Chemical Equilibrium

3       1.         IIT–JEE Syllabus       2.         Introduction 3.         Law of Mass action &                                   Equilibrium Constant         4.         Reaction Quotient   5.         Le Chatilier’s Principle 5.1.       Effect of addition of inert                               gas at Equilibrium 5.2.       Dependence of Kp on                                  Temperature   6.         Degree of Dissociation (a) 6.1.       Degree of dissociation from                                   density measurement   7.         Thermodynamics of Chemical                   Equilibrium   8.            Solution to Exercises         9.         Solved Problems 7.1. Subjective 7.2. Objective                         10.       Assignments                                     (Subjective Problems)                                  11.       Assignments                                     (Objective Problems)                                   12.       Answers to Subjective Assignments                            11.       Answers to Objective Assignments

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		Whenever we hear the word ‘Equilibrium’ immediately a picture arises in our mind – ‘an object under the influence of two opposing forces’. For chemical reactions also this is true. A reaction also can exist in a state of equilibrium balancing forward and backward reactions. Never be surprised if I say all reactions are reversible. It’s a fact. Those reactions for which the forward reaction occurs to a much greater extent, we consider them to be unidirectional reactions. Keep in mind equilibrium always exists in closed systems only. This chapter deals with the concepts of chemical equilibrium, how the equilibrium is reached, what will be its value and unit, what are the factors that affect chemical equilibrium, how do they affect and all such related questions and ultimately predicts the outcome in which direction the reaction will be favoured.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1.    IIT–JEE syllabus

Law of mass action; equilibrium constant; exothermic and endothermic reactions;  Le – Chatelier’s Principle and its applications 

2.    Introduction

Chemical equilibrium is the most important characteristic property of a reversible reaction. It is the state of reversible reaction, at given temperature, characterised by constancy of certain observable properties such as pressure, conc., density etc. The equilibrium state is dynamic and not static in nature. A reaction is said to have attained equilibrium when the rate of forward reaction equals that of backward reaction.

A + B  C + D

Characteristics of Equilibrium State

1. i) It can be attained only if the reversible reaction is carried out in closed vessel.
2. ii) It can be attained from either side of the reaction.

iii)   A catalyst can hasten the approach of equilibrium but does not alter the state of equilibrium.

1. iv) It is dynamic in nature i.e. reaction does not stop but both forward and backward reactions take place at equal rate.
2. v) Change of pressure, concentration or temperature favours one of the reaction (forward or backward) resulting in shift of equilibrium point in one direction.

3.    Law of Mass Action and Equilibrium Constant

According to the law of mass action, the rate of reaction, at given temperature is directly proportional to the product of the active masses of the reacting substances each raised to power equal to corresponding stoichiometric coefficient appearing in balanced chemical equations.

Active mass of reactant (a) = Conc. × activity coefficient

i.e. a = Molarity × f                        for dilute solution f = 1

Applying Law of mass action for general reversible reaction

aA + bB  mC + nD

Rate of forward reaction µ [A]^a[B]^b

or R_f = K_f [A]^b[B]^b

Similarly for backward reaction

R_b = K_b[C]^m[D]ⁿ

At equilibrium K_f[A]^a[B]^b = K_b[C]^m[D]ⁿ

\  = K_C

K_C = equilibrium constant in terms of concentration

In terms of partial pressures equilibrium constant is denoted by K_p

K_p =  (P = CRT, Pµ C at constant temperature)

Where P terms represent partial pressures at equilibrium

In terms of mole fraction equilibrium constant is denoted by K_­X

K_X =

Where ‘X’ terms represent mole fractions at equilibrium

The equilibrium constants are related as

K_p=                                                                             …(1)

K_P =                                                         …(2)

Where ‘P’ is total pressure at equilibrium and T-temperature

Dn_g = (m + n) – (a + b)

When Dn_g = 0, K_p= K_C = K_X

Also K_p =

Where n_A, n_B … etc. are respective number of moles at equilibrium, ån is total number of moles at equilibrium P is total pressure at equilibrium.

Equilibrium Constant of a reaction involving Condensed Phase: The expression for the equilibrium constant of a reaction involving condensed phase (solid or liquid) and gaseous state is obtained by considering the concentrations or partial pressures of only the gaseous species. The concentration of condensed phases being constant are merged with the equilibrium constant.

Some facts about equilibrium constant

1. a) A very large value of K_C or K_P signifies that the forward reaction, as written, goes to completion or very nearly so.
2. b) A very small numerical value of K_C or K_P signifies that the forward reaction, as written, does not go to any significant extent.
3. c) A reaction is most likely to reach a state of equilibrium in which both reactants and products are present if numerical value of K_C or K_P is neither very large nor very small.
4. d) The equilibrium constant of a forward reaction and that of its backward reaction are reciprocal of each other.

If A + B  C + D    K = x

Then C + D  A + B          K¢ =

1. e) If a chemical equation is multiplied by certain factor, its equilibrium constant must be raised to a power equal to that factor in order to obtain the equilibrium constant for the new reaction.

If, for NO + O₂  NO₂ K =

Then for 2NO + O₂  2NO₂ K ¹ =  = K²

If K₁, K₂ and K₃ are step-wise equilibrium constant for

A  B                      (1)

B  C                     (2)

C D         (3)

Then for A  D K = K₁ × K₂ × K₃

4. The Reaction Quotient ‘Q’

At any point in a reversible reaction, ratio of the concentration terms  in the same form as in the equilibrium constant expression can be formulated. This ratio is referred to as reaction quotient or mass action ratio (Q)

For general reaction, aA + bB  mC + nD

Q_C =  (in terms of partial pressure)

1. a) A net reaction proceeds from left to right (the forward reaction) if Q_C < K_C.
2. b) The backward reaction takes place if

Q > K_C

1. c) The reaction is at equilibrium if

Q = K_C

Illustration-1:    At 1000 K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation of I₂(g) into I. Had there been no dissociation, the pressure would have been 0.074 atm. Calculate the value of K­_p for the reaction:
I₂(g)  2I(g).

Solution:        analysing in terms of pressure directly

Partial pressure	I2	2I
Initial	0.074	0
at equilibrium	0.074- p	2p

Þ total pressure at equilibrium

= (0.074 – p) + 2p = 0.112 (given)

Þ p = 0.038 atm

K_p =

Substituting value of p      Þ K_p = 0.16 atm

Illustration-2:    In an evacuated vessel of capacity 110 litres, 4 moles of Argon and 5 moles of PCl₅ were introduced and equilibriated at a temperature at 250^oC. At equilibrium, the total pressure of the mixture was found to be 4.678 atm. Calculate the degree of dissociation, a of PCl₅ and K_P for the reaction

                        PCl₅   PCl₃ + Cl₂  at this temperature.

Solution:                                PCl₅(g)  PCl₃(g)   +   Cl₂(g)

Initial moles        5                   0                0

At equilibrium     5-x               x                x

Total moles           = 5 + x + 4  (including moles of Argon) = 9 + x

Since total moles =  = 11.99 12

\ x = 3

\  a =  = 0.6

K_P = = 1.75 atm.

The degree of dissociation of PCl₅ would have been 0.6 even in the absence of Argon. As one can see, the total pressure of gases  constituting equilibrium is equal to 3.11 atm. The observed equilibrium pressure is 4.678 atm which means by the addition of 4 moles of Argon, the total pressure increases. This implies that addition of Argon has been done at constant volume which does not result in any change in degree of dissociation.

5.    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 I 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)  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,

= K_P  =

Where,

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

Þ  = K_P

Here,  = mole fraction of N₂

= mole fraction of H₂

= mole fraction NH₃

Þ  = K_P

Since P_T =   ( assuming all gases to be ideal)

\ = K_P   ———— (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 Q_P ( as it is disturbed from equilibrium) and we can see that  Q_P > K_P

The reaction therefore moves backward to  make Q_P = K_P.

If we increase c, Q_P < K_P 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), Q_P < K_P 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 (Q_P) and  therefore does it mean that the reaction goes backward (since Q_P > K_P)?. 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 K_P also changes with temperature. Therefore, we need to know the effect of temperature on both Q_P and K_P to decide the course of the reaction.

      5.1.      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

We know,

 

Where,

n_C n_D, n_A, n_B denotes the no. of moles of respective components and PT is the total pressure and ån = total no. of moles of reactants and products.

Now, rearranging,

 

Where Dn = (c + d) – (a + b)

Now, Dn can be = 0, < 0 or > 0

Lets take each case separately.

1. a) Dn = 0 : No effect
2. b) Dn = ‘+ve’ :

Addition of inert gas increases the ån i.e.  is decreased and so is . So products have to increase and reactants have to decrease to maintain constancy of Kp. So the equilibrium moves forward.

1. c) Dn = ‘–ve’ :

In this case  decreases but  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 ån also increases, they almost counter balance each other. So  can be safely approximated as constant. Thus addition of inert gas has no effect at constant volume.

      5.2.      Dependence of  K_P on Temperature

Now we will derive the dependence of  K_P on temperature.

Starting with Arrhenius equation of  rate constant

——————– (i)

Where, k_f = rate constant for forward reaction, A_f = Arrhenius constant of forward reaction,

= Energy of activation of forward reaction

——————– (ii)

Dividing (2) by (3) we get,

We know that  (equilibrium constant )

\  K =

At temperature T₁

————– (iii)

At temperature T₂

——————– (iv)

Dividing  (5) by (4) we get

Þlog

The enthalpy of a reaction is defined in terms of activation energies  as  = DH

\

\log—————- (v)

For an exothermic reaction , DH would be negative. If we increase the temperature of the system ( T₂>T₁), the right hand side of the equations  (V) becomes negative.

\ , that is, the equilibrium 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 K_P would either increase or decrease or may remain same. Let us assume it increases.

Now, Q_P can also increase, decrease or remain unchanged. If K_P increases and Q_P decreases, than , therefore the reaction moves forward. If K_P increase and Q_P remains same, then  . Again, the reaction moves forward. What, if K_P increase and Q_P also increases?

Will  or  or ? This can be answered by simply looking at the dependence of Q_P and K_P on temperature. You can see from the equation (6)  that K_P depends on temperature exponentially. While Q’s dependence  on T would be either to the power  g,l,t…….. Therefore the variation in K_P due to T would be more than in Q_P due to T.

\ K_P would still be greater than Q_P and the reaction moves forward again.

Therefore, to see the temperature effect, we need to look at K_P only. If it increases the reaction moves forward, if it decreases, reaction moves backward and if it remains fixed, then, no change at all.

6. Degree of Dissociation (a)

Let us consider the reaction,

2NH₃ (g)  N₂ (g) + 3H₂ (g)

Let the initial moles of NH₃(g) be ‘a’. Let  x moles of NH₃ dissociate at equilibrium.

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

Initial moles         a                              0                     0

At equilibrium    a–x

Degree of dissociation (a) of NH₃ is defined as the number of moles of NH₃ dissociated per mole of NH₃.

\ if x moles dissociate from ‘a’ moles of NH₃, then, the degree of dissociation of NH₃ would be .

We can also look at the reaction in the following manner.

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

Initial moles       a                                0                      0

At equilibrium  a(1–a)

or                     a–2x¢                           x ¢                    3 x ¢

where  a=

Here total no. of moles at equilibrium is

a – 2x¢ + x¢ + 3x¢ = a + 2x¢

Mole fraction of NH₃ =

Mole fraction of N₂ =

Mole fraction of H₂=

The expression of K_p is

K_p =

=

In this way you should calculate the basic equation. So my advice to you is that, while solving problem follow the method given below:

1. Write the balanced chemical reaction (mostly it will be given)
2. Under each component write the initial no. of moles.
3. Do the same for equilibrium condition.
4. Then derive the expression.

Do it and you are the winner.

Exercise-1:       H₂ and I­₂ are mixed at 400°C in a 1.0 L container and when equilibrium established, the following concentrations are present: [HI] = 0.49 M, [H₂] = 0.08 M and [I₂] = 0.06 M. If now an additional 0.3 mol of HI are added, what are the new equilibrium concentrations, when the new equilibrium
H₂ + I₂  2HI is re-established?

6.1       Dependence of Degree of Dissociation from Density Measurements

The following is the method of calculating the degree of dissociation of a gas using vapour densities. This method is valid only for reactions whose K_P exist, i.e., reactions having at least one gas and having no solution.

Since   PV = nRT

PV =

M =

\ VD =

Since P =

\ VD =

For a reaction at eqb., V is a constant and r is a constant.   \ vapour Density

\ =

(Q molecular weight = 2 ´ V.D)

Here    M = molecular weight initial

m = molecular weight at equilibrium

 

Let us take a reaction

PCl₅        PCl₃    +   Cl₂

Initial moles        C                        0              0

At eqb.             C(1-a)                  Ca            Ca

\  =

Knowing D and d, a can be calculated and so for M and m.

Illustration-3:    In an experiment 5 moles of HI were enclosed in a 5 litre container. At 717 K equilibrium constant for the gaseous reaction 2HI (g)  H₂ (g) + I₂ (g) is 0.025. Calculate the equilibrium concentrations of HI, H₂ and I₂. What is the fraction of HI that decomposes.

Solution:        Let 2n be the number of moles of HI which is decomposed, the number of moles of H₂ and I₂ produced will be n moles each. Then molar concentrations of various species at equilibrium are –

[HI] =  mol/l,         [H₂] =  mol/l,      [I₂] =  mol/l

Also, K_C =  =

0.025 =

Solving for n, we get n = 0.6

\ [HI] =  = = 0.76 mol/l

[H₂] =  = 0.12 mol /l

[I₂] =  = 0.12 mol /l

Fraction of HI decomposed =  = 0.24 or 24%

Illustration-4:    1 mole of N₂ and 3 moles of PCl₅ are placed in a 100 litre container heated to 227°C. The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate the degree of dissociation of PCl₅ and value of K_p for its dissociation.

Solution:        Dissociation of PCl­₅ is written as

PCl₅(g)  PCl₃ + Cl₂(g)

Let x be the no. of moles of PCl₅ decomposed at equilibrium

Moles	PCl5	PCl3	Cl2
Initial	3	0	0
At equilibrium	3-x	x	X

Now total gaseous moles in the container = n_T

n_T = moles of (PCl₅ + PCl₃ + Cl₂) + moles of N₂

n_T = 3 – x + x + x + 1 = 4 + x

The mixture behaves ideally, hence PV = n_TRT

Let us calculate no. of moles by using gas equation

Þ n_T =

Þ n_T = 5

now equating the two values of n_T, we have

4 + x = 5 Þ x = 1

Þ degree of dissociation = 1/3 = 0.333

Now K_p =

P = equilibrium pressure

=  = 0.82 atm

atm

Þ K_p =  = 0.205 atm

K_p = 0.205 atm

                        Note 1:      The inert gases like N₂ or noble gases (He, Ne etc.) though do not take part in the reaction, but still they affect the degree of dissociation and equilibrium concentrations for the reactions in which Dn ¹ 0. They add to the total pressure of the equilibrium mixture (p a n).

Exercise -2:      Determine K_c for the reaction ½N₂(g) + ½O₂(g) + ½Br₂(g)  NOBr(g) from the following information (at 298^oK)

K_c = 2.4 ´ 10³⁰  for  2NO(g)  N₂(g) + O₂(g) ;

= 1.4 for NO(g) + ½Br₂(g)   NOBr(g)

 

 

 

7. Thermodynamics of chemical equilibrium

DG = DG⁰ + 2.303 RT logQ                      …(1) (Q = Reaction Quotient)

Under equilibrium condition Q =  K_C or K_P

The Gibbs free energy G for a system is defined as G = H – TS

DG = DH – TDS                                         …(2)

DG⁰ = DH⁰ – TDS                                      …(3) (standard condition)

For a process occurring at constant T and P if (1) DG < O, the process is spontaneous.

(2) DG > O, the process is non-spontaneous

(3) DG = 0, the process is at equilibrium

\ At equilibrium

0 =DG⁰ + 2.303 RT logK

\DG⁰ = –2.303 RT logK                           …(4)

\logK = – Þ lnK =

From equation (3) and (4)

– 2.303 RT logK = DH⁰ – TDS⁰

Þ logK =

logK₁ =  (At temperature T₁)

logK₂ =  (At temperature T₂, T₁ < T₂)

(Assuming DH⁰ to be constant in temperature range T₁ and T₂)

\ lok

Þlog

Also

Graph between logK and

K is referred to as thermodynamic equilibrium constant and can be replaced by K_C or K_p for general reaction.

aA + bB  mC + nD

K =  (Thermodynamic equilibrium constant)

1. i) For pure solids and liquids a = 1
2. ii) For gases ideal behaviour is considered and activity is equal to its pressure in (atm)

iii)   For components in solution a = molar concentration.

Activation energies for forward and backward reaction: Both forward and backward reactions follow the same reaction path and involve the same activated complex. However, the activation energies of both reactions are different.

 

 

DH = DE = E_a(f) – E_a(b)  (at constant volume)

For exothermic reaction Ea_(f) < E_a(b)

Hence, DE is negative

For endothermic reaction, E_a(f) > E_a(b)

Hence DE is positive

8. Solution to Exercises

Exercise-1:     First determine the equilibrium constant

K_C for H₂ + I₂  2HI

K_C =

When 0.3 mol of HI are added, equilibrium is disturbed. At the instant,

[HI] = 0.49 + 0.3 = 0.79 M

Þ  Q > K_C, since K_C =

Þ  backward reaction dominate and the equilibrium shifts to the left.

Let 2x = concentration of HI consumed (while going left) then concentration of each of H₂ and I₂ formed = x

Þ  [HI] = 0.79 – 2x, [H₂] = 0.08 + x, [I₂] = 0.06 +x and K_C = 50

Þ  K_C =

Þ  46x² + 10.2x – 0.35 = 0

Þ  x = 0.033 or – 0.25 (neglecting the –ve value)

Finally, the equilibrium concentrations are:

[HI] = 0.79 – 2x =0.79 – 0.033 ´ 2 = 0.724 M

[H₂] = 0.08 + x = 0.08 + 0.033 = 0.113 M

[I₂] = 0.06 + x = 0.06 + 0.033 = 0.093

 

Exercise-2:     N₂(g) + O₂(g)     2NO (g)

K_c^/  =

1. i) ½N₂(g) + ½O₂(g )  NO(g)

K_c^// =

1. ii) NO(g) + ½Br₂(g)   NOBr(g)  = 1.4.

(i) + (ii) gives the net reaction:

½N₂(g) + ½O₂(g) + ½Br₂(g)  NOBr(g)

K =

=  ´   = 0.6455 ´ 10^-15 ´ 1.4  = 9.037 ´ 10^-16

9.    Solved Problems

9.1.      Subjective 

Problem -1:      At 540 K, 0.10 mol of PCl₅ are heated in a 8 litre flask. The pressure of equilibrium mixture is found to be 1.0 atm. Calculate K_p and K_C for the reaction.

Solution:                                            PCl₅          PCl₃     +    Cl₂

Initial moles                 0.1                   0                0

Moles at equilibrium    (0.1 – a)          a                a

(a = degree of dissociation)

Applying (PV = nRT) for the equilibrium mixture at equilibrium

1 × 8 = (0.1 + a) × 0.082 × 540

\ a = 0.08

\ K_C =  = 4 × 10^–2 mol/litre

K_p = K_C(RT)^Dn

= 4 × 10^–2 × (0.082 × 540)

= 1.77 atm

Problem- 2:      For a gaseous phase reaction A + 2B  AB₂ K_C = 0.3475 lt² mol^–2 200°C. When 2 mol of B are mixed with one mole of A, what total pressure is required to convert 60% of A in AB₂.

Solution:                                      A          +    2B                   AB₂

Initial moles                 1                2                                  0

Moles at eqm        (1 – a)       (2 – 2a)                 a

Total moles at equilibrium (ån) = 1 – a + 2 – 2a  + a

K_p =

Þ K_p =

=

Þ K_p =                …(1)

Also K_p = K_C(RT)^Dn

Þ K_p = 0.3475 × (0.0821 × 473)^–2           …(2)

From equation (1) and (2)

P = 181.5 atm

Problem- 3:      A sample of air consisting of N₂  and O₂ was heated to 2500 K until the equilibrium

                        N_2(g) + O_2(g)  2NO_(g)

                        was established with an equilibrium constant K_C = 2.1 × 10^–3. At equilibrium mole % of NO was 1.8. Estimate the initial composition of air in mole fraction of N₂  and O₂

 

Solution:                                N₂        +    O₂                  2NO Let total no. of mole = 100

Initial mole       n                (100 – n)                0       Moles of N₂

At equilibrium  (n – x)        (100 – n – x)          2x

ån = (n – x) + (100 – n – x) + 2x = 100

At equilibrium mole % of NO =  × 100 = 1.8

\ x = 0.9

As Dm_g = 0, \ K_p= K_C

\K_p – K_C =  =  = 2.1 × 10^–3

\n = 79%, (10 – n) = 21%

Problem- 4:      For the reaction [Ag(CN)₂] ^–  Ag⁺ + 2CN^– the equilibrium is constant 4.0 × 10^–19. Calculate the silver ion concentration in a solution which is originally 0.10 M in KCN and 0.03 M in AgNO₃.

 

Solution:        [Ag(CN)₂]^–  Ag⁺ + 2CN^– K_­C = 4 × 10^–19

Hence for Ag⁺ + 2CN^–  [Ag(CN)₂]^– ¢_C = 0.25 × 10¹⁹

Very large value of K¢_C shows that complex forming equilibrium is spontaneous and almost all the Ag⁺ ion would have reacted leaving xM in solution.

Ag⁺            +    2CN^–               [Ag(CN)₂]^–

Initial mole       0.03           0.1 M                     0

At equilibrium  xM             (0.1 – 2 × 0.03)M  0.03 M

K¢_C =  = 0.25 ×10¹⁹

\ 0.25 × 10¹⁹ =

\ [Ag⁺] = x = 7.5 × 10^–18M

 

Problem- 5:      DG⁰ = 77.77 kJ/mol at 100 K for the reaction N_2(g) + O_2(g)  NO_(g). What is the partial pressure of NO under equilibrium at 1000K for air at 1 atm containing 80% N₂ and 20% O₂ by volume.

Solution:        DG⁰ = – 2.303 RT log₁₀K_C

Þ 77.77 × 10³ = – 2.303 × 8.314 × 1000 × logK_C

\ K_C= 8.67 × 10^–5

 

N₂    +    O₂         NO

Initial pressure       0.8             0.2

At eqm. pressure      x

\ 8.67 × 10^–5=

\ x = 3.47 × 10^–5 atm

Problem- 6:      NH₃ is heated at 15 atm from 27°C to 347°C assuming volume constant. The new pressure becomes 50 atm at the equilibrium of the reaction
2NH₃  N₂ + 3H₂. Calculate % of mole of NH₃ actually decomposed.

 

Solution:                          2NH₃        N₂        +    3H₂

Initial mol   a                      0                0

(a – 2a)           a                3a

Initial pressure of NH₃ of a mol = 15 atm, at 27°C

The pressure of a mol of NH₃ = P atm at 347°C

\ P = 31 atm

At constant volume at 347°C mole µ pressure

a µ 31 atm (before equilibrium)

(a + 2x) µ 50 atm (after equilibrium)

\

\ x = a

\ % of NH₃ decomposed =  = 61.3%

Problem -7:      K_p for the reaction PCl₅  PCl₅ + Cl₂ at 250°C is 0.82. Calcualte the degree of dissociation at given temperature under a total pressure of 5 atm. What will be the degree of dissociation if the equilibrium pressure is 10 atm at same temperature.

 

Solution:        Let 1 mole of PCl₅ be taken initially. If ‘x’ moles of PCl₅ dissociate at equilibrium, its degree of dissociation = x

Moles	PCl5	PCl3	Cl2
Initially	1	0	0
at equilibrium	1-x	x	X
Total moles	1 – x + x + x = 1 + x

P = 5 atm K_p = 0.82

=

K_p =

K_p =

 

x = 0.375 (0r 37.5%)

Now the new pressure P = 10 atm

Let y be the new degree of dissociation. As the temperature is same (250°C), the value of K_p will remain same. i.e. the same manner. Proceeding in the same manner.

K_p =

Þ y =  or y = 0.275 (or 27.5%)

Note 1:      by increasing pressure, degree of dissociation has decreased, i.e., the system shifts to reverse direction. Compare the result by applying Le Chatelier’s principle.

Problem-8:       The value of K_C for the reaction: A₂(g) + B₂(g) ] 2AB(g) at 100°C is 50. If 1.0 L flask containing one mole of A₂ is connected with a 2.0 L flask containing two moles of B₂, how many moles of AB will be formed at 100°C?

Solutions:      A₂(g)  2AB(g)

As the two vessels are connected, the final volume for the contents is now 3.0 L. Let x mole each of A₂ and B_­2 react to form 2x moles of AB₂ (from stoichiometry of reaction)

Moles	A2	B2	2AB
Initially	1	2	0
at equilibrium	1-x	2-x	2x

K_C = ; concentration of species at equilibrium are:

[A₂] = (1 – x)/3, [B₂] = (2-x)/3, [AB] = 2x/3

K_C =

Þ 46x² + 150x + 100 = 0

Þ x = 0.93 or 2.33 (neglecting this value)

Þ moles of AB(g) formed at equilibrium = 2x = 1.86

 

Problem -9:      Variation of equilibrium constant K with temperature T is given by Van’t Hoff equation

                  log K = log A –

                        A graph between log  K and  T^–1 was a straight line as shown and having OP = 10 and tan  q = 0.5. Calculate

(i)    equilibrium constant at 298 K, and (ii)   and equilibrium constant at 798 K, assuming DH° to be independent of temperature.

 

Solution:        To calculate equilibrium constant, we need to know A and DH°, which are calculated as –

The given equation represents a straight line of slope =
= –tan q = –0.5

\ DH° = 2.303 ´8.314 ´0.5 = 9.574 J/mol

Intercept = log A = OP = 10

\ log K = log A–

                        \ K = 9.96 ´10⁹

Now, to calculate equilibrium constant at some other temperature, we will use the expression

=

log  =

                        \ K₂ (equilibrium constant at 798) = 9.98 ´ 10⁹

Problem-10:           The  equilibrium constant K_P of the reaction

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

                        is 900 atm^–1 at 800°C. A mixture containing SO₃ and O₂ having initial partial pressures of 1 atm  and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at 800°C.

Solution:        It can be seen that as SO₂ is not present initially, so equilibrium cannot be established in the forward direction. Therefore it is established from reverse direction. Let n be the increase in partial pressure of O₂. Then at equilibrium the partial pressures of SO₂, O₂ and SO₃ are (0+2n), (2+n) and (1–2n) in atm respectively.

2SO₂(g)    +    O₂(g)        2SO₃(g)

(0+2n)             (2+n)               (1–2n)

Also,

As n is small (because equilibrium constant for the reverse reaction is very small i.e., 1/900), it can be neglected in comparison to 2 and also
1–2n can be taken approximately to 1.

\ 900 =

Solving for n, we get  n = 0.0118

                        Hence,           =    2+n      =    2.0118 atm

                                              =    2n =    0.0236 atm

                            =    1–2n    =    1–0.0236 = 0.9764 atm

 

Problem- 11:    At a given temperature and a total pressure of 1.0 atm for the homogenous gaseous reaction N₂O₄  2NO₂, the partial pressure of NO₂ is 0.5 atm

1. a) Find the value of K_p
2. b) If the volume of the vessel is decreased to half of its original volume, at constant temperature, what are the partial pressures of the components of the equilibrium mixture.

 

Solution:        For equilibrium system, N­₂O₄  2NO₂, the total pressure is 1.0 atm

Þ the total pressure =

Þ

1. i)  = 0.5 atm
2. ii) as volume is decreased to half its original volume, equilibrium is disturbed and the new initial conditions for the re-establishment new equilibrium are

atm &  = 1.0 atm

According to Le Chatelier principle, when volume is decreased, the system moves in that direction where there is decrease in number of moles. Hence, the system (here will move in reverse direction, as there is a decrease in mole (Dn = 2 – 1 = 1) i.e. the NO₂ will be converted to N₂O₄.

Let the decrease in pressure of NO₂ be x atm.

Pressure	N2O4	NO2
Initial	1.0	1.0
At equilibrium	1 + x/2	1 – x

Þ K_p =

Þ 4x²  – 9x + 2  = 0

Þ x = 2 or 0.25 (x ¹ 2 as initial pressure = 1.0)

Þ x = 0.25

&

= 0.75 atm

 

Problem-12:           Ammonium hydrogen sulphide dissociated according to the equation

NH₄HS(s)  NH₃(g) + H₂S(g).  If the observed pressure of the mixture is 1.12  atm at 106^oC, what is the equilibrium constant K_p of the reaction ?

 

Solution:              The reaction is

NH₄HS(s)  NH₃(g) + H₂S(g)

If a is the degree of dissociation of equilibrium,

Total moles of NH₃ + H₂S = 2a.

Partial pressure =

\ p_NH3 = ´ P = 0.5P;    p_H2S = ´ P = 0.5P

K_p = p_NH3 ´ p_H2S = 0.5P ´ 0.5P = 0.25P²

Substituting the value of P = 1.12 atm.,

K_p = 0.25 ´ 1.12 ´ 1.12 = 0.3136.

 

Problem -13:          In a mixture of N₂ and H₂ initially in  a mole ratio of 1:3 at 30 atm and 300^oC , the percentage of ammonia by volume under the equilibrium is 17.8. Calculate the  equilibrium constant (K_P) of the mixture, for the reaction

N₂(g) + 3H₂(g)  2NH₃(g)

 

Solution:              Let the initial moles of N₂ and H₂  be 1 and 3 respectively (this assumption is valid  as K_P will not depend on the  exact no. of moles of N₂ and H₂. One can even start  with x and 3x).

 

Alternatively

N₂(g)  +    3H₂(g)    2NH₃

Initial    1                      3                 0

At eqb  1-x                  3-3x            2x

Since % by volume of a gas is same as % by mole,

\  =0.178

\

\  Mole fraction of H₂ at equilibrium  =

Mole fraction of N₂ at equilibrium   = 1 – 0.6165 – 0.178 = 0.2055

\K_P =

 

= 7.31 ´ 10^-4

N₂(g)   +  3H₂(g)    2NH₃(g)

Initia       x               3x                   0

At eqb.   x-a       3x-3a                 2a

 

\  =0.178

\  = 0.177

\   = 0.302

Similarly we can calculate the mole fraction of N₂(g) and H₂(g) at equilibrium.

 

Problem -14:          The density of an equilibrium mixture of N₂O₄ and NO₂ at 1 atm. and 348 K is  1.84 g dm^-3. Calculate the equilibrium constant of the reaction

N₂O₄(g)  2NO₂(g).

 

Solution:             Let us assume that we start with C moles of N₂O₄(g) initially.

N₂O₄(g)    2NO₂(g)

Initial                      0                            0

At equilibrium      C(1-a)                    2Ca

Where a is the degree of dissociation of N₂O₄(g)

Since

=

Initial vapour density   =

 

Since vapour density and actual density are related by the equation, V.D.   = =      =  26.25

\ 1 + a =

\  a = 0.752

• K_p =
• = 2 atm.

 

Problem-15:                 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 with unity. Deduce the expression for x in terms of the equilibrium constant, K_p and the total pressure P.

 

Solution:                    Let the initial pressure of AB₂(g) be P₁

2AB₂(g)  2AB(g) + B₂(g)

P₁(1-x)      P₁x

                                                            The total pressure is = P₁ << 1)

• P₁ = P
• So, K_P =

 

 

 

 

 

 

 

 

 

 

 

      9.2       Objective

 

Problem -1:     For the gaseous phase reaction

2A + B C+ D,

The value of K_p is . What is the value of DG° at 27°C.

(A) – 600 K cals                                (B) – 6 Kcals

(C) 10³ Kcals                                    (D) 10 Kcals

(E) None

 

Solution:       DG° = – 2.303 RT log K

= –2.303 ´ 2 cal. K^–1 ´ 300 K log 10

= – 2.303 ´ 2 ´ 300 cals ´

= – 600 Kcals

\(A)

 

Problem-2:      For the reaction

A + B  2C + D

The initial concentration of A and B are 1M each. The value of K_c is 10⁸. What is equilibrium concentration of A?

(A) 2 ´ 10^–4 M                                   (B) 2 ´ 10⁴ M

(C) .005 M                                        (D) .0025 M       (e) None

 

Solution:.                          A                      +          B            2C       +          D

t= 0             1M                               1M                   0                      0

–(1–x)                          (1–x)                (2–2x)              (1–x)

——————————————————————————

xM                   x                      (2–2x)              (1–x)

K =  »  = 10⁸

\ x = 2 ´ 10^–4

                       \(A)

 

Problem- 3:     The solid dissociates as  H₂NCOONH₄(s)  2NH₃(g) + CO₂(g)

The equilibrium total pressure is 0.3 atm. What is the value of K_p

(A) 4 ´ 10^–3                                       (B) 4

(C) 10^–3                                             (D) 8 ´ 10³

 

Solution:       NH₂COONH₄(s)      2NH₃   +          CO₂

2x                    x

3x = .3

x = .1

K_p = (2x)² ´ x = 4x³ = 4 ´ .1³  = 4 ´ 10^–3

                              \(A)

Problem-4:       For the gas phase reaction 2NO(g)  N₂(g) + O₂(g) , DH = –43.5 kcal, which one of the following is true for N₂(g) + O₂(g)  2NO(g)      

                        (A) K is independent of T                     (B) K decreases as T decreases

                        (c) K increases as T decreases             (d) K varies with addition of NO

 

Solution:        The given reaction N₂(g) + O₂(g)  2NO(g) is endothermic. Therefore, according to Le–Chatlier’s  Principle, high temperature favours forward reaction and hence K increases as T increases or K decreases as T decreases.

      \ (b)

 

Problem-5:       For the reaction PCl₃ (g) + Cl₂(g) PCl₅(g), the value of K_P at 250°C is 0.61 atm^–1. The value of K_C at this temperature will be

                        (A) 15 (mol/l)^–1                                      (B) 26 (mol/l)^–1

                        (C) 35 (mol/l)^–1                                      (D)  52 (mol /l)^–1

 

Solution:        PCl₃(g)      +    Cl₂(g)        PCl₅(g)

Dn = –1, K_P = 0.61 atm^–1.

K_C = K_P (RT)^–Dn

= 0.61 (0.0821 ´523)⁺¹ = 26 mol /l.

                        \(b)

 

Problem-6:       In the reaction A₂(g) + 4B₂ (g)  2AB₄(g) , DH > 0. The decomposition of AB₄ (g) will be favoured at

                        (A) low temperature and high pressure     

                        (B) high temperature and low pressure

                        (C) low temperature and low pressure

                        (D) high temperature and high pressure

 

Solution:        2AB₄(g)  A₂(g) + 4B₂(g) DH = –ve

It is an Exothermic reaction and hence  favoured at low temperature. Dn for the reaction is +3. Therefore low pressure  will favour the forward reaction

\(C)

 

Problem -7:      For the reaction SO₂(g) + O₂(g)  SO₃(g) K_p = 1.7 ´ 10¹² at 20°C and 1 atm pressure. Calculate K_c.

                        (A)  1.7 ´ 10¹²                                  (B)  0.7 ´ 10¹²

                        (C)  8.33 ´ 10¹²                                (D)  1.2 ´ 10¹²

 

Solution:        For the reaction SO₂(g) + O₂(g)  SO₃(g)

Given K_P = 1.7 ´ 10¹² at 20°C and 1 atm pressure

K_P = K_c (RT)^Dn; Dn = 1–

1.7 ´ 10¹² = K_c (0.0821 ´ 293)^–1/2

K_c =

K_c = 7 ´ 10¹² ´ (0.0821 ´ 293)^+1/2

= 8.338 ´ 10¹²

                        \(C)

 

Problem-8:       Applying the law of mass action to the dissociation of hydrogen Iodide

                        2HI(g)  H₂ + I₂

                        We get the following expression

                       

                        Where a is original concentration of H₂, b is original concentration of I₂, x is the number of molecules of H₂ and I₂ reacted with each other. If the pressure is increased in such a reaction then

                        (A) K =                      (B) K >

                        (C) K <                      (D) none of these

Solution:        For the reaction H₂ + I₂  2HI (g) since Dn = 0. Equilibrium constant K_C will be uneffected by change in pressure.

                        \ (A)

 

Problem-9:       For a gaseous equilibrium

                        2A(g)  2B(g) + C(g) , K_p has a value 1.8 at 700°K. What is the value of K_c for the equilibrium 2B(g) + C(g) 2A at that temperature

                        (A) » 0.031                                      (B) » 32

                        (C) » 44.4                                       (D) » 1.3  ´ 10^–3

 

Solution:        For 2A(g)  2B(g) + C(g)

K_P = 1.8 at 700°C

\ For 2B(g) + C(g)  2A(g) at 700°C

K_P =  Dn = 2 – (2+1) = –1

K_P = K_c(RT)^Dn

K_c =

K_c = K_P  ´ (0.0821 ´ 973)¹

=  ´ 0.0821 ´ 973

K_c = 44.3796

                        \ (C)

 

Problem-10:     The following equilibrium when methyl mercaptan   (CH₃SH – the smell in onions) is dissolved in water, which statement is false?

                        CH₃SH + H₂O  CH₃S^– + H₃O⁺

                        (A) CH₃SH is weak acid                 

                        (B) H₂O is a weak base

                        (C) CH₃S^– is the conjugate base

                        (D)  H₃O⁺ is the conjugate acid of H₂O

 

Solution:        Since CH₃SH is acid, CH₃S^– will be its conjugate base.

                        \ (C)

 

Problem-11:     In a closed container at 1 atm pressure 2 moles of SO₂(g) and 1 mole of O₂(g) were allowed to react to form SO₃(g) under the influence of catalyst . Reaction 2SO₂(g) + O₂(g)  2SO₃(g) occurred

At equilibrium it was found that 50% of SO₂(g) was converted to SO₃(g). The formal pressure of O­₂ (g) at equilibrium will be

                        (A) 0.66 atm                                    (B) 0.493 atm

                        (C) 0.33 atm                                    (D)  0.20 atm

 

Solution:        For                         H₂(g)   +    O₂(g)                       2SO₃(g)

t = 0                       2 moles     1 moles                             0

t = equilibrium             2(1 – 0.5)   1(1– 0.5)                                  1 mole

2–1 + 0.5 +1  = 2.5

P¢ = P ´ mF

=  = 1 ´  = 0.20 atm

\ (D)

 

Problem-12:     For the reaction N₂O₄(g)  2NO₂(g) the reaction connecting the degree of dissociation (a) of N₂O₄(g) with its equilibrium constant K_P is

                        (A) a =                           (B) a =

                        (C) a =                    (D) a =

Solution:                          N₂O₄(g)       2NO₂(g)

t = 0                 1                            0

t = equilibrium 1–a                        2a

ån = 1 + a

K_p =

K_p =

\(C)

Problem-13:     40% of a mixture of 0.2 mol of N₂ and 0.6 mol of H₂ react to give NH₃ according to the equation: N₂(g) + 3H₂(g)  2NH₃(g) at constant  temperature  and pressure. Then the ratio of the final volume to the initial volume of gases are

      (a) 4:5                                                  (b) 5:4

      (c) 7:10                                                (d) 8:5

Solution:                          N₂ +    3H₂          2NH₃

t = 0           0.2       0.6                   0

at = t          0.2–n   0.6–3x             2x

40% of N₂ = 0.2 ´0.4 = 0.08

40% of H₂= 0.6 ´0.4 = 0.24

\ number of moles of N₂ remaining = 0.2 – 0.08 = 0.12

number of moles of H₂ remaining  = 0.6–0.24 = 0.36

number of moles of NH₃ formed = 0.16

Total number of moles = 0.12 + 0.36 + 0.16 = 0.64

\  =

                        \(a)

       

Problem- 14:    One mole of N₂O₄(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when N₂O₄ (g) decomposes to NO₂(g). If the resultant pressure is 2.4 atm, the percentage dissociation by mass of N₂O₄ (g) is

                        (A) 10%                                                      (B) 20%

                        (C) 30%                                                      (D) 40%

 

Solution:                                N₂O₄(g)          2NO₂(g)

Equilibrium      1–a                  2a

Where a is the degree of dissociation

= ( Q V is constant )

\ P₂ = 2 atm

Ater dissociation of N₂O₄ at 600K,

= 2 (1–a) = 2–2a

= 2 ´ 2a = 4 a

Total pressure = 2–2a+4a = 2+2a

2+2a = 2.4      (Given)

a = 0.2

\ Percentage dissociation  = 20%

                        \(b)

Problem- 15:    K for the synthesis of HI (g) is 50. The degree of dissociation of HI is

                        (A) 0.10                                                      (B) 0.14

                        (C) 0.18                                                      (D) 0.22

 

 

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

1–a                                  Where a is the degree  of dissociations

 

 

 

a = 2–2a

2+5 a = 2

a =  = 0.22

                        \(d)

 

 

10. Assignments (Subjective Problems)

 

Level – I

 

1. The value of K_p for the reaction

2H₂O(g) + 2Cl₂(g)  4HCl(g) + O₂(g)

is 0.035 atm at 400^oC, when the partial pressures are expressed in atmosphere. Calculate K_c for the reaction,

Cl₂(g) + H₂O(g)

2. K_p for the reaction N₂O₄ (g) 2NO₂ (g) is 0.66 at 46°C. Calculate the percent dissociation of N₂O₄ at 46°C and a total pressure of 0.5 atm. Also calculate the partial pressure of N₂O₄ and NO₂ at equilibrium.

 

3. The density of an equilibrium mixture of N₂O₄ and NO₂ at 1 atm. and 348 K is 1.84 g dm^-3. Calculate the equilibrium constant of the reaction,
   N₂O₄(g) 2NO₂(g).
4. Ammonium hydrogen sulphide dissociated according to the equation

NH₄HS(s)  NH₃(g) + H₂S(g).  If the observed pressure of the mixture is 1.12 atm at 106^oC, what is the equilibrium constant K_p of the reaction ?

 

5. In a mixture of N₂ and H₂ initially in a mole ratio of 1:3 at 30 atm and 300^oC, the percentage of ammonia by volume under the equilibrium is 17.8. Calculate the equilibrium constant (K_P) of the mixture, for the reaction,
   N₂(g) + 3H₂(g)  2NH₃(g)

 

6. 25 c.c. of H₂ and 18 c.c. of I₂ vapour were heated in a sealed tube at 456°C and at equilibrium 30.8 c.c. of HI were formed. Calculate the degree of dissociation of pure HI at 456°C.

 

7. The pressure of iodine gas at 1273 K is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation
   I₂ 2I. Calculate K_P.

 

8. 0.0755 gm of Selenium vapour occupying a volume of 114.2 ml. at 700°C exerts a pressure of 185 mm. The Selenium is in a state of equilibrium according to the reaction ; Se₆(g)  3Se₂(g). Calculate (i) the degree of dissociation of Selenium (ii) K_P and (iii) K_C . [At. weight of Se = 79]

 

9          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 K_C = 2.1 ´10^–3. At equilibrium, the mole percentage of NO was 1.8. Estimate the initial composition of air in terms of mole fraction of O₂ and N₂  assuming the moles of N₂ were greater than of O₂, initially.

10. To 500 ml of 0.150 M AgNO₃ solution we add 500 ml of 1.09 M Fe²⁺ solution and the reaction is allowed to reach equilibrium at 25°C.

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 under acidic conditions. Calculate K_C for the reaction.

 

11. N₂O₄ (g) dissociates as follows: N₂O₄(g) 2NO₂ (g)

The density of equilibrium mixture is found to be 3.62 gL^–1 at 298K and at 1 atmospheric pressure. Calculate the equilibrium constant for the reaction.

 

12. When a–D–glucose is dissolved in water, it undergoes a partial conversion to
    b–D–glucose to exhibit mutarotation. This conversion stops when 63.6% of glucose is in b–form. Assuming that equilibrium has been attained,
    calculate K_C for mutarotation.

13.       The K_P for the reaction, H₂ + I₂  2HI at 460°C is 49. If the initial partial pressure of H₂ and I₂ each is 0.5 atm respectively, determine the partial pressure of each gas at equilibrium.

 

14. N₂O₄ is 25% dissociated at 37°C and one atmospheric pressure. Calculate (i) K_P and (ii) percent dissociation at 0.1 atmospheric pressure and 37°C.

 

15. 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 K_C and K_P for the reaction at 27°C. (ii) What would  happen to the equilibrium when more solid  NH₄HS is introduced into the flask?

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Level– II

 

1. Two solid compounds A and C dissociates 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.

 

2. For the reaction NH₃ (g)  N₂ (g) + H₂ (g)

Show that the degree of dissociation of NH₃ is given as

a =

where P is the equilibrium pressure. If K_P of the above reaction is 78.1 atm at 400°C, determine the value of K_C.

 

3. Calculate the total pressure at 400°C that must be applied to a mixture of three parts of hydrogen and one part of nitrogen to give a mixture containing 10% ammonia at equilibrium.

Given: N₂  (g) + 3H₂ (g)  2NH₃  (g);   K_p =1.60 ´ 10^-4  bar^-2  at 400°C.

 

4. Given the following equilibrium constant at 1476 K

CO (g) +  O₂ (g)  CO₂ (g);      K_P = 2.5 ´10⁵

C (graphite) + CO₂ (g)  2CO (g);      K_P = 1.67 ´10³

Find K_P for the reaction C (graphite) + O₂   CO₂ (g)

 

5. When one mole of benzoic acid (C₆H₅COOH) and three moles of  ethanol (C₂H₅OH) are mixed and kept at  200°C until equilibrium is reached, it is found that 87% of the acid is consumed by the reaction.

C₆H₅COOH(l) + C₂H₅OH (l)  C₆H₅COOC₂H₅(l) + H₂O(l)

Find out the percentage of the acid consumed when one mole of the benzoic acid is mixed with four moles of ethanol and  treated in the same way.

 

6. For the reaction, A+3B 2C, the reaction was carried out at a temperature of 400°C and pressure 1 atm in a closed container with the molar ratio of A:B as 1:3. At equilibrium the mole percent of C was found to be 20%. Calculate K_P and K_C.

 

7. 15 g sample of BaO₂ is heated to 794°C in a closed evacuated vessel of 5 litre capacity. How many gm of peroxide are converted to BaO (s)?

2BaO₂ (s)  2BaO (s) + O₂ (g); K_P = 0.5 atm

 

 

 

8. Consider the vapour phase dissociation of an oxyacid HXO₃ according to the equation:

4HXO₃ (g)   4XO₂ (g) + 2H₂O (g) + O₂(g)

Derive the expression : K_P = , where P is the total pressure  and  is the partial pressure of O₂.

 

9. A mixture of 0.373 atm of NO(g) and 0.310 atm Cl₂(g) is prepared at 500K.
   The reaction 2NO(g) + Cl₂ (g)  2NOCl (g)  takes place. The total pressure at equilibrium is 0.544 atm. Determine K_P for the reaction.

 

10. For the reaction I₂ (g) 2I (g), K_C = 1.0 ´10^–2  mol lit^–1. What volume of the vessel should be taken so that at equilibrium 1 mole of I₂ and 0.5 mole of I are  present at equilibrium?

 

11. Show that K_P for the reaction

2H₂S(g)  2H₂ (g)  + S₂ (g) is given by the expression

K_P =

Where a is the degree of dissociation and P is the total equilibrium pressure. Calculate K_C of the reaction if a at 298 K and 1 atm pressure is 0.055.

 

12. The density of an equilibrium mixture of N₂O₄ and NO₂ at 101.325k Pa is 3.62 g dm^–3 at 288K and 1.84 g dm^–3 at 348K. What is the heat of reaction for
    N₂O₄(g)  2NO₂(g)?

 

13*.      For the equilibrium

LiCl.3NH₃ (s)  LiCl. NH₃ (s) + 2NH₃,K_P = 9 atm² at 40°C. A five litre vessel contains 0.1 mole of LiCl.NH₃. How many minimum moles of NH₃  should be added to the flask at this temperature to derive the backward reaction for completion?

 

14. What is the percent decomposition of H₂S if 1.0 mole of H₂S is introduced into 1.1L vessel at 1000K? K_C for the reaction:

2H₂S(g)  2H₂(g) + S₂(g) is 1.0 ´10^–6.

 

15. At a certain temperature, equilibrium constant (K_C) 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?                             

 

 

 

 

 

 

LEVEL – III

 

1. A mixture of SO₃, SO₂ and O₂ gases is maintained in a 10 litre flask at a temperature at which the equilibrium constant (K_C) for the reaction: 2SO₂(g) + O₂(g) 2SO₃(g) is 100.

• If the number of moles of SO₂ and SO₃ in the flask are equal, how many moles of O₂ are present?
• If the number of moles of SO₃ in the flask is twice the number of moles of SO₂, how many moles of O₂ are present?

 

2. 0.15 mole of CO taken in a 2.5L flask is maintained at 750 K along with a catalyst so that the following reaction can take place.

CO(g)+2H₂(g)CH₃OH(g)

Hydrogen is introduced until the total pressure of the system is 8.5 atmosphere at equilibrium and 0.08 mole of methanol is formed. Calculate  (a) K_P  and K_C and (b) the final pressure if the same amount of CO and H₂ as before are used, but with no catalyst so that reaction does not take place.

 

3. At temperature T, a compound AB₂(g) dissociates according to the reaction, 2AB₂(g) 2AB(g) + B₂(g) with degree of dissociation, a, which is   small compared to unity. Deduce the expression for a in terms of the equilibrium constant K_P and the total pressure P.

 

4. The equilibrium constant of the reaction A₂(g) + B₂(g)  2AB (g) at 100°C is 50. If one litre flask containing one mole of A₂ is connected  to a two litre flask containing two moles of B₂, how many moles of AB will be formed  at 373K.                                                                                                                                [IIT–JEE 1985]
5. A saturated solution of I₂ in water contains 0.33 g of I₂ L^–1. More than this can be dissolved in a KI solution because of the following equilibrium :
   I₂ + I^–

A 0.1 M KI solution (0.1M I^–‑) actually dissolves 12.5 g I₂/ litre, most of which is converted to , assuming that the concentration of I₂ in all saturated solution is the same, calculate the equilibrium constant (K_C) for the above reaction. What is  the effect of adding water to a clear saturated solution of I₂ in the KI solution?

6. 3. 0.1 mole of ethanol and 0.1 mole of butanoic acid are allowed to react. At equilibrium, the mixture is titrated with 0.85 M NaOH solution and the titre value was 100 ml. Assuming that no ester was hydrolysed by the base, calculate K for the reaction.

C₂H ₅OH + C₃H₇COOH C₃H₇COOC₂H₅ + H₂O

 

7. COF_2(g) passed over catalyst at 1000°C comes to equilibrium, thus

2COF₂(g) CO₂(g) + CF₄(g)

Analysis of the equilibrium mixture (after quick cooling to ‘freeze’ the equilibrium) shows that 500 c.c of the equilibrium mixture (STP) contains 300 c.c. (STP) of (COF₂ + CO₂) taking the total pressure to be 10 atm. Calculate K_p?

8. Ammonia under a pressure of 15 atm at 27° C is heated 347°C in a closed vessle in the presence of catalyst. Under these conditions, NH₃ partially decomposes to H₂ and N₂. The vessel is such that the volume remains effectively constant, whereas the pressure increases to 50 atm. Calculate the % of NH₃ actually decomposed.

 

9. Solid Ammonium carbamate dissociates as: NH₂COONH₄(s) 2NH₃(g) +CO₂(g). In a closed vessel solid ammonium carbamate is in equilibrium with its dissociation products. 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.

 

10. Two solids X and Y dissociates into gaseous products at a certain temperature as follows

X(s)  A(g) + C(g), and Y(s)  B(g) + C(g). At a given temperature,             pressure over excess solid X is 40 mm and total pressure over solid Y is 60 mm.   Calculate.

1. a) the values of K_p for two reactions
2. b) the ratio of mole sof A and B in the vapour state over a mixture of X and Y
3. c) the total pressure of gases over a mixture of X and Y

 

11. Calculate the % age dissociation of H₂S(g) if 0.1 mole of H₂S is kept in a 0.4 L vessel at 1000 K. The value of K_C for the reaction 2H₂S (g) 2H­₂(g) + S₂(g) is 1.0 ´ 10⁴

 

12. Consider the equilibrium: P(g) + 2Q(g) R(g). When the reaction between P and Q is carried out at a certain temperature, the equilibrium concentration of P and Q are 3 M and 4 M respectively. When the volume of the vessel is doubled and the equilibrium is allowed to re-established, the concentration of Q is found to be 3 M. Find the
13. a)   value of K_C
14. b)   concentration of R at new equilibrium stages

 

13. The degree of dissociation of HI at a particular temperature is 0.8. Find the volume of 1.5 M sodium thiosulphate solution required to react completely with the iodine present at equilibrium in acidic condition, when 0.135 mol each of H₂ and IO₂ are heated at 440 K in a closed vessel of capacity 2.0 L.

 

14. At 817°C, K_p for chemical reaction between CO₂ and excess of hot graphite form 2CO_(g) is 10.
15. i) What is the fraction of the gases at equilibrium at 817°C and a totral pressure of          4 atm? What is the partial pressure of CO₂ at equilibrium?
16. ii) At what total pressure will the gas mixture have 6% CO₂ by volume?

 

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

N₂O₄(g)  2NO₂(g)

If it is 35%, dissociated at a given condition,

Find: i) The percent dissociation at same temperature if total pressure is 0.2 atm

1. ii) The percent dissociation (at same temperature) in a 80 gm 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.

11. Assignments (Objective Problems)

 

Level – I

 

Instruction:    In each of the following objective questions, 4/5 alternatives are given. In each one, one or more than one may be correct. It is important to do these questions in one attempt and under a time limit. Therefore you are required  to do this assignment in 20 minutes. 

 

1. In the reaction C(s)+ CO₂(g) 2CO(g) the following amounts of substances were formed in  0.2 litre flask at equilibrium: C = 0.1 mole CO = 0.05 mole CO₂ = 0.06 mole. The equilibrium constant is

(A) 0.208                                                   (B) 4.10

(C) 0.30                                                     (D) 0.416

 

2. The vapour density of PCl₅ is 104.25 but when heated to 230°C, 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%

 

3. For the reaction,

PCl₅(g)PCl₃(g) + Cl₂(g), the forward reaction at constant temperature is favoured by

(A) introducing an inert gas at constant volume

(B) introducing chlorine gas at  constant volume

(C) introducing an inert gas at constant pressure

(D) increasing the volume of the container

 

4. For a reversible reaction, if the concentrations of the reactants are doubled, then the equilibrium constant will
5. A) be halved (B) be doubled
6. C) remain the same (D) will become one fourth

 

5. The vapour density of fully dissociated NH₄Cl would be

(A) less than half of the vapour density of pure NH₄Cl

(B)  double of the vapour density of pure NH₄Cl

(C) half of the vapour density of pure NH₄Cl

(D) one fourth of the vapour density of pure NH₄Cl

 

6. vant Hoff’s equation giving the effect of temperature on chemical equilibrium may be represented as

(A)  ln K_P =                                (B)  ln K_C =

(C)  ln K_P =                              (D) ln K_P =

7. For the reaction A + 2B 2C at equilibrium [C] = 1.4 M , [A]_O = 1M, [B]_O = 2M, [C]_O = 3M, The value of K_Cis

(A) 0.084                                                   (B) 8.4

(C) 84                                                        (D) 840

 

8. The equilibrium constant K_C of the reaction : A₂ (g) + B₂ (g) 2AB (g) is 50. If 1 mol of A₂ and 2 mol of B₂ are mixed, the amount of AB at equilibrium would be

(A) 0.467 mol                                            (B) 0.934 mol

(C) 1.401 mol                                            (D) 1.866 mol

 

9. In an aqueous solution of volume 500 ml, when the reaction of
   2Ag⁺ + Cu  Cu²⁺ + 2Ag reached equilibrium the [Cu²⁺] was x M. When 500 ml of water is further added, at the equilibrium [Cu²⁺] will be

(A)  2 x  M                                                 (B) x  M

(C) between x  M and x /2 M                    (D) less than x/2 M

 

10. A 10 litre box contains O₃ and O₂ at equilibrium at 2000K. K_P = 4.17 ´ 10¹⁴ for 2O₃ 3O₂. Assume that  >> and if total pressure is 7.33 atm, then partial pressure of O₃ will be

(A) 9.71 ´10^–5 atm                                    (B) 9.71 ´10^–7 atm.

(C) 9.71 ´10^–6 atm                                    (D) 9.71 ´10^–2 atm

 

11. In the system, LaCl₃(s) + H₂O (g)+heat¾®LaClO(s)+2HCl (g), equilibrium is established. More water vapour is added to reestablish the equilibrium. The pressure of water vapour is doubled. The factor by which pressure of HCl is changed is

(A) 2                                                          (B)

(C)                                                       (D)

 

12. Which one of the following statements is correct?

In a reversible reaction, catalyst

(A) increases the rate of the forward reaction

(B) decreases the rate of the reverse reaction

(C) increases the rate of the forward and reverse reactions equally

(D) alters the equilibrium constant of the reaction

 

13. In what manner will increase of pressure affect the following equation?

C(s) + H₂O  (g)  CO(g) + H₂(g)

(A) shift in the forward direction         (B) shift in the reverse reaction

(C) increase in the yield of H₂            (D) no effect

 

14. The reaction A + B C + D proceeds to right hand side upto 99.9%. The equilibrium constant K of the  reaction will be

(A) 10⁴                                                 (B) 10⁵

(C) 10⁶                                                      (D) 10⁸

 

15. A chemical reaction is said to attain equilibrium when

(A) reactants get completely converted into products

(B) equal amounts of reactants and products are formed

(C) the rate of forward reaction becomes equal to the rate of reverse reaction

(D) the concentration of reactants and products are same

 

16        Chemical equilibrium is dynamic in nature because

(A) the equilibrium is maintained rapidly

(B) the concentration of reactants and products become same at equilibrium

(C) the concentration of reactants and products are constant but different

(D) both forward and reverse reaction occur at all times with same speed

­

17        For an equilibrium reaction involving gases, the forward reaction is Ist order while the reverse reaction is IInd  order. The units of K_P for the forward equilibrium is

(A) atm                                                (B) atm²

(C) atm^–1                                             (D) atm^–2

 

18. In a reversible reaction, two substances are in equilibrium. If the concentration of each one is doubled, the equilibrium constant will be

(A) reduced to half its original value

(B) reduced to one fourth of its original value

(C) doubled

(D) constant

 

19        In a slow reaction, the rate of reaction generally ——————– with time

(A) decreases

(B) increases

(C) sometimes increases and sometimes decreases

(D) remains constant

 

20        The yield of product in the reaction A₂(g) + 2B (g)  C(g) + Q kJ would
be high at

(A) high temperature and high  pressure

(B) high temperatue and low pressure

(C) low temperature and high pressure

(D) low temperature and low pressure

 

 

Level – II

 

Instruction:    In each of the following objective questions, 4/5 alternatives are given. In each one, one or more than one may be correct. It is important to do these questions in one attempt and under a time limit. Therefore you are required  to do this assignment in 20 minutes. 

 

1. In which of the following equilibrium, the value of K_P is less than K_C?

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

(C) N₂ + 3H₂  2NH₃                                (D) 2SO₂ + O₂  2SO₃

 

2. For the reaction PCl₃ (g) + Cl₂(g) PCl₅(g), the value of K_P at 250°C is 0.61 atm^–1. The value of K_C at this temperature will be

(A) 15 (mol/l)^–1                                               (B) 26 (mol/l)^–1

(C) 35 (mol/l)^–1                                                      (D)  52 (mol /l)^–1

 

3. In the reaction A₂(g) + 4B₂ (g) 2AB₄(g) , DH <  0. The decomposition of AB₄ (g) will be favoured at

(A) low temperature and high pressure

(B) high temperature and low pressure

(C) low temperature and low pressure

(D) high temperature and high pressure

 

4. The equilibrium constant for the reaction

N₂(g) + O₂(g)  2NO(g) is 4 ´10^–4 at 200 K. In the presence of a catalyst the equilibrium is attained 10 times faster. Therefore the equilibirum constant in presence of the catalyst at 200 K is

(A) 4 ´10^–3                                                     (B) 4 ´10^–-4

(C) 4 ´10^–5                                                            (D) None

 

5. In a system A (s) 2B (g) + 3C (g)

If the conc. of C at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of B to change to

(A) two times the original value                      (B) one half of its original value

(C)   times the original value                   (D)  times the original value

 

6. For the decomposition reaction

NH₂COONH₄ (s)  2NH₃(g) + CO₂(g)

The K_P = 2.9 ´10^–5 atm³. The total pressure of gases at  equilibrium when 1 mole of NH₂COONH₄ (s) was taken to start with would be

(A) 0.0194 atm                                                (B) 0.0388 atm

(C) 0.0582 atm                                                (D) 0.0766 atm

 

7. For the gas phase reaction C₂H₄(g) + H₂(g) C₂H₆(g); DH = –32.7 kcal carried out in a vessel, the equilibrium concentration of C₂H₄ can be increased by

(A) increasing temperature                             (B) decreasing pressure

(C) removing some H₂                                   (D) adding some C₂H₆

8. K_P for a reaction at 25°C is 10 atm. The activation energy for forward and reverse reactions are 12 and 20 kJ / mol respectively. The K_C for the reaction at 40°C will be

(A) 4.33 ´10^–1 M                                             (B ) 3.33 ´10^–2 M

(C) 3.33 ´10^–1 M                                             (D) 4.33 ´10^–2 M

 

9. For the reaction CaCO₃(s)CaO(s) +CO₂ (g), the pressure of CO₂ depends on

(A) the mass of CaCO₃(s)

(B) the mass of CaO(s)

(C) the masses of both CaCO₃(s) and CaO(s)

(D) temperature of the system

 

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

(A) total pressure                                (B) catalyst

(C) the amount of H₂ and I₂ present   (D) temperature

[IIT–JEE 1981]

11. An example of reversible reaction is

(A) Pb(NO₃)₂(aq) + 2NaI (aq) Þ PbI₂(s) + 2NaNO₃ (aq)

(B) AgNO₃(aq) + NaCl (aq)  Þ AgCl (s)  + NaNO₃ (aq)

(C) 2Na(s) + 2H₂O(l) Þ 2NaOH (aq) + H₂(g)

(D) KNO₃ (aq) + NaCl (aq) Þ KCl (aq) + NaNO₃ (aq)

[IIT–JEE 1985]

12. Pure ammonia is placed in a vessel at a temperature where its dissociation constant (a) is appreciable. At equilibrium

(A) K_P does not change significantly with pressure

(B) a does not  change with pressure

(C) concentration of NH₃ does not change with pressure

(D) concentration of H₂ is less than that of N₂

[iit–JEE 1984]

13. When NaNO₃ is heated in a closed vessel, O₂ is liberated and NaNO₂ is left behind. At equilibrium.

(A) addition of NaNO₂ favours reverse reaction

(B) addition of NaNO₃ favours forward reaction

(C) increasing temperature favours forward reaction

(D) increasing pressure favours forward reaction                            [IIT–JEE 1986]

 

14.For the reaction, 2NO₂ (g)  2NO(g) + O₂ (g), K_C = 1.8 ´10^–6 at 185°C. At 185°C,
the value of K_C for the reaction:

NO(g) + O₂ (g)  NO₂ (g) is

(A) 0.9 ´ 10⁶                                              (B) 7.5 ´10²

(C) 1.95 ´10^–3                                           (D) 1.95 ´10³

15. The 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

 

12. Answers to Subjective Assignments

 

Level – I

 

1. 39.7 (mol l^–1)^–1/2

 

2. (i) Partial pressure of N₂O₄ = 0.167 atm (ii) Partial pressure of NO₂ = 0.333 atm

 

3. K_p = 5.2 atm

 

4. K_p = 0.3136 atm²

 

5. K_p = 7.31 ´ 10^–4 atm^–2

 

6. Degree of dissociation of HI = 0.245

 

7. 0.16

 

8. (i) 2 = 0.59 (ii) K_P = 0.1687 (iii) 2.645 ´ 10^–5

 

9. Mole fraction of N₂ = 0.788

Mole of fraction of O₂ = 0.212

 

10. K_c = 3.178

 

11. K_p = 6.41 ´ 10^–3 atm

 

12. K_c = 1.747

 

P_HI = 0.776

 

14. (i) K_p = 0.266 atm

(ii) 2  = 63.27%

 

15. (i) K_c = 8.1 ´ 10^–5 mol² litre^–2

K_p = 4.90 ´ 10^–2 atm²

(ii)  Addition of more NH₄HS will cause no effect on this equilibrium because concentration of NH₄HS is not involved in the expression of K_C or K_P

 

 

 

 

 

 

 

Level – II

1. 84.34 atm
2. K_c = 1.413 mol / lit
3. 30 bar
4. 1.04 ´ 10¹⁴
5. 90%

6.

7. 9.65 gm
8. 50.07
9. 25 L
10. 3.71 ´ 10^–6
11. DH = 75.17 ´ 10³ J
12. 0.7837
13. 1.3%
14. [NO] = 1.6 mol / lit, [NO₂] = 0.4 mol / lit

 

 

Level – III

 

1. (i) 0.10 mole (ii) 0.40 mole

 

2. (a) K_c = 178.57 M^–2

K_p = 0.047 atm^–2

(b)  12.546 atm

 

3. a =

 

4. Moles of AB = 1.868

 

5. K_c = 707 on adding water, reverse reaction is favoured

 

6. 0.0311

 

7. 4 atm

 

8. % NH₃ decomposed = 61.2%

 

 

10. (a) K_p = 900 mm²

(b)  =

(c) 72.15

 

11. 2% dissociation of H₂S

 

12. (a) K_c =

(b) Conc.  of R = 1.5 M

 

13. V = 144 ml

 

14. (i) X_CO = 0.7655, = 0.938 atm

(ii) P = 0.68 atm

 

15. K_P = 0.56 (i) 64% (ii) 21.42% (iii) 12.998 ml

 

 

 

 

13. Answers to Objective Assignments

Level – I

 

1. A                                           2.         B
2. C,D                                     4.         C
3. A,C                                     6.         A
4. A                                           8.         D
5. D                                           10.       B
6. B                                           12.       C
7. B                                           14.       C
8. C                                           16.       D
9. A                                                             18.       D
10. D                                           20.       C

Level – II

 

1. C,D                                     2.        B   
2. C                                           4.         B
3. D                                           6.         C  
4. A,B,C,D                               8.         C
5. D                                                                      D
6. D                                           12.       A
7. A,B,C 14.       B
8. B