl Cl and 2(g) (g) 2(g)COCl CO Cl

are simultaneously in equilibrium in an equilibrium box at constant volume. A few moles of

(g)CO are later introduced into the vessel. After some time, the new equilibrium concentration of

(A) 5PCl will remain unchanged (B) 2Cl

will be greater

(C) 5PCl will become less (D) 5PCl

will become greater.

Q.27 For the reaction : (g) 2(g) 2(g)2HI H I ; the degree of dissociation () of HI(g) is related to

equilibrium constant by the expression

(A)

p1 2 K

2

(B)

p1 2K

2

(C)

p

p

2K

1 2K (D)

p

p

2 K

1 2 K.

Q.28 For which of the following reactions, the degree of dissociation cannot be calculated from the

vapour density data

I. (g) 2(g) 2(g)2HI H I II. 3(g) 2(g) 2(g)2NH N 3H

III. (g) 2(g) 2(g)2NO N O IV. 5(g) 3(g) 2(g)PCl PCl Cl

(A) I and III (B) II and IV (C) I and II (D) III and IV.


Ph.: 0291-3291970.

44

Exercise – 2: Subjective Type Questions

Q.1 For the reaction A + B 3C at 25ºC, a, 3 litre vessel contains 1,2,4 mole of A, B and C

respectively. Predict the direction of reaction if:

(a) KC for the reaction is 10.

(b) KC for the reaction is 15.

(c) KC for the reaction is 10.66.

Q.2 The equilibrium constant KC for A(g) B(g) is 1.1. Which gas has a molar concentration

greater than 1?

Q.3 For a gaseous phase reaction, A + 2B AB2, KC = 0.3475 litre2 mol–2 at 200ºC. When 2

mole of B are mixed with one mole of A, what total pressure is required to convert 60% of A in

AB2?

Q.4 For a gaseous phase reaction, 2HI H2 + I2, at equilibrium 7.8 g, 203.2 g and 1638.4 g of

H2, I2 and HI respectively were found in 5 litre vessel. Calculate KC. If all the reactants and

products are transferred to a 2 litre vessel., what will be the amount of reactants and product at

equilibrium?

Q.5 One mole of H2, two mole of I2 and three mole of HI are injected in one litre flask. What will be

the concentration of H2, I2 and HI at equilibrium at 500ºC? KC for reaction:

H2 + I2 2HI is 45.9.

Q.6 The degree of dissociation of HI at a particular temperature is 0.8. Calculate the volume of 2M

Na2S2O3 solution required to neutralise the iodine present in a equilibrium mixture of a reaction

when 2 mole each of H2 and I2 are heated in a closed vessel of 2 litre capacity.

Q.7 A mixture of one mole of CO2 and one mole of H2 attains equilibrium at a temperature of 250ºC

and a total pressure of 0.1 atm for the change CO2(g) + H2(g) CO(g) + H2O(g). Calculate

KP if the analysis of final reaction mixture shows 0.16 volume per cent of CO.

Q.8 At a certain temperature, KC for

SO2(g) + NO2(g) SO3(g) + NO(g)

is 16. If we take one mole of each of the four gases in one litre container, what would be the

equilibrium concentration of NO and NO2 ?

Q.9 At 627ºC and one atmosphere pressure SO3 is partially dissociated into SO2 and O2 by SO3(g)

SO2(g) + 1

2O2(g). The density of the equilibrium mixture is 0.925 g/litre. What is the

degree of dissociation?

Q.10 The equilibrium mixture for 2SO2(g) + O2(g) 2SO3(g)

present in 1 litre vessel at 600ºC contains 0.50, 0.12 and 5.0 mole of SO2, O2 and SO3

respectively.

(a) Calculate KC for the given change at 600ºC

(b)Also calculate KP

(c) How many mole of O2 must be forced into the equilibrium vessel at 600ºC in order to increase

the concentration of SO3 to 5.2 mole?


Ph.: 0291-3291970.

45

Q.11 At 273 K and one atm, ‗a‘ litre of N2O4 decomposes to NO2 according to equation N2O4(g)

2NO2(g). To what extent has the decomposition proceeded when the original volume is

25% less than that of existing volume?

Q.12 At 25ºC and 1 atmospheric pressure, the partial pressure in equilibrium mixture of gaseous N2O4

and NO2 are 0.7 and 0.3 atm respectively. Calculate the partial pressures of these gases when

they are in equilibrium at 25ºC and a total pressure of 10 atm.

Q.13 In a mixture of N2 and H2 in the ratio 1 : 3 at 30 atm and 300ºC, the % of NH3 at equilibrium is

17.8. Calculate KP for N2 + 3H2 2NH3.

Q.14 A reaction carried out by 1 mole of N2 and 3 mole of H2 shows at equilibrium, the mole fraction

of NH3 as 0.012 at 500ºC and 10 atm pressure. Calculate KP. Also report the pressure at which

mole % of NH3 in equilibrium mixture is increased to 10.4.

Q.15 At 450ºC, the equilibrium constant KP for the reaction N2 + 3H2 2NH3 was found to be

1.6 × 10–5 at a pressure of 200 atm. If N2 and H2 are taken in 1 : 3 ratio, what is the % of NH3

formed at this temperature?

Q.16 KP for the reaction 2O3(g) 3O2(g); is 1.3 × 1057 atm. How many O3 molecules are

present in equilibrium in 10 million cubic metres of air at 25ºC and 720 mm Hg pressure?

Q.17 The density of an equilibrium mixture of N2O4 and NO2 at 1 atm is 3.62 g/litre at 288 K and

1.84 g /litre at 348K. Calculte the enthalpy of reaction : N2O4 2NO2. Also calculate the

entropy change during the reaction at 348 K.

Q.18 The amount of CO2 in a gaseous mixture of CO2 and CO is determined by passing the mixture

into an aqueous solution that contains excess of Ba(OH)2. The CO2 reacts yielding a precipitate

of BaCO3 but CO does not react. This method is used to analyses the equilibrium composition of

gas when 1.77 g CO2 reacted with 2.0 g graphite in a 1.0 litre container at 1100 K. The analysis

yielded 3.41 g of BaCO3. Use this data to calculate Kp at 1100 K for the reaction

CO2(g) + C(s) 2CO

Q.19 At a certain temperature, KP for dissociation of solid CaCO3 is 4 × 10–2 atm and for the reaction

C(s) + CO2 2CO is 2.0 atm respectively. Calculate the pressure of CO at this temperature

when solid C, CaO, CaCO3 are mixed and allowed to attain equilibrium.

Q.20 Would 1% CO2 in air be sufficient to prevent any loss in weight when M2CO3 is heated at

120ºC?

Q.21 Calculate KC for the reaction, KI + I2 KI3. Given that initial weight of KI is 1.326 g.

Weight of KI3 is 0.105 g and no. of mole of free I2 is 0.0025 at equilibrium and the volume of

solution is one litre. The solubility of I2 in water is 0.0013 M.

Q.22 Sulphide ions in alkaline solution react with solid sulphur to form polyvalent sulphide ions. The

equilibrium constant for the formation of S22– and S3

2– from S and S2– ions are 1.7 and 5.3

respectively. Calculate equilibrium constant for the formation of S32– from S2

2– and S.


Ph.: 0291-3291970.

46

Exercise – 3: Objective Type Questions

Q.1 The equation = ( 1)

D d

n d

is correctly matched for :

(A) A (n/2)B + (n/3)C (B) A (n/3)B + (2n/3)C

(C) A (n/2)B + (n/4)C (D) A (n/2)B + C

Q.2 For a reaction nA An, degree of dissociation when A trimerises is:

(A)1

2

D d

d

(B) 3

2

D d

d

(C) 4

3

D d

d

(D) 2D d

d

Q.3 Ammonium hydrogen sulphide is contained in a closed vessel at 313 K when total pressure at

equilibrium is found to be 0.8 atm. The value of Kp for the reaction,

NH4HS(s) NH3(g) + H2S(g) is :

(A) 0.16 (B) 1.6 (C) 0.016 (D) 16

Q.4 In the melting of ice, which one of the conditions will be more favourable ?

(A) high temperature and high pressure (B) low temperature and low pressure

(C) low temperature and high pressure (D) high temperature and low pressure

Q.5 A cylinder provided with a piston has some PCl5 which is in equilibrium with PCl3 and Cl2. The

system is compressed with the help of piston. Indicate the correct statement :

(A) some more PCl5 will decompose (B) the system remains unaffected

(C) PCl3 and Cl2 will combine to form PCl5 (D) explosion occurs

Q.6 XY2 dissociates as : XY2(g) XY(g) + Y(g)

Initial pressure of XY2 is 600 mm Hg. The total pressure at equilibrium is 800 mm Hg. Assuming

volume of system to remain constant, the value of Kp is :

(A) 50 (B) 100 (C) 200 (D) 400

Q.7 At temperature T, a compound AB2(g) dissociates according to the reaction

2AB2(g) 2AB(g) + B2(g)

with a degree of dissociation ‗x‘ which is small as compared to unity. The expression for Kp, in

terms of ‗x‘ and total pressure ‗P‘ is

(A) 3

2

Px (B)

2

3

Px (C)

3

3

Px (D)

2

2

Px

Q.8 Two systems


and COCl2 (g) CO(g) + Cl2(g)

are simultaneously in equilibrium in a vessel at constant volume. If some CO(g) is introduced in

the vessel at constant volume, then at new equilibrium the concentration of :

(A) PCl5 is greater (B) PCl3 remains unchanged (C) PCl5 is less (D) Cl2 is greater

Q.9 To the system, LaCl3(s) + H2O(g) LaClO(s) + 2HCl(g) – heat

already at equilibrium, more water vapour is added without altering T or V of the system. When

equilibrium is re-established, the pressure of water vapour is doubled. The pressure of HCl

present in the system increases by a factor of –

(A) 2 (B) 21/2 (C) 3 (D) 4


Ph.: 0291-3291970.

47

Q.10 It is known that heat is needed to dissociated ammonia into N2 and H2. For the reaction

N2 + 3H2 2NH3, Kf is velocity constant for forward reaction and Kb is velocity constant

for back reaction, KC is equilibrium constant for the reaction shown. Then fdk

dT (where T is

symbol for absolute temp.)

(A) is greater than bdk

dT (B) is less than bdk

dT

(C) is equal to bdk

dT (D) cannot be compared with bdk

dT

Q.11 For the reaction, PCl5 PCl3 + Cl2, the equilibrium is attained at t = t1 assuming the

dissociation starts at t = 0. Select the correct alternative :

(A) dRf/dt < 0 for t [0, (2n/3)t1] (B) dRf/dt 0 for t [(n/3)t1, (2n/3)t1]

(C) dRb/dt > 0 for t [(3n/4)t1, (4n/5)t1] (D) dRb/dt 0 for t [(4n/5)t1, (7n/8)t1]

where n is least prime number.

Q.12 At t = 0 a vessel (volume 1 litre) contains 1 mole N2, 3 mole H2 and 2 mole NH3. The value of

KC for N2 + 3H2 2NH3 is 17.5 L2 mol–2. Then :

(A) total gaseous weight at equilibrium is more than 68 g

(B) total number of moles (gaseous) at equilibrium are more than 6 mol

(C) dM

dt is +ve for t (0, Teq.) where M is average molar mass of the reaction mixture

(D) total gaseous weight at equilibrium is less than 68 g.

Q.13 A vessel (volume = 2L) contains 60 g water gas. When steam is passed through the vessel the

reaction CO + H2O CO2 + H2 occurs and equilibrium is attained :

(A) 2( )

( )

n H

n CO < 1 (at equilibrium)

(B) n(H2) + n(CO) = 4 (at any instant)

(C) dM

dt > 0 (where M is average molar mass of gas mixture before the attainment of

equilibrium)

(D) dM

dt < 0

Q.14 A vessel contains CO2 and CO with pressure 2 atm and 3 atm respectively at 27°C. At a

temperature of 2727°C the reaction 2CO(g) CO2(g) + C(s) occurs and equilibrium is

attained. If the equilibrium pressure is 45 atm, then

(A) Kp = 2/9 atm–1 (B) PCO : PCO2 = 4 : 5 at equilibrium

(C) Kp = 16 atm–1 (D) % dissociation of CO = 0.333

Q.15 A vessel (volume 8.2 L) contains H2(g) at 2 atm. pressure. When H2S(g) at a pressure of 4 atm. is

introduced into the vessel the reaction 8H2S(g) 8H2(g) + S8(s) occurs at a temperature of

2000 K. It is found that:

2

2

( )

( )at equilibrium

n H

n H S

= 2

2 0

( )

( )at t

n H S

n H

(A) maximum weight of solid formed is 32 g(B) maximum weight of solid formed is 0.32 g

(C) Kp = KC(RT) (D) KC = 256


Ph.: 0291-3291970.

48

Q.16 N2(g) and H2(g) are taken in a vessel in mass ratio of 7 : 1. The only reaction,

N2 + 2H2 N2H4(g) occurs. Pressure due to N2H4 at equilibrium is 0.2 times of total

pressure ‗P‘. Then at equilibrium :

(A) Partial pressure of N2 = 2P/15 (B) partial pressure of H2 = 8P/25

(C) 2

2 Np = 2Hp (D)

2Np = 2

2 Hp

Q.17 In the equilibrium SO2Cl2 SO2 + Cl2 at 2000 K and 10 atm pressure, % Cl2 = % SO2 =

40 by volume. Then

(A) Kp = 2 atm (B) 2 2

2

( )

( )

n SO Cl

n SO=

1

4 at equilibrium

(C) Kp = 8 atm (D) n(SOCl2) = n(SO2) = n(Cl2)

Q.18 A 20 litre box contains O3 and O2 at equilibrium at 202 K. Kp = 2 × 1014 for 2O3 3O2.

Assume that 2Op >>

3Op and total pressure is 8 atm.; the partial pressure of O3 is :

(A) exactly 1.6 × 10–6 atm (B) sufficiently less than 1.6 ×10–6 atm

(C) slightly more than 1.6 × 10–6 atm (D) very slightly less than 1.6 × 10–6 atm

Q.19 A 2B, Kp; C D + E; '

pK . If degrees of dissociation of A and C are same and Kp =

2 '

pK , then the ratio of total pressure p/ 'p = ?

(A) 1/2 (B) 1/3 (C) 1/4 (D) 2

Q.20 For the reaction N2O4 2NO2, if degrees of dissociation of N2O4 are 25%, 50% 75% and

100%, the gradation of observed vapour densities is :

(A) d1 > d2 > d3 > d4 (B) d4 > d3 > d2 >d1 (C) d1 = d2 = d3 = d4 (D) none

Q.21 2 4 2 cN O 2NO , K 4 . This reversible reaction is studied graphically as shown in figure.

Select the correct statements out of I, II and III.

Ti me

Con

cen

trat

ion

A D E

F G

C

B I. Reaction quotient has maximum value at point A

II. Reaction proceeds left to right at a point when 2 4 2[N O ] [NO ] 0.1M

III. CK Q when point D or F is reached :

(A) I, II (B) II, III (C) I, III (D) I, II, III.

Q.22 When 20g of CaCO3 were put into 10 litre flask and heated to 800°C, 35% CaCO3 remained

unreacted at equilibrium, Kp for decomposition of CaCO3 is

(A) 1.145 atm (B) 0.145 atm (C) 2.145 atm (D) 3.145 atm

Q.23 Two moles of an equimolar mixture of two alcohols, 1R — OH and 2R — OH

are esterified

with one mole of acetic acid. If 80% of the acid is consumed and the quantities of ester formed

under equilibrium are in the ratio of 3 : 2, the value of the equilibrium constant for the

esterification of 1R — OH with acetic acid is

(A) 3.3 (B) 3.7 (C) 3.5 (D) 3.9 .


Ph.: 0291-3291970.

49

Q.24 A 1 M solution of glucose reaches dissociation equilibrium according to the equation given

below: 6 12 66HCHO C H O . hat is the concentration of HCHO at equilibrium, if equilibrium

constant is 6 × 1022

(A) 1.6 × 10–8M (B) 3.2 × 10–6M (C) 3.2 × 10–4M (D) 1.6 × 10–4M

Q.25 In a reversible reaction, study of its mechanism says that both the forward and backward reactions

follow first order kinetics. If the half life of forward reaction 1/ 2 f(t ) is 400 sec. and that of

backward reaction 1/ 2 b(t ) is 250 sec. The equilibrium constant of the reaction is

(g) 2(g) 2(g) (g)CO NO CO NO

(A) 1.6 (B) 0.433 (C) 0.625 (D) 1.109.

Q.26 At 27°C NO and 2Cl gases are introduced in a 10 litre flask such that their initial partial pressures

are 20 and 16 atm respectively. The flask already contains 24 g of magnesium. After some time,

the amount of magnesium left was 0.2 moles due to the establishment of following two equilibria

(g) 2(g) (g)2NO Cl 2NOCl

–1

2(g) (s) 2(s) pCl Mg MgCl ; K 0.2 atm

The final pressure of NOCl would be

(A) 7.84 atm (B) 18.06 atm (C) 129.6 atm (D) 64.8 atm.

Q.27 The 3CaCO is heated in a closed vessel of volume 1 litre at 600 K to form CaO and 2CO

. The

minimum weight of 3CaCO required to establish the equilibrium

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

(A) 2g (B) 4.57 g (C) 10g (D) 100 g.

Q.28 One mole of N2O(g) is kept in a closed container along with gold catalyst at 450 K under one

atmosphere. It is heated to 900 K when it dissociates to N2(g) and O2(g) giving an equilibrium

pressure of 2.4 atm. The degree of dissociation of N2O(g) is

(A) 20% (B) 40% (C) 50% (D) 60%.

Exercise – 4: Objective Type Questions

Q.1 Ammonium hydrogen sulphide dissociates as follows :

NH4HS(s) H2S(g) + NH3(g)

If solid NH4HS is placed in an evacuated flask at a certain temperature, it will dissociated until

the total gas pressure is 500 torr. Calculate the following :

(a) Equilibrium constant for the dissociation reaction

(b) If the additional NH3 is introduced into the equilibrium mixture without change in

temperature until the partial pressure of ammonia is 700 torr.

(i) What is the partial pressure of H2S under these conditions

(ii) What is the total pressure in the flask?

Q.2 The freezing point of an aqueous saturated solution of I2 is –0.0024°C. More than this can

dissolve in a KI solution because of the following equilibrium:


Ph.: 0291-3291970.

50

I2(aq) + I–(aq) I3– (aq), .1 M KI solution dissolves 12.5 g/L of I2. Calculate the

equilibrium constant KC for the above equilibrium. Also calculate the freezing point of the

resulting solution. Assume molarity to be equal to molality and also assume that conc. of I2 in all

saturated solutions is same [Kf for water = 1.86 K kg mol–1]

Q.3 Two solid compounds A and C dissolve into gaseous product at temperature T as follows :

(i) A(s) B(g) + D(g)

(ii) C(s) E(g) + D(g)

At 20°C pressure over excess solid A is 50 atm and that over excess solid C is 68 atm. Find the

total pressure of the gases over the solid mixture.

Q.4 At 1200°C, the following equilibrium is established between chlorine atoms and molecules

Cl2(g) 2Cl(g)

The composition of the equilibrium mixture may be determined by measuring the rate of effusion

of the mixture through a pin-hole. It is found that at 1200°C and 1 atm pressure the mixture

effuses 1.16 times as fast as krypton effuses under the same condition. Calculate the equilibrium

constant KC.

Q.5 For the equilibrium

CO2(g) + C(s) 2CO(g)

Kp at 817°C is 10 atm. Calculate the following

(a) Analysis of the gases at equilibrium at 817°C and a total pressure of 4.0 atm ?

(b) Partial pressure of CO2 at equilibrium.

(c) At what total pressure will the gas mixture analyze 6% CO2 by volume.

Q.6 Solid NH4I on rapid heating in a closed vessel at 357°C develops a constant pressure of 275 mm

of Hg owing to the partial decomposition of NH4I into NH3 and HI but the pressure gradually

increases further (when excess solid residually remain in the vessel) owing to the dissociation of

HI. Calculate the final pressure developed under equilibrium.

NH4I(s) NH3(g) + HI(g)

2HI(g) H2(g) + I2(g), KC = 0.015 at 357°C.

Q.7 Suppose you start with a mixture of 1.00 mol CO and 4.00 mol H2 in a 10.00 L vessel. Find the

moles of substances present at equilibrium at 1200 K for the reaction.

CO(g) + 3H2(g) CH4(g) + H2O(g); KC = 3.92

You will get an equation of the form

f(x) = 3.92

where f(x) is an expression in the unknown x (the amount of CH4). Solve this equation by

guessing the value of x, then computing value of f(x). Find values of x such that the values of f(x)

bracket 3.92. Then choose values of x to get a smaller bracket around 3.92. Obtain x to two

significant figures.

Q.8 A 250 ml flask, and 100 ml flask are separated by a stop cock. At 350 K, the nitric oxide in the

larger flask exerts a pressure of 0.46 atm. and the smaller one contains oxygen at 0.86 atm. The

gases are mixed by opening the stop cock. The reaction occurring are

2NO + O2 2NO2 N2O4

The first reaction is complete while the second one is at equilibrium. Assuming all the gases to

behave ideally, calculate the Kp for the equilibrium reaction if the final total pressure is 0.37 atm.

Q.9 Variation of equilibrium constant K with temperature T is given by

van‘t Hoff equation,


Ph.: 0291-3291970.

51

log K = log A –2.303

H

RT

A graph between log K and T–1 was a straight line as shown in the

figure and having = tan–1(0.5) and

OP = 10. Calculate:

(a) H° (standard heat of reaction) when T = 298 K,

(b) A (pre-exponential factor),

(c) Equilibrium constant K, at 298 K,

(d) K at 798 K if H° is independent of temperature.

Q.10 The moisture content of a gas is some times expressed in terms of dew point; it may be defined as

the temperature to which the gas must be cooled before the gas becomes saturated with water

vapour. At this point, water or ice will deposit on solid surface. Dew point of H2O is –43°C at

which vapour pressure of water is 0.07 torr. Assuming that CaCl2 owes its desicating property to

form CaCl2.2H2O. Calculate Kp and G° for

CaCl2.2H2O(s) CaCl2(s) + 2H2O(g)

Q.11 At 1000 K, the pressure of CO2 in equilibrium with CaCO3 and CaO is equal to 3.9 × 10–2 atm.

The equilibrium constant for the reaction

C(s) + CO2(g) 2CO(g)

is 1.9 at the same temperature when pressures are in atmospheres. Solid carbon, CaO and CaCO3

are mixed and allowed to come to equilibrium at 1000 K in a closed vessel. What is the pressure

of CO at equilibrium.

Q.12 The following equilibrium was studied by analyzing the equilibrium mixture for the amount of

HCl produced

LaCl3(s) + H2O(g) LaOCl(s) + 2HCl(g)

A vessel whose volume was 1.25 L was filled with 0.0125 mol of lanthanum (III) chloride,

LaCl3, and 0.0250 mol H2O. After the mixture came to equilibrium in the closed vessel at 619°C,

the gaseous mixture was removed and dissolved in more water. Sufficient silver (I) ion was added

to precipitate the chloride ion completely as silver chloride. If 3.59 g AgCl was obtained, what is

the valve of KC at 619°C ?

Q.13 The following equilibrium was studied by analyzing the equilibrium mixture for the amount of

H2S produced.

Sb2S3(s) + 3H2(g) 2Sb(s) + 3H2S(g)

A vessel whose volume was 2.50 L was filled with 0.0100 mol of antimony (III) sulfide, Sb2S3

and 0.0100 mol H2. After the mixture came to equilibrium in the closed vessel at 440°C, the

gaseous mixture was removed, and the hydrogen sulfide was dissolved in water. Sufficient

lead(II) ion was added to react completely with the H2S to precipitate lead (II) sulfide PbS. If

1.02 g PbS was obtained, what is the value of KC at 440°C ?

Q.14 The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO2), is partially regulated

by concentration of H3O+ and dissolved CO2 in blood. Although the equilibrium is complicated,

it can be summarised as :

HbO2 +H3O+ + CO2 H+ – Hb – CO2 + O2 + H2O

Explain why the production of lactic acid and CO2 in a muscle during exertion simulates release

of O2 from the oxyhemoglobin in the blood passing through the muscle.


Ph.: 0291-3291970.

52

Q.15 Equilibrium constants are given (in atm) for the following reactions at 0°C :

(a) SrCl2.6H2O(s) SrCl2.2H2O(s) + 4H2O(g) , Kp = 6.89 × 10–12

(b) Na2HPO4.12H2O(s) Na2HPO4.7H2O(s) + 5H2O(g), Kp = 5.25 × 10–13

(c) Na2SO4.10H2O(s) Na2SO4(s) + 10H2O(g), Kp = 4.08 × 10–25

The vapour pressure of water at 0°C is 4.58 torr.

(i) Calculate the pressure of water vapour in equilibrium at 0°C with each of (a), (b) and (c).

(ii) Which is most effective drying agent at 0°C ?

(iii)At what relative humidities will Na2SO4.10H2O be efflorescent when exposed to air at 0°C?

(iv) At what relative humidities will Na2SO4 be deliquescent (i.e. absorb moisture) when exposed

to air at 0°C?

Q.16 In the manufacture of Na2CO3 by the solvay process, NaHCO3(s) is decomposed by heating

2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g) ; Kp = 0.23 at 100°C

(a) If a sample of NaHCO3(s) is brought to a temperature of 100°C in a closed container, what

will be the total gas pressure (in atm) at equilibrium ?

(b) A mixture of 1.00 mol each of NaHCO3(s) and Na2CO3(s) is introduced into a 2.5 L flask in

which PCO2 = 2.10 atm and PH2O = 715 mm Hg. When equilibrium is established at 100°C, will

the partial pressure of CO2(g) and H2O(g) be greater or less than their initial pressures ? Explain.

(c) Also calculate PCO2 and PH2O by taking value of (b).

Q.17 A reaction system in equilibrium according to the equation 2 SO2 + O2 2SO3 in 1 litre

reaction vessel at a given temperature was found to contain 0.11 mol of SO2, 0.12 mol of SO3

and 0.05 mol of O2. Another 1 litre reaction vessel contains 64 g of SO2 at the same temperature.

What mass of O2 must be added to this vessel in order that at equilibrium half of SO2 is oxidised

to SO3?

Q.18 A mixture of hydrogen & iodine in the mole ratio 1.5 : 1 is maintained at 450°C. After the

attainement of equilibrium H2(g) + I2(g) 2HI (g), it is found on analysis that the mole ratio of

I2 to HI is 1 : 18. Calculate the equilibrium constant & the number of moles of each species

present under equilibrium, if initially, 127 grams of iodine were taken.

Q.19 A gas mixture of volumetric composition 45% CO, 50% CO2 & 5% N2 was passed at 1380 K

over a briquet of ZnO so that Zn vapour was produced according to the reaction ZnO(s) + CO(g)

Zn(g) + CO2(g). Under steady state conditions the rate of input gas flow (STP) is 0.224 dm3

min–1. The rate of loss of mass of ZnO is 5 × 10–4 mol. min–1. P = 770 torr. Find Kp (atm) at

1380 K. (assume ideal gas behaviour)

Q.20 When 1 mol of A(g) is introduced in a closed 1 L vessel maintained at constant temperature, the

following equilibria are established.

A(g) B(g) + 2C(g) K1 = ?

C(g) 2D(g) + 3B(g) K2 = ?

The pressure at equilibrium is 13

6

times the initial pressure. Calculate k1 & k2 if [ ]

[ ]

eq

eq

C

A =

4

9.

Q.21 When NO & NO2 are mixed, the following equilibria are readily obtained:

2NO2 N2O4 Kp = 6.8 atm–1 NO + NO2 N2O3 Kp = ?

In an experiment when NO & NO2 are mixed in the ratio of 1 : 2, the total final pressure was 5.05

atm & the partial pressure of N2O4 was 1.7 atm. Calculate

(a) the equilibrium partial pressure of NO. (b) Kp for NO + NO2 N2O3


Ph.: 0291-3291970.

53

Q.22 Consider the equilibriums: P (g) + 2Q(g) R(g).

When the reaction is carried out at a certain temperature, the equilibrium concentration of P and

Q are 3M and 4M respectively. When the volume of the vessel is doubled and the equilibrium is

allowed to be reestablished, the concentration of Q is found to be 3M. Find (a) Kc (b)

concentration of R at two equilibrium stages.

Q.23 Two solid compounds A & C dissociates into gaseous products at temperature T as follows

A(s) N2O + H2O (g)

C(s) N2 (g) + H2O (g)

At 20°C pressure over excess solid A is 50 atm & that over excess solid C is 68 atm. Find the

total pressure of gases over the solid mixture. Also identify A & C.

Q.24 A saturated solution of iodine in water contains 0.33 g I2/L. More than this can dissolve in a KI

solution because of the following equilibrium : I2(aq) + I– I3–. A 0.10 M KI solution actually

dissolves 12.5 g of iodine/L, most of which is converted to I3–. Assuming that the concentration

of I2 in all saturated solutions is the same, calculate the equilibrium constant for the above

reaction. What is the effect of adding water to a clear saturated KI solution.

Q.25 The equilibrium p-Xyloquinone + methylene white p –Xylohydroquinone + methylene blue

may be studied conveniently by observing the difference in color methylene white and methylene

blue. One mole of methylene blue was added to 1L of solution that was 0.24 M in p–

Xylohydroquinone and 0.012 M in p–Xyloquinone. It was then found that 4% of the added

methylene blue was reduced to methylene white. What is the equilibrium constant of the above

reaction? The equation is balanced with one mole each of 4 substance.

Q.26 A certain gas A polymerizes to a small extent at a given temperature & pressure, nA An. Show

that the gas obeys the approx. equation PV

RT=

1

( 1)1 c

n

n K

V

where Kc=

n

n

A

A & V is the

volume of the container. Assume that initially one mole of A was taken in the container.

Q.27 To 500 ml of 0.150 M AgNO3 solution were added 500 mL of 1.09 M Fe2+ solution and the

reaction is allowed to reach and equilibrium at 25ºC.

Ag+ (aq.) + Fe2+ (aq.) Fe3+ (aq.) + Ag(s)

For 25 mL of the solution , 30 mL of 0.0832 M KMnO4 were required for oxidation. Calculate

equilibrium constant for the reaction at 25ºC.

Q.28 Following two equilibria are established on mixing two gases A2 and C.

(i) 3A2 (g) A6(g) KP = 1.6 atm–2

(ii) A2(g) + C(g) A2C(g)

If A2 and C are mixed in 2 : 1 molar ratio, calculate the equilibrium partial pressures of A2, C,

A2C and KP for the reaction (ii). Given that the total pressure to be 1.4 atm and partial pressure

of A6 to be 0.2 atm at equilibrium.

Q.29 N2 and O2 combine at a given temperature to produce NO. At equilibrium the yield of NO is ‗x‘

per cent by volume. If x = ( )

. .4

K a bK a b

, where K is the equilibrium constant of the given

reaction at the given temperature and a and b are the volume percentage of N2 and O2


Ph.: 0291-3291970.

54

respectively in the initial pure mixture, what should be the initial composition of the reacting

mixture in order that maximum yield of NO is ensured? Also report the maximum value of K at

which ‗x‘ is maximum.

Q.30 In a reaction at equilibrium ‗X‘ mole of the reactant A decompose to give 1mole each of C and D.

It has been found that the fraction of A decomposed at equilibrium is independent of initial

concentration of A. Calculate X.

Q.31 A vessel of 2.50 litre was filled with 0.01 Mole of Sb2S3 and 0.01 mole of H2 to attain the

equilibrium at 440ºC as

Sb2S3(s) + 3H2(g) 2Sb(s) + 3H2S(g)

After equilibrium the H2S formed was analysed by dissolving it in water and treating with excess

of Pb2+ to give 1.029 g of PbS as precipitate. What is value of KC of the reaction at 440ºC?

(At.weight of Pb= 206]

Q.32 Gas A (mol. mass 16) and gas B (mol. mass 4) are confined into a vessel at partial pressure of 2

atm each at 27ºC. The gases are then allowed to effuse out at 27ºC through a valve till the

pressure inside the vessel becomes 2 atm after which the valve is closed and the gases are heated

to 127ºC and allowed to react as A + B C, Kp = 0.1 atm–1. Calculate the partial pressure of

C at equilibrium assuming C is not formed till the effusing process is completed at 27ºC.

Exercise – 5 : Objective Multiple Choice

Q.1 Which of the following statements are incorrect?

(A) For a closed system, S is always maximum at equilibrium.

(B) The addition of a reactant gas to an ideal-gas reaction mixture shifts the equilibrium in such a

way that some of the added gas is used up

(C) In any closed system, G is always minimum at equilibrium.

(D) In the limit T 0, Gº approaches Hº.

Q.2 The reactions in which the yield of the products cannot be increased by the application of high

pressure

(A) 2SO2(g) + O2(g) 2SO3(g) (B) NH4HS(s) NH3(g) + H2S(g)

(C) N2O4(g) 2NO2(g) (D) N2(g) + 3H2(g) 2NH3(g)

Q.3 The condition for spontaneity in a chemical reaction is

(A) (G)T, p 0 (B) (U)S, V 0 (C) (H)S, p 0 (D) (S)U, V 0

Q.4 In which of the following reactions is Kp < Kc ?

(A) CO(g) + Cl2(g) COCl2(g) (B) CO(g) + 3H2(g) CH4(g) + H2O(g)

(C) 2BrCl(g) Cl2(g) + Br2(g) (D) I2(g) 2I(g)

Q.5 The dissociation of phosgene, which occurs according to the reaction

COCl2(g) CO(g) + Cl2(g)

is an endothermic process. Which of the following will increase the degree of dissociation of

COCl2?

(A) Adding Cl2 to the system

(B) Adding helium to the system

(C) Decreasing the temperature of the system

(D) Reducing the total pressure of the system


Ph.: 0291-3291970.

55

Q.6 The equilibrium of which of the following reactions will not be disturbed by the addition of an

inert gas at constant volume ?

(A) H2(g) + I2(g) 2HI(g) (B) N2O4(g) 2NO2(g)

(C) CO2(g) + 2H2(g) CH3OH(g) (D) C(s) + H2O(g) CO(g) + H2(g)

Q.7 A box contains CO(g), Cl2(g) and COCl2(g) in equilibrium at 1000 K. The removal of CO(g)

will

(A) decrease the concentration of COCl2 (B) increase the concentration of Cl2

(C) increase the concentration of COCl2 (D) reduce the concentration of CO as well as Cl2

Q.8 An industrial fuel, ‗water gas‘, which consists of a mixture of H2 and CO can be made by passing

steam over red-hot carbon. The reaction is

C(s) + H2O(g) CO(g) + H2(g) , H = +131 kJ

The yield of CO and H2 at equilibrium would be shifted to the product side by

(A) raising the relative pressure of the steam (B) adding hot carbon

(C) raising the temperature (D) reducing the volume of the system

Q.9 For the equilibrium 2SO2(g) + O2(g) 2SO3(g), H = -198 kJ, the equilibrium

concentration of SO3 will be affected by

(A) doubling the volume of the reaction vessel

(B) increasing the temperature at constant volume

(C) adding more oxygen to the reaction vessel

(D) adding helium to the reaction vessel at constant volume

Q.10 The following reaction attains equilibrium at high temperature

N2(g) + 2H2O(g) + heat 2NO(g) + 2H2(g)

The yield of NO is affected by

(A) increasing the nitrogen concentration (B) decreasing the hydrogen concentration

(C) compressing the reaction mixture (D) none of these

Q.11 N2(g) + 3H2(g) Catalyst

500ºC 2NH3 + heat

In this reaction, the direction of equilibrium will be shifted to the right by

(A) increasing the concentration of nitrogen (B) compressing the reaciton mixture

(C) removing the catalyst (D) decreasing the concentration of ammonia

Q.12 The dissociation of ammonia carbonate may be represented by the equation

NH4CO2NH2(s) 2NH3(g) + CO2(g)

Hº for the forward reaction is negative. The equilibrium will shift from right to left is there is

(A) a decrease in pressure (B) an increase in temperature

(C) an increase in the concentration of ammonia

(D) an increase in the concentration of carbon dioxide

Q.13 For which of the following reactions is the equilibrium constant called an acidity constant?

(A) 2H2O H3O+ + OH– (B) H2S + H2O H3O+ + HS–

(C) HONO + H2O ONO– + H3O+ (D) H3O+ + CH3NH2 CH3NH3 + H2O

Q.14 Which of the following statements about the reaction quotient Q are correct?

(A) the reaction quotient, Q and the equilibrium constant always have the same numerical value.

(B) Q may be >, < , = Keq .

(C) Q(numerical value) varies as reaction proceeds

(D) Q = 1 at equilibrium


Ph.: 0291-3291970.

56

Q.15 Which of the following factors will increase solubility of NH3(g) in H2O?

NH3(g) + H2O(aq) NH4OH(aq)

(A) increase in pressure (B) addition of water.

(C) liquefaction of NH3 (D) decrease in pressure

Q.16 Which of the following factors will affect solubility of CaO in H2O?

(A) pressure (B) temperature (C) addition of water (D) volume

Q.17 For the following equilibrium

NH4HS(s) NH3(g) + H2S(g)

partial pressure of NH3 will increase

(A) if NH3 is added after equilibrium is established

(B) if H2S is added after equilibrium is established

(C) temperature is increased

(D) volume of the flask is decreased

Q.18 Formation of Y from X is through the graph. It can be concluded:

A

B

C

DE

Y

temperature (A) reaction A to B is endothermic (B) reaction A to B as well as C to D is exothermic

(C) reaction B to C is exothermic (D) H = 0 for the stage D to E

Q.19 Which are true statements for the following equilibrium?

H2O(l) H2O(g)

(A) increase in pressure will result in the formation of more liquid water

(B) increase in pressure will increase b.p.

(C) decrease in pressure will vaporise H2O(l) to a greater extent

(D) increase in pressure will liquefy steam.

Q.20 Which is/are correct relation (s) for thermodynamic equilibrium constant

(A) Gº = –2.303 RT log K (B) G = Gº + 2.303 RT log K

(C) Eºcell =0.0591

n log K (D) E = Eº –

0.0591

n logK

Q.21 Which is/are correct relation (s) for equilibrium constant K?

(A) Gº = – 2.303 RT log K (B) Ecellº = 2.303 RT log K

nF

(C) K = Σ[product]

Σ[reactant] (D) log K = log A –

H

2.303RT

Q.22 Volume of the flask in which species are transferred is double of the earlier flask. In which of the

following cases, equilibrium constant is disturbed

(A) N2(g) + 3H2(g) 2H3(g) (B) N2(g) + O2(g) 2NO(g)

(C) PCl5(g) PCl3(g) + Cl2(g) (D) 2NOBr(g) N2(g) + O2(g)

Q.23 For the gas phase reaction carried out in a vessel,

C2H6 C2H4 + H2

the equilibrium concentration of C2H4 can be increased by -


Ph.: 0291-3291970.

57

(A) increasing the temperature (B) decreasing the pressure

(C) removing some H2 (D) adding some C2H6

Q.24 For the reaction

2A(g) + B(g) 2C(g) ; H = +13.6 kJ

which of the following will increase the extent of the reaction at equilibrium

(A) increasing the temperature (B) increasing the pressure

(C) addition of catalyst (D) removing C

Assertion & Reason

This section contains FIVE questions. Each question contains STATEMENT- 1

(Assertion) and STATEMENT - 2 (Reason). Each question has 4 choices (A), (B), (C)

and (D) out of which ONLY ONE is correct.

(A) „S‟ is correct but E is wrong

(B) „S‟ is wrong but E is correct

(C) Both „S‟ and „E‟ are correct and „E‟ is correct explanation of „S‟.

(D) Both „S‟ and „E‟ are correct but „E‟ is not correct explanation of „S‟.

Q.26 Statement-1 : When the following equilibrium is studied in a vessel of twice the volume the

earlier one, A(g) B(g) + C(g) equilibrium constant is decreased.

Statement-2 : Equilibrium constant Kc =

2x

(1 x) V. (where x is degree of dissociation of A(g)

and V is the volume of the container. )

Q.27 Statement-1: A(s) B(g) + C(g) + D(l), on addition of A(s), the equilibrium

shifts to forward direction and there will be increase in equilibrium constant.

Statement-1: Equilibrium constant changes only with temperature.

Q.28 Statement-1: In a reaction initially free energy decreases and attains a stage, the reaction

becomes dynamic in either sides.

Statement-2: 0G and equilibrium is attained.

Q.29 Statement-1: When an inert gas is added at constant volume the equilibrium state is

not disturbed. But when inert gas is added at constant pressure dissociation of PCl 5

increases

Statement-1: To maintain the constant value of Kc, the no.of moles of PCl5

dissociated decreases.

Q.30 Statement-1: kp can be equal to or less than or even grater than the value of kc.

Statement-2: kp = kp (RT)–n Relation between kp and kc depends on the change in

the number of moles of gaseous reactants and products.

Q.31 Statement-1: Ice water, if pressure is applied to the system then more water

will form;

Statement-2: Increase of pressure pushes the equilibrium towards the side where

volume will decrease.

Q.32 Statement-1: 1% CO2 in air is sufficient to prevent any loss in weight when Ag2CO3 is dried

at 1200C. Given that 2 3 (B) 2(g)Ag CO AgO CO Kp = 0.0095 at 120oC

Statement-2: Loss in weight is due to liberation of CO2


Ph.: 0291-3291970.

58

Q.33 Statement-1: When an inert gas is passed into a reaction at equilibrium at constant volume,

there will not be any change in equilibrium position.

Statement-2: In the above process reaction quotient (Q) doesn‘t change its value

Paragraph type PARAGRAPH–I

When 306 g of NH4HS is introduced in a 4.1 litre evacuated flask and heated to

227ºC then solid NH4HS decomposes into gaseous ammonia and hydrogen sulphide.

At equilibrium, it is found that 170 gm of H2S are present in the container.

Q.34 Mass of solid remaining in the container is -

(A) 51 g (B) 136 g (C) 68 g (D) 100 g

Q.35 Kp for the reaction at 227ºC is -

(A) 50 (B) 100 (C) 2500 (D) 1500

Q.36 If after establishment of equilibrium, H2S is added in the container in such a way

that partial pressure of H2S now equals to original total pressure, then partial

pressure of NH3 at new equilibrium is -

(A) 50 atm (B) 25 atm (C) 100 atm (D) 75 atm

Q.37 When more solid NH4HS is added into the flask at equilibrium then -

(A) reaction moves in forward direction(B) reaction moves in backward direction

(C) reaction remains at equilibrium (D) none of these

Q.38 If at 227ºC, ammonia also starts decomposing into nitrogen and hydrogen and

establishes equilibrium (2NH3 N2 + 3H2) then Kp for the reaction NH4HS(s)

NH3(g) + H2S(g) and total pressure developed in the container will -

(A) decreaes, increases (B) increases, decreases

(C) no change, no change (D) no change, increases

PARAGRAPH–II

SrCl2.6H2O(s) SrCl2.2H2O(s) + 4H2O(g)

KP = 6.89 × 10–12

Na2HPO4.12H2O(s) Na2HPO4.7H2O(s) + 5H2O(g)

KP = 5.25 × 10–13

Na2SO4.10H2O(s) Na2SO4(s) + 10H2O(g)

KP = 4.08 × 10–25

The vapour pressure of water at 0ºC is 4.58 Torr.

Q.39 Which is most effective drying agent at 0ºC

(A) SrCl2.2H2O(s) (B) Na2HPO4.7H2O(s)

(C) Na2SO4(s) (D) All equally

Q.40 Na2SO4.10H2O(s) will be efflorescent when exposed to the air at 0ºC at :

(A) pressure below 2.77 Torr (B) pressure above 2.77 Torr

(C) pressure above 4.58 Torr (D) pressure equal to 2.77 Torr


Ph.: 0291-3291970.

59

Q.41 At what relative humidity will Na2SO4 be deleiquescent (absorb moisture) when exposed to the

air at 0ºC

(A) below 60.5% (B) above 60.5% (C) above 39.5% (D) below 39.5%

Q.42 Drying capacity of the substance mentioned above willbe in order:

(A) Na2SO4 = Na2HPO4.7H2O < SrCl2.2H2O

(B) Na2SO4 > Na2HPO4.7H2O > SrCl2.2H2O

(C) Na2SO4 < Na2HPO4.7H2O < SrCl2.2H2O

(D) Na2HPO4.7H2O < Na2SO4 < SrCl2.2H2O

PARAGRAPH–III

If a stress is applied to a reaction mixture at equilibrium, reaction occurs in that

direction that ―relieves the stress‖.

Stress means a change in concentration, pressure, volume (or) temperature that

disturbs the original equilibrium. The direction that the reaction takes is the one that

reduces the stress.

If you increase concentration of reactants , reaction goes in a direction that tends to

decrease concentration.

If you increase pressure, reaction goes in a direction that tends to decrease pressure.

If you increase temperature, reaction goes in a direction that tends to decrease the

temperature.

Q.43 In the Ice-water system the favourable condition for melting of ice are :

Ice + Heat Water

(A) Low temperature (B) High volume

(C) High pressure (D) Both (B) and (C)

Q.44 For the equation, P + Q R + S, at equilibrium number of moles of P, Q,

R, S are 2, 3, 3, 4 M . Calculate the number of moles of P which should be

added to increase the concentration of S to 5 M.

(A) 5 M (B) 8 M (C) 6 M (D) 4 M

Q.45 When NaNO3 is heated in a closed vessel , oxygen is liberated and NaNO2 is left

behind. At equilibrium, which is correct?

(A) addition of NaNO2 favours reverse reaction

(B) addition of NaNO3 favours forward reaction

(C) increasing temperature favours forward reaction

(D) increasing pressure favours reverse reaction

PARAGRAPH–IV

Physical and chemical equilibria respond to stress, e.g., change in pressure, temperature and

concentration of reactants and products. According to Le Chatelier‘s principle a system at

equilibrium, when subjected to a disturbance (stress), responds in a way that to minimize the

effect of the disturbance (stress), responds in a way that tends to minimize the effect of the

disturbance.

Effect of pressure: The principle implies that if a system at equilibrium is compressed, the

reaction will adjust so as to minimize the increase in pressure. Thus, if volume of gas phase

reaction system is reduced, the equilibrium shifts in the direction in which number of moles

decreases. Although equilibrium constant is independent of pressure (at a fixed temperature), the

various concentrations are changed on reducing the volume of the system, i.e. increasing the


Ph.: 0291-3291970.

60

pressure. In case of physical equilibria on increasing pressure the system shifts in the direction of

species of higher density as to reduce the effect of increased pressure.

However, if increase in pressure effected by introducing an inert gas to reaction vessel of

constant volume, partial pressures or concentrations of various species remain unchanged as they

continue to occupy the same volume. But at constant pressure condition, introduction of inert gas

into the reaction system will lead to increase in volume and in consequence to the change in

concentrations of various species. Under this condition, the equilibrium shifts in the direction in

which number of moles of gaseous species increases.

Effect of temperature: Le-Chatelier‘s principle predicts a system at equilibrium will tend to shift

in the endothermic direction when temperature is raised, for then energy is absorbed as heat and

the rise in temperature is opposed. Conversely, an equilibrium will shift in the exothermic

direction if the temperature is lowered, for then that energy is released and the reduction in

temperature is opposed. Vant Hoff equation shows the dependence of equilibrium constant K on

temperature as:

d

dT lnK = 2

Hº

RT

or ln K = constant –

Hº

R

.

1

T .

Effect of addition of reactants or products: on addition of one or more reactants the

equilibrium will shift to the products causes the equilibrium to shift in backward direction..

Q.46 Consider the equilibrium, 2CO(g) + O2(g) 2CO2(g) + heat

If O2 is added and volume of the reaction vessel is reduced, the equilibrium will

(A) shift in forward direction (B) Shift in reverse direction

(C) remains unchanged (D) be unpredictable

Q.47 The volume of the systems describing the equilibria

(I) heat + MgCO3(s) MgO(s) + CO2(g)

(II) 2C(s) + O2(g) 2CO(g) + heat

is decreased. The equilibrium will shift

(A) to the right in both (i) and (II)

(B) to the left in both (I) and (II)

(C) to the left in (I) and to the right in (II)

(D) to the right in (I) and to the left in (II)

Q.48 To the system X(s) + 2A(g) + heat 2Y(s) + 3B(g) at equilibrium more A(g) is added with

temperature volume constant. If the new steady pressure of A(g) becomes 2.828 times of the

previous steady pressure, then the no. of moles of B(g) are increased by a factor of

(A) 31/3 (B) 3 (C) 2 (D) 2

Match the Column Q.49 Match the following

List I List II

(a) N2(g) + 3H2(g) 2NH3 (P) KP = KC

(b) H2(g) + S(s) H2S(g) (Q) KP > KC

(c) H2(g) + I2(g) 2HI(g) (R) KP < KC

(d) 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)

(A) a-Q, b-P, c-P, d-R (B) a-P, b-P, c-Q, d-R

(C) a-Q, b-R, c-P, d-R (D) a-R, b-P, c-P, d-Q


Ph.: 0291-3291970.

61

Q.50 Match the following

List I List II

(a) Keq < 1 (P) Affected by temperature

(b) Degree of dissociation (Q) Affected by pressure

(c) Equilibrium constant (R) Kf < Kb

(d) Melting of ice (S) Kf > Kb

(A) a-R, b-P, c-S, d-Q (B) a-R, b-P,Q, c-P, d-P,Q

(C) a-R, b-S c-P, d-Q (D) a-Q, b-R, c-S, d-P


List I (Reaction List II (units)

(a) N2 + O2 2NO (P) Kc = m/l

(b) N2 + 3H2 2NH3 (Q) no unit

(c) PCl5 PCl3 + Cl2 (R) kp = atm–2

(d) A(l) B(g) (S) Kp = atm

(A) a-Q, b-R,c-P,S, d-P,S (B) a-Q, b-R,c-P, d-S

(C) a-S, b-Q,c-P,R, d-P,S (D) a-R, b-Q,c-P,S, d-S


List I (Reaction) List II (degree of dissociation)

(a) 2NH3 2N2 + 3H2 (P) 2

D d

d

(b) N2 O4 2NO2 (Q) D d

d

(c) 4PH3 P4 + 6H2 (R) 3

D d

d

(d) 5A 6B + 9C (S)2(D d)

3d

(A) a-Q, b-P, c-R, d-S (B) a-P, b-Q, c-R, d-S

(C) a-Q, b-Q, c-S, d-P (D) a-S, b-R, c-Q, d-P

Q.53 match List I with List II

(P is partial pressure of prdouct having stiochiometric coefficient unity)

List I (Reaction) List II (expression of Kp)

(a) NH4Cl(s) NH3(g) + HCl(g) (P) 6P6

(b) NH2COONH4(s) 2NH3(g)+ CO2(g) (Q) P2

(c) A(s) 2B(g) + C(g) + 3D(g) (R) 4P3

(d) A(s) B(g) + C(g) (S) 108P6

(A) a-Q, b-R, c-S, d-Q (B) a-P, b-Q, c-R, d-S

(C) a-S, b-R, c-Q, d-P (D) a-S, b-P, c-R, d-Q


Ph.: 0291-3291970.

62

Exercise – 6: Other Exams Questions

Q.1 Consider the reacton equilibrium [AIEEE 2003]

2SO2(g) + O2(g) 2SO3(g); Hº = –198 kJ.

On the basis of Le Chatelier‘s principle, the condition favourable for the forward reaction is:

(A) Lowering of temprature as well as pressure

(B) increasing temperature as well as pressure

(C) lowering the temperature and increasing the pressure

(D) any value of temperature and pressure.

Q.2 For the reaction equilibrium, N2O4 (g) 2NO2(g) the concentration of N2O4 and NO2 at

equilibrium are 4.8 × 10–2 and 1.2 × 10–2 mol L–1 respectively. The value of KC for the reacion is:

[AIEEE 2003]

(A) 3.3 × 102 mol L–1(B) 3 × 10–1 mol L–1 (C) 3 × 10–3 mol L–1 (D) 3 × 103 mol L–1

Q.3 What is the equilibrium expression for the reaction: [AIEEE 2004]

P4 (s) + 5O2 (g) P4O10 (s) ?

(A) KC = [P4O10]/[P4] [O2]5 (B) KC = 1/[O2]5

(C) KC = [O2]5 (D) KC = [P4O10]/5[P4] [O2]

Q.4 For the reaction CO(g) + Cl2(g) COCl2(g) then Kp/Kc equal to [AIEEE 2004]

(A) 1/RT (B) 1.0 (C) RT (D) RT

Q.5 The equilibrium constant for the reaction [AIEEE 2004]

N2(g) + O2(g) 2NO(g)

at temperature T is 4 × 10–4. The value of Kc for the raction

NO(g) 1

2N2(g) +

1

2O2(g)

at the same temperatue is

(A) 2.5 × 102 (B) 0.02 (C) 4 × 10–4 (D) 50

Q.6 For the reaciton [AIEEE 2005]

2NO2(g) 3 NO(g) + O2(g)

(KC = 1.8 × 10–6 at 184ºC)

(R = 0.0831 kJ/mol. K

When Kp and Kc are compared at 184ºC it is found that

(A) Whether Kp is greater than, less than or equal to Kc depends upon the total gas pressure

(B) Kp = Kc

(C) Kp is less than KC

(D) Kp is greater than Kc.

Q.7 The exothermic formation of ClF3 is represented by the equation: [AIEEE 2005]

Cl2(g) + 3F2 (g) 2ClF3(g) ; rH = –329 J

which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2, F2

and ClF3.

(A) Adding F2 (B) Increasing the volume of container

(C) Removing Cl2 (D) Increasing the temperature


Ph.: 0291-3291970.

63

Q.8 An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain

temperature at 0.50 atm pressure. Ammonium hydrogen sulphide decompose to yield NH3 and

H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total presssure

in the flask rises to 0.84 atm? The equilibrium constant for NH4HS decomposition at this

temperature is: [AIEEE 2005]

(A) 0.11 (B) 0.17 (C) 0.18 (D) 0.30

Q.9 Phosphorus petachloride dissociates as follows, in a closed reaction vessel, [AIEEE 2006]


If total pressure at equilibrium of the reaction mixture is P and degree of dissocation of PCl5 is x,

the partial pressure of PCl3 will be

(A) 1

x

x

P (B)

1

x

x

P (C) x

x 1

P (D)

2

1

x

x

P

Q.10 For the following three reactions a, b and c, equilibrium constants are given: [AIEEE 2008]

(1) CO(g) + H2O(g) CO2(g) + H2(g); K1

(2) CH4(g) + H2O(g) CO(g) + 3H2(g); K2

(3) CH4(g) + 2H2O(g) CO2(g) + 4H2(g); K3

Which of the following relations is correct ?

(A) K2 K3 = K1 (B) K3 = K1K2 (C) K3 K23 = K1

2 (D) K1 = K3

Exercise – 7: IIT JEE Flash Back (Objective)

Q.1 The oxidation of SO2 by O2 to SO3 is an exothermic reaction. The yield of SO3 will be

maximum if [in 1981]

(A) temperature is increased and pressure is kept constant

(B) temperature is reduced and pressure is increased

(C) both temperature and pressure are increased

(D) both temperature and pressure are reduced

Q.2 For the reaction : H2(g) + I2 (g) 2HI (g) the equilibrium constant Kp changes with

(A) pressure (B) catalyst [in 1981]

(C) the amounts of H2 and I2 present (D) temperature

Q.3 If equilibrium constant for the reaction, A2 + B2 2AB, is K, then for the backward reaction

AB ½A2 + ½B2, the equilibrium constant is 1/K. [in 1984]

Q.4 When a liquid and its vapour are at equilibrium and the pressure is suddenly decreased, cooling

occurs. [in 1984]

Q.5 A liquid is in equilibrium with its vapour at its boiling point. On the average, the molecules in the

two phases have equal: [in 1984]

(A) inter-molecular forces (B) potential energy

(C) total energy (D) kinetic energy

Q.6 Pure ammonia is placed in a vessel at a temperature where its dissociation constant (a) is

appreciable at equilibrium [in 1984]

(A) Kp does not change significantly with pressure

(B) a does not change with pressure


Ph.: 0291-3291970.

64

(C) concentration of NH3 does not change with pressure

(D) concentration of hydrogen is less than that of nitrogen

Q.7 For the gas phase reaction : C2H4 + H2 C2H6 (H = –32.7 kcal) [in 1984]

carried out in a vessel, the equilibrium concentration of C2H4 can be increased by

(A) increasing the temperature (B) decreasing the pressure

(C) removing some H2 (D) adding some C2H6

Q.8 An example of reversible reaction is [in 1985]

(A) Pb(NO3)2 + 2NaI = PbI2 + 2NaNO3 (B) AgNO3 + HCl = AgCl + HNO3

(C) 2Na + 2H2O = 2NaOH + H2 (D) KNO3 + NaCl = KCl + NaNO3

Q.9 When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At

equilibrium: [in 1987]

(A) Addition of NaNO2 favours reverse reaction

(B) Addition of NaNO2 favours forward reaction

(C) Increasing temperature favours forwards

(D) Decreasing pressure favours reverse reaction

Q.10 The equilibrium : SO2Cl2 (g) SO2(g) + Cl2(g)

is attained at 25ºC in a closed container and an inert gas, helium is introduced. Which of the

following statements are correct ? [in 1989]

(A) concentration of SO2, Cl2 and SO2Cl2 do not change

(B) more chlorine is formed

(C) concentration of SO2 is reduced

(D) more SO2Cl2 is formed

Q.11 The equilibrium constant for the synthesis of HI at 490°C is 50.0. The value of K of the

dissociation of HI will be: [in 1990]

(A) 0.02 (B) 50.0 (C) 0.50 (D) 0.20

Q.12 In the reaction PCl5(g) PCl3(g) + Cl2(g), the amounts of PCl5, PCl3 and Cl2 are 2 mole

each at equilibrium and the total pressure is 3 atmospheres. The equilibrium constant Kp, is :

(A) 1 atm (B) 2 atm (C) 3 atm (D) 1.5 atm [in 1991]

Q.13 For the reaction PCl5(g) PCl3(g) + Cl2(g) [in 1991]

The forward reaction at constant temperature is favoured by:

(A) Introducing inert gas at constant volume (B) Introducing chlorine gas at constant volume

(C) introducing an inert gas at constant pressure (D) None

Q.14 For the reaction, C(s) + CO2(g) 2CO(g), the partial pressure of CO2 and CO are 2.0 atm

4.0 atm, respectively, at equilibrium. The Kp of the reaction is – [in 1992]

(A) 0.5 (B) 4.0 (C) 32.0 (D) 8.0

Q.15 A ten-fold increase in pressure on the reaction N2(g) + 3H2(g) 2NH3(g) at equilibrium

results in ..........in Kp. [in 1996]

Q.16 One mole of N2O4 (g) at 300 K is kept in a closed container under one atmosphere. It is heated to

600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is –

(A) 1.2 atm (B) 2.4 atm (C) 2.0 atm (D) 1.0 atm [in 1996]

Q.17 For a gaseous reaction 2B A, the equilibrium constant Kp is ................ to/than Kc. [in 1997]

Q.18 8 moles of a gas AB3 are introduced into a 1.0 dm3 vessel. It dissociates as [in 1997]

2AB3(g) A2(g) + 3B2(g)


Ph.: 0291-3291970.

65

At equilibrium, 2 moles of A2 are found to be present. The equilibrium constant of the reaction is

(A) 2 mol2 L–2 (B) 3 mol2 L–2 (C) 27 mol2 L–2 (D) 36 mol2 L–2

Q.19 For the reaction, CO(g) + H2O (g) CO2 (g) + H2(g) , at a given temperature the

equilibrium amount of CO2 (g) can be increased by [in 1998]

(A) adding a suitable catalyst (B) adding an inert gas

(C) decreasing the volume of the container (D) increasing the amount of CO(g)

Q.20 For the chemical reaction, 3X(g) + Y(g) X3Y(g) the amount of X3Y at equilibrium is

affected by : [in 1999]

(A) temperature and pressure (B) temperature only

(C) pressure only (D) temperature, pressure and catalyst

Q.21 The partial pressures of CH3OH, CO and H2 in the equilibrium mixture for the reaction

CO + 2H2 CH3OH

at 427°C are 2.0, 1.0 and 0.1 atm. respectively. The value of Kp for the decomposition of CH3OH

into CO and H2 is [in 1999]

(A) 1 × 102 atm (B) 2 × 102 atm–1 (C) 50 atm2 (D) 5 × 10–3 atm2

Q.22 For the reversible reaction N2(g) + 3H2(g) 2NH3(g) at 500º , the value of Kp is

1.44 × 10–5 when partial pressure is measured in atmospheres. The corresponding value of KC,

with concentration in mole litre–1, is [in 2000]

(A)5

2

1.44 10

(0.082 500)

(B)

5

2

1.44 10

(8.314 773)

(C)

5

2

1.44 10

(0.082 773)

(D)

5

2

1.44 10

(0.082 773)

Q.23 When two reactants, A & B are mixed to give product C & D, the reaction quotient question at

the initial stages of the reaction [in 2000]

(A) is zero (B) decrease with time

(C) is independent of time (D) increase with time

Q.24 At constant temperature, the equilibrium constant (Kp) for the decomposition reaction

N2O4 2NO2

is expressed by Kp = 2

2

4

(1 )

x P

x [in 2000]

where P = pressure, x = extent of decomposition. Which one of the following statements is true?

(A) Kp increases with increase of P (B) Kp increases with increase of x

(C) Kp increases with decrease of x (D) Kp remains constant with change in P and x

Q.25 Consider the following equilibrium in a closed container

N2O4(g) 2NO2(g)

At a fixed temperature, the volume of the reaction container is halved, for this change, which of

the following statements holds true regarding the equilibrium constant (Kp) and degree of

dissociation () : [in 2002]

(A) neither Kp nor ‗‘ changes (B) both Kp and ‗‘ changes

(C) Kp changes but ‗‘ does not (D) Kp does not change but ‗changes

Q.26 The Haber‘s process for the formation of NH3 at 298K is: [in 2006]

N2 + 3H2 2NH3; H = – 46.0 kJ; Which of the following is the correct statement


Ph.: 0291-3291970.

66

(A) The condition for equilibrium is 2NG +

23 HG =

32 NHG where G is Gibbs free energy per mole

of the gaseous species measured at that partial pressure.

(B) On adding N2, the equilibrium will shift to forward direction becasue according to IInd law of

thermodynamics the entropy must increasse in the direction of spontaneous reaction.

(C) The catalyst will increase th rate of forward reaction by 2 times and that of backward reaction

by 1.5 times

(D) None of these

27. The value of log10K for a reaction A B is [in 2007]

(Given: rHº298K = – 54.07 kJ mol–1, rSº298K = 10 JK–1 mol–1 and R = 8.314 JK–1 mol–1;

2.303 × 8.314 × 298 = 5705)

(A) 5 (B) 10 (C) 95 (D) 100

28. Statement-1: For every chemical reaction at equilibrium, standard Gibbs energy of

reaction is zero.

Statement-2: At constant temperature and pressure, chemical reactions are

spontaneous in the direction of decreasing Gibbs energy. [in 2008]

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a CORRECT

explanation for Statement-1

(B)Statemen -1 is True, Statement-2 is True; Statement - 2 is a NOT CORRECT

explanation for Statement - 1

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

Exercise – 8 : IIT JEE Flash Back (Subjective)

Q.1 One mole of nitrogen is mixed with three moles of hydrogen in a 4 litre container. If 0.25 percent

of nitrogen is converted to ammonia by the following reaction

N2(g) + 3H2(g) 2NH3(g)

calculate the equilibrium constant (KC) in concentration units. What will be the value of KC for

the fallowing equilibrium ? [in 1981]

1/2 N2(g) + 3/2 H2(g) NH3(g).

Q.2 One mole of Cl2 and 3 moles of PCl5 are placed in a 100 litre vessel heated to 227ºC. The

equilibrium pressure is 2.05 atmosphere. Assuming ideal behaviour, calculate the degree of

dissociation for PCl5 and Kp for the reaction : PCl5(g) PCl3(g) + Cl2(g). [in 1984]

Q.3 The equilibrium constant of the reaction

A2(g) + B2(g) 2AB(g)

at 100ºC is 50. If a one litre flask containing one mole of A1 is connected to a two litre flask

containing two mole of B2, how many mole of AB will be formed at 373ºC? [in 1985]

Q.4 A mixture of SO3, SO2 and O2 gases is maintained at equilibrium in 10 litre flask at a

temperature at which KC for the reaction, 2SO2(g) + O2(g) 2SO3(g) is 100 mol–1 litre. At

equilibrium. [in 1987]

(a) If number of mole of SO3 and SO2 in flask are same, how many mole of O2 are present ?


Ph.: 0291-3291970.

67

(b) If number of mole of SO3 in flask are twice the number of mole of SO2, how many mole of

O2 are present?

Q.5 At a certain temperature, equilibrium constant (KC) is 16 for the reaction:

SO2(g) + NO2(g) SO3(g) + NO(g)

If we take one mole of each of the four gases in one litre container, what would be the

equilibrium concentration of NO and NO2? [in 1987]

Q.6 One mole of H2, two mole of I2 and three mole of HI are injected in one litre flask. What will be

the concentration of H2, I2 and HI at equilibrium at 500°C. KC for reaction. [in 1988]

H2 + I2 2HI is 45.9.

Q.7 N2O4 is 25% dissociated at 37°C and one atmospheric pressure. Calculate : [in 1988]

(a) Kp for N2O4 2NO2

(b) % dissociation at 37°C and 0.1 atm.

Q.8 At 700 K, CO2 and H2 react to form CO and H2O. For this purpose KC is 0.11. If a mixture of

0.45 mole of CO2 and 0.45 mole of H2 is heated to 700 K. [in 1989]

(a) Find out amount of each gas at equilibrium.

(b) When equilibrium has been reached, another 0.34 mole of CO2 and 0.34 mole of H2 are

added to the reaction mixture. Find composition of mixture at new equilibrium.

Q.9 The equilibrium constant Kp of the reaction.

2SO2 + O2 2SO3

is 900 atm–1 at 800 K. A mixture containing SO3 and O2 having initial pressure of 1 atm and 2

atm respectively is heated at constant volume to equilibrate. Calculate partial pressure of each gas

at 800 K at equilibrium. [in 1989]

Q.10 When S in the form of S8 is heated at 900 K, the initial pressure of 1 atmosphere falls by 29% at

equilibrium. This is because of conversion of some S8 to S2. Find the Kp for reaction. [in 1990]

Q.11 For the reaction:

CO(g) + 2H2(g) CH3OH(g):

H2 is introduced into a five litre flask at 327°C, containing 0.2 mole of CO(g) and a catalyst till

the pressure is 4.92 atmosphere. At this point 0.1 mole of CH3OH(g) is formed. Calculate KC

and Kp. [in 1990]

Q.12 An equilibrium mixture at 300 K contains N2O4 and NO2 at 0.28 and 1.1 atmosphere

respectively. If the volume of container is doubled, calculate the new equilibrium pressure of two

gases. [in 1991]

Q.13 The equilibrium constant for the reaction:

CO(g) + H2O(g) CO2(g) + H2(g) at 986°C is 0.63. A mixture of 1 mole H2O(g) and 3 mole

CO(g) is allowed to react to c ome to an equilibrium. The equilibrium pressure is 2.0 atm.

(a) How many mole of H2 are present at equilibrium ?

(b) Calculate partial pressure of each gas at equilibrium. [in 1992]

Q.14 0.15 mole of CO taken in a 2.5 litre flask is maintained at 750 K along with a catalyst so that the

following reaction can take place

CO(g) + 2H2(g) CH3OH(g)


Ph.: 0291-3291970.

68

Hydrogen is introduced until the total pressure of the system is 8.5 atm at equilibrium and 0.08

mole of methanol is formed. Calculate (i) Kp and KC.

(ii) The final pressure if the same amount of CO and H2 as before are used but no catalyst so that

the reaction does not take place. [in 1993]

Q.15 A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into

CO on addition of graphite. Calculate the value of Kp if total pressure at equilibrium is 0.8 atm.

[in 1993]

Q.16 For a given reversible reaction equilibrium constant Kp and Kc are related by .......... [in 1984]

Q.17 At temperature T, a compound AB2(g) dissociates according to the reaction :

2AB2(g) 2AB(g) + B2(g)

with a degree of dissociation ‗x‘ which is small compared to unity. Deduce the expression for ‗x‘

in terms of the equilibrium constant Kp and the total pressure P. [in 1984]

Q.18 Calculate the percent dissociation of H2S(g) if 0.1 mole of H2S is kept in 0.4 litre vessel at 1000

K. for the reaction 2H2S(g) 2H2(g) + S2(g) the value of KC is 1.0 × 10–6. [in 1984]

Q.19 For the reaction Ag(CN2)– Ag+ + 2CN–, the KC at 25°C is 4 × 10–19. Calculate [Ag+] in

solution which was originally 0.1 M in KCN and 0.03 M in AgNO3. [in 1984]

Q.20 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 mole of H2 and 0.2 mole of Br2 is heated to 700 K. [in 1985]

Q.21 At some temperature and under a pressure of 4 atm, PCl5 is 10% dissociated. Calculate the

pressure at which PCl5 will be 20% dissociated, temperature remaining same. [in 1996]

Q.22 A sample of air consisting of N2 and O2 was heated to 2500 K until the equilibrium

N2(g) + O2(g) 2NO(g)

was established with an equilibrium constant KC = 2.1 × 10–3. At equilibrium, the mole % of NO

was 1.8. Estimate the initial composition of air in mole fraction of N2 and O2. [in 1997]

Q.23 The Kp for the reaction N2O4 2NO2 is 640 mm at 775 K. Calculate the percentage

dissociation of N2O4 at equilibrium pressure of 160 mm. At what pressure, the dissociation will

be 50% ? [in 1997]

Q.24 At 540 K, 0.10 mole of PCl5 are heated in a 8 litre flask. The pressure of the equilibrium mixture

is found to be 1.0 atm. Calculate Kp and KC for reaction. [in 1998]

Q.25 The Kp value for the reaction : H2 + I2 2HI, at 460°C is 49. If the initial pressure of H2 and

I2 is 0.5 atm respectively, determine the partial pressure of each gas at equilibrium. [in 1999]

Q.26 Calculate the value of log Kp for the reaction, [in 1999]

N2(g) + 3H2(g) 2NH3(g)

at 25°C. The standard enthalpy of formation of NH3(g) is –46 kJ and standard entropies of N2(g),

H2(g) and NH3(g) are 191, 130, 192 J K–1 mol–1 respectively. (R = 8.3 × JK–1mol–1)


Ph.: 0291-3291970.

69

Q.27 When 3.06 g of solid NH4HS is introduced into a two litre evacuated flask at 27°C, 30% of the

solid decomposes into gaseous ammonia and hydrogen sulphide (i) calculate Kc and Kp for the

reaction at 27°C. (ii) what would happen to the equilibrium when more solid NH4HS introduced

into the flask . [in 1999]

Q.28 At 817°C, Kp for the reaction between CO2(g) and excess hot graphite C (s) is 10 atm.

(a) What are the equilibrium concentration of the gases at 817 °C and a total pressure of 5 atm?

(b) At what total pressure, the gas contains 5% CO2 by volume? [in 2000]

Q.29 The value of Kp is 1 × 10–3 atm–1 at 25°C for the reaction : 2NO + Cl2 2NOCl. A flask contains

NO at 0.02 atm and at 25°C. Calculate the mol of Cl2 that must be added if 1% of the NO is to be

converted to NOCl at equilibrium. The volume of the flask is such that 0.2 mol of gas produce 1

atm pressure at 25°C. (Ignore probable association of NO to N2O2.) [in 2000]

Q.30 When 1-pentyne (A) is treated with 4 N alcoholic KOH at 175°C, it is converted slowly into an

equilibrium mixture of 1.3% 1-pentyne(A), 95.2% 2-pentyne (B) and 3.5% of 1,2-pentadiene (C).

The equilibrium was maintained at 175°C. Calculate G° for the following equilibria :

B A ; G1° = ? [in 2000]

B C ; G2º = ?

From the calculated value of G1º and G2º indicate the order of stability of (A), (B) and (C).

Write a reasonable reaction mechanism showing all intermediates leading to (A), (B) and (C).

Q.31 (a) In the following equilibrium : N2O4(g) 2NO2(g)

When 5 moles of each are taken, the temperature is kept at 298 K the total pressure was found to

be 20 bar. Given that – [in 2004]

Gf0 (N2O4) = 100 kJ

Gf0 (NO2) = 50 kJ

(i) Find G of the reaction. (ii) The direction in which the equilibrium shifts.


Ph.: 0291-3291970.

70

Answer Key of Exercises Exercise- 1

1. A 2. C 3. C 4. B 5. D 6. B 7. B 8. D

9. B 10. D 11. A 12. D 13. D 14. B 15. C 16. A

17. A 18. A 19. A 20. B 21. D 22. C 23. C 24. B

25. C 26. C 27. D 28. A 29. 30. 31. 32.

Exercise- 2

1. (a) backward direction (b) forward direction (c) equilibrium constant

2. only B > 1, both A & B > 1 3. 181.5 atm

4. 0.019 5. 0.316 mol litre–1, 1.316 mol litre–1 , 4.368 mol litre–1

6. 1.6 litre 7. 1.03 × 10–5 8. 0.4 mol litre–1, 1.6 mol litre–1

9. 0.3408, 34.08% 10. (a) 833.33 mol–1 litre1, (b) 11.62 (atm)–1 (c) 0.34 mole

11. 0.33 or 33% 12. 1.07 atm, 8.93 atm 13. 7.29 ×10–4 atm–2

14. 1.431 × 10–5 atm–2, 105.41 atm 15. 17.76%

16. 12.58 × 102 17. 18.196 kcal, 55.62 cal 18. 6.14 atm

19. 0.28 20. no decompostion of M2CO3

21. 24.62 22. 3.11

Exercise- 3 1. B 2. B 3. A 4. A 5. C 6. B 7. A 8. C

9. B 10. B 11. B 12. C 13. B 14. B 15. D 16. C

17. C 18. D 19. A 20. A 21. B 22. A 23. B 24. D

25. C 26. B 27. B 28. B 29. 30. 31. 32.

Exercise- 4

1. (a) Kp =0.108 atm2 (b) (i) 0.117 atm (ii) 789 torr

2. (i) KC = 707.2 (ii) freezing point of resulting solution = 272.6256 K

3. total pressure = 84.34 atm 4. KC = 6.36 × 10–4 mol/lit

5. (a) XCO2 = 23% (by vol.) XCO = 77% (by vol.) (b) PCO2 = 0.92 atm

(c) P = 0.68 atm

6. Total P = 314.64 mm Hg

7. CH4 0.52 mole, H2O 0.52 mole , CO(g) 0.48 mole, H2 2.44 mole

8. Kp = 0.64 atm–1

9. a = 9.574 J mol–1; b = A = 1010; c = 9.96 × 109; d = 9.98 × 109

10. Kp = 8.48 × 10–9 atm2, G° = 35.53 kJ mol–1 11. PCO = 0.272 atm


Ph.: 0291-3291970.

71

12. 0.04 13. 0.43

14. H3O+ produces by lactic acid shifts the euilibrium in forward direction according to

Lechatlier‘s principle. Thus oxygen gas will be released.

15. (i) (a) PH2O = 1.62 × 10–3 atm, (b) PH2O = 3.5 ×10–3 atm; (c) PH2O = 3.64 × 10–

3 atm (ii) SrCl2.2H2O is most effective drying reagent because it has least vapour pressure or

water. (iii) Na2SO4.10H2( will efforesce (loose water) when its pressure is below3.64 × 10–3

atm, i.e., 2.77 torr. Relative humidity at 0°C = 2.77

4.58 × 100 = 60.5%

16. (a) 0.96 atm; (b) less, (c) PCO2 = 1.33 atm, PH2O = 0.169 atm 17. 9.43 g

18. Kc = 54, nHI = 0.9 mol, nI2 = 0.05 mol, nH2 = 0.3 mol

19. 0.0663 atm 20. K1 = 0.0488 ; K2 = 357.777

21. (a) 1.05 atm, (b) 3.43 atm-1

22. Kc = 1/12, [R] = 4(initial), = 1.5 (final)

23. ANH4NO3; BNH4NO2; Total pressure = 84.34 atm

24. K = 707.2, backward reaction is favoured 25. Kc = 480

27. 3.1420 28. 0.7 atm, 1.5 atm–1 29. 4 k > K2, 0 < K < 4

30. 2 31. 4.3 × 10–1 32. 0.126 atm

Exercise- 5

1. A,B,C 2. B,C 3. A,B,C,D 4. A,B

5. D 6. A,B,C,D 7. A,B 8. A,C

9. A,B,C 10. A,B,C 11. A, B, D 12. B,C,D

13. B, C 14. B, C 15. A, C 16. B, C

17. A, C, D 18. A,C, D 19. A, B, C, D 20. A, C

21. A, B, D 22. A, C 23. A,B,C,D 24. A, B,D

26. D 27. D 28. A 29 C 30. A 31. A

32. A 33. A 34. A 35. C 36. B 37. C

38. D 39. A 40. A 41. B 42. C 43. C

44. D 45. CD 46. A 47. B 48. D 49. D

50. B 51. A 52. C 53. A

Exercise- 6

1. C 2. C 3. B 4. A 5. D 6. D

7. A 8. A 9. C 10. B

Exercise- 7

1. B 2. D 3. F 4. T 5. C 6. A

7. ABCD 8. D 9. C 10. A 11. A 12. A


Ph.: 0291-3291970.

72

13. C 14. D 15. no change 16. B 17. less 18. C

19. D 20. A 21. D 22. B 23. D

24. D 25. D 26. A 27. B 28. D

Exercise- 8

1. 1.48 ×10–5 l2 mol–2, 3.82 ×10–3 litre mol–1 2. 33.3%, 0.41 3. 1.886

4. (a) 0.1 (b) 0.4 5. (a) 1.6 moles, 0.4 moles

6. [H2] = 0.316 mol lit–1 [I2] = 1.316 mol lit–1 [HI] = 4.368 mol lit–1

7. (a) 0.267 atm, (b) 63.26 %

8. (a) Mole of CO2 = mole of H2 = 0.338 mol ; mole of CO = mole of H2O = 0.112 mole

(b) mole of CO2 = mole of H2 = 0.593 mole; mole of CO = mole of H2O = 0.197 mole

9. PSO3 = 0.9764 atm; PSO2 = 0.0236 atm; PO2 = 2.0118 atm

10. Kp = 2.55 (atm)3 11. KC = 279.64 lit2mol–2; Kp = 0.115 atm–2

12. PN2O4 = 0.095 atm PNO2 = 0.64 atm

13. (a) 0.681 (b) PH2 = PCO2 = 0.34 atm; PCO = 1.16 atm ; PH2O = 0.16 atm

14. (i) Kp = 0.05 atm–2; KC = 187.85 mol–2 lit2 (ii) P = 12.438 atm

15. Kp = 1.8 atm 16. Kp = Kc (RT)n 17. X = p2K3

P 18. 2%

19. 7.5 × 10–18 20. HBr = 0.4 mole, H2 = 0.4 mole, Br2 = 0.0

21. 0.969 atm 22. a = 79%, 100 – a = 21% 23. n = 70.7%, P = 480 mm

24. KC = 4 × 10–2 mol lit–1 Kp = 1.77 atm

25. PH2 = 0.111 atm ; PI2 = 0.111 atm; PHI = 0.778 atm

26. log Kp = 5.845 27. Kp = 4.90 × 10–2 atm2 KC = 8.1 × 10–5 mol2 lit–2

28. (a) CO = 0.041 mol lit–1 (b) P = 0.145 atm; CO2 = 0.015 mol lit–1

29. 0.02042 30. G1° = 16.178 kJ; G2° = 12.282 kJ

31. (i) 56.0304 (ii) G = positive so reverse reaction will take place

Education

chemical equilibrium for iit jee