INTRODUCTION

It is an experimental fact that most of the process including chemical reactions, when carried out in a closed vessel, do not go to completion. They proceed to some extent leaving considerable amounts of reactants & products. When such stage is reached in a reaction, it is said that the reaction has attained the state of equilibrium. Equilibrium represents the state of a process in which the properties like temperature, pressure, concentration etc of the system do not show any change with passage of time. In all processes which attain equilibrium, two opposing processes are involved. Equilibrium is attained when the rates of the two opposing processes become equal. 

If the opposing processes involve only physical changes, the equilibrium is called Physical Equilibrium. If the opposing processes are chemical reactions, the equilibrium is called Chemical Equilibrium.

Some common physical equilibria are:

Solid Liquid

Liquid Gas

Gas Solution

Generally, a chemical equilibrium is represented as

aA + bB xX + yY

Where A, B are reactants and X, Y are products.

Note: 

The double arrow between the left hand part and right hand part shows that changes is taking place in both the directions.

On the basis of extent of reaction, before equilibrium is attained chemical reactions may be classified into three categories.

(a) Those reactions which proceed to almost completion.

(b) Those reactions which proceed to almost only upto little extent.

(c) Those reactions which proceed to such an extent, that the concentrations of reactants and products at equilibrium are comparable.

EQUILIBRIUM IN PHYSICAL PROCESS   

The different types of physical Equilibrium are briefly described below 

(a) Solid – liquid Equilibrium

The equilibrium that exist between ice and water is an example of solid – liquid equilibrium. In a close system, at ice and water attain equilibrium. At that point rate of melting of ice is equal to rate of freezing of water. The equilibrium is represented as 

(b) Liquid-Gas Equilibrium

Evaporation of water in a closed vessel is an example of liquid – gas equilibrium. Where rate of evaporation is equal to rate of condensation. The equilibrium is represented as 

 

(c) Solid – solution equilibrium

If you add more and more salt in water taken in a container of a glass and stirred with a glass rod, after dissolving of some amount. You will find out no further salt is going to the solution and it settles down at the bottom. The solution is now said to be saturated and in a state of equilibrium. At this stage, many molecule of salt from the undissolved salt go into the solution (dissolution) and same amount of dissolved salt are deposited back (Precipitation). 

Thus, at equilibrium rate of dissolution is equal to rate of precipitation. 

(d) Gas –Solution equilibrium

Dissolution of a gas in a liquid under pressure in a closed vessel established a gas – liquid equilibrium. The best example of this type of equilibrium is cold drink bottles. The equilibrium that exists with in the bottle is 

Equilibrium in Chemical Process (Reversible and irreversible reactions)

A reaction in which not only the reactants react to form the products under certain conditions but also the products react to form reactants under the same conditions is called a reversible reaction.

Examples are

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

(ii) CaCO3(s)  CaO(s) + CO2(g)

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

If a reaction cannot take place in the reverse direction, i.e. the products formed do not react to give back the reactants under the same condition it is called an irreversible reaction.

Examples are:

(i) AgNO3(aq) + NaCl(aq) ⎯→ AgCl(s) + NaNO3(g)

(ii) 2M(g) + O2(g) ⎯→ 2MgO(s)

Note: 

If any of the product will be removed from the system, reversible reaction will become irreversible one.

CONCEPT OF CHEMICAL EQUILIBRIUM

Let we have a general reversible reaction,

  A + B C + D

at time t = 0 a0 b0 0 0

as time will pass, A and B will be converted to C and D. As soon as C and D will be formed, reaction will start in back direction also. A time will come when rate of decomposition of A and B will be equal to the rate of formation of A & B, i.e. the rate of forward reaction will be equal to the rate of reverse reaction. At this stage the reaction is said to be in a state of Chemical Equilibrium.

If we see variation of reactants and products with time on graph we will be having following graph.

When the equilibrium is reacted, the concentration of reactant as well as product remains constant. At this stage reaction seems to be stopped but in actual the reaction is going in forward as well as reverse direction with same rate and hence this equilibrium is called as dynamic equilibrium.

Characteristics of Chemical Equilibrium

• The equilibrium is dynamic i.e. the reaction continues in both forward and reverse directions. 
• The rate of forward reaction equals to the rate of reverse reaction.
• The observable properties of the system such as pressure, concentration, density remains invariant with time.
• The chemical equilibrium can be approached from either side.
• A catalyst can hasten the approach of equilibrium but does not alter the state of equilibrium.

 

Types of Equilibria

There are mainly two types of equilibria:

 

(a) Homogeneous: Equilibrium is said to be homogeneous if reactants and products are in same phase.

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

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

N2O4(g) 2NO2(g)

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

 

(b) Heterogeneous: Equilibrium is said to be heterogeneous if reactants and products are in different phases 

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

NH4HS(s) NH3(g) + H2S(g)

NH2CO2NH4(s) 2NH3(g) + CO2(g)    

Note:

To write the equilibrium constant expression, the concentration of pure liquid and pure solid assumed to be unity, as the concentration of such substances remain constant, i.e., 

concentration = mole/litre ∝ density.

LAW OF MASS ACTION

Guldberg and Waage established a relationship between rate of chemical reaction and the concentration of the reactants or, with their partial pressure in the form of law of mass action.

According to this law, “The rate at which a substance reacts is directly proportional to its active mass and rate of a chemical reaction is directly proportional to product of active masses of reactants each raised to a power equal to corresponding stoichiometric coefficient appearing in the balanced chemical equation”.

rate of reaction ∝ [A]a.[B]b

rate of reaction = K[A]a[B]b

where K is rate constant or velocity constant of the reaction at that temperature.

Unit of rate constant (K) (where n is order of reaction.)

Note:

For unit concentration of reactants rate of the reaction is equal to rate constant or specific reaction rate.

At equilibrium the rate of both forward and backward reactions become equal and after achieving equilibria, the concentration of reactants and products remains constant as shown in figure.

 

Note: 

Active mass is the molar concentration of the reacting substances actually participating in the reaction.

Hence 

Active mass =

Active mass of solid is taken as unity.

Illustration 1. Calculate the partial pressure of each component in the following equilibria

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

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

at t = 0 a b 0

at equilibrium a – x b – 3x 2x

ntotal = a – x + b – 3x + 2x = (a + b – 2x)

Partial pressure:

Exercise 1.

Calculate the equilibrium pressure of each reactant of given reaction with the help of following data.

A + B C + D

Initial pressure of A and B are 6 atm and 5 atm. The equilibrium pressure of C and D are 2 atm in each case.

Law of Chemical Equilibrium

According to this law, the ratio of product of concentration of products to the product of concentration of reactants, with each concentration term is raised to the power by its coefficient in overall balanced chemical equation, is a constant quantity at a given temperature and it is called equilibrium constant.

Derivation of law of chemical equilibrium

Let us consider for the following equilibrium

aA + bB cC + dD

then, from Law of mass action

rate of forward reaction r1 ∝ [A]a [B]b

or r1 = Ka [A]a [B]b and rate of reverse reaction r2 ∝ [C]c [D]d

r2 = K2 [C]c [D]d

at equilibrium, r1 = r2

⇒ K1 [A]a[B]b = K2 [C]c[D]d

⇒

Kc = K1/K2, an equilibrium constant in terms of active masses of reacting species.

For the reaction

SO2Cl2   SO2 + Cl2

at t = 0 a       0 0

at equilibrium a − x       x x

equilibrium conc.       (V is volume of container)

So,

So,  

CHARACTERISTICS OF EQUILIBRIUM CONSTANT (Kc)

(i) Kc for a particular reaction at given temperature has a constant value.

(ii) Value of Kc always depends on nature of reactants and the temperature, but independent of presence of catalyst or, of inert material.

(iii) Its value is always independent of the initial concentration of reactants as well as the products.

(iv) The value of Kc indicates the proportion of products/product formed at equilibrium. Large Kc value means large proportions of product.

(v) When the reaction is reversed, equilibrium constant for reverse reaction will also be inversed.

Let us have

A + B C + D

By reversing the reaction,

C + D A + B

(vi) If the coefficients of reactants of products are halved or, doubled then accordingly, value of K′c will change.

A + B C + D

  

for 2A + 2B 2C + 2D

(vii) When a number of equilibrium reactions are added, the equilibrium constant, for overall reaction is the product of equilibrium constants of respective reactions.

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

N2O(g) N2(g) + 1/2O2(g);

N2O(g) + 1/2O2(g) 2NO(g);

i.e.

Equilibrium constant in terms of partial pressures

Let us consider a general reaction

aA + bB cC + dD

PA   PB           PC     PD

From law of mass action

at equilibrium

r1 = r2    or         K1 (PA)a (PB)b = K2(PC)c(PD)d

RELATION BETWEEN KP AND KC

For a general equation,

aA + bB cC + dD

Where, a , b, c and d are coefficients of the reacting substance

From gas equation

PV = nRT       

So, PA = [A]RT’ PB = [B]RT

PC = [C]RT; PD = [D]RT

Hence,

or,

Where Δng = {(c + d) – (a + b)} = change in the numbers of gaseous moles.

Hence,

When Δn = 0; Kp = Kc

Δn > 0; Kp > Kc

Δn < 0; Kp < Kc

Note: 

Similarly we can find equilibrium constant (Kx) interms of mole fraction and can find out its relation with Kp and Kc.

 

Illustration 2. At 27°C Kp value for reaction is 0.1 atm, calculate its Kc value.

 

Solution: KP = Kc(RT)Δn 

Δn = 1

Kc = = 4×10-3

Illustration 3. Solid NH4I dissociates according to the reaction at 400 K 

NH4I(s) NH3(g) + HI(g) ; Kp = 16 atm. In presence of catalyst HI dissociates in H2 and I2 as 2HI H2 + I2. If partial pressure of H2 at this temp is 1 atm in the container when both the equilibrium exist simultaneously, calculate Kp value of second equilibrium (for the dissociation of HI).

 

Solution: NH4I(s)  NH3 (g) + HI (g)  

2HI H2 + I2 

P – x    

Here = 1 atm

∴  x = 2 atm 

P(P -2) = 16

∴ P = 5.1

 

Illustration 4. 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. What is the value of Kp if the total pressure at equilibrium is 0.8 atm?

 

Solution: CO2(g) + C(s) ⎯⎯→ 2CO(g)

  0.5   2n

    or, 0.5-n           0.5+n = 0.8

n = 0.3

Kp =

Illustration 5. Given that the equilibrium constant for the reaction, H2(g) + I2(g) 2HI(g) is 50 at 700K. Calculate the equilibrium constant for the reaction.

HI(g)

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

Kc =

The new equilibria is

HI(g)

Illustration 6. One mole of N2 is mixed with 3 moles of H2 in a 4 litre container. If 25% of N2 is converted into NH3 by the following reaction

N2(g) + 3H2(g) 2NH3(g). Calculate Kc and Kp of the reaction.
(Temperature = 227°C and R = 0.08231).

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

Percentage N2 reacted, 25%

∴ x = 0.25

Now, (a – x) = 1 – 0.25 = 0.75

b – 3x = 3 – 0.75 = 2.25

∴

= 1.48 × 10–5 L2 mol–2

Now, Kp =

 

= 8.78 × 10–9

 

Exercise 2.

The equilibrium constant for the reaction

PCl5(g) PCl3(g) + Cl2(g) is 2.4 × 10–3. Determine the equilibrium constant for the reverse reaction at same temperature.

 

Exercise 3.

At 21.5°C and a total pressure of 0.0787 atm, N2O4 is 48.3% dissociated into NO2. Calculate Kp and Kc for the reaction.

N2O4(g) 2NO2(g)

 

Exercise 4.

(i) Write the expression for Kp and Kc for the following reactions.

N2 + 3H2    2NH3

(ii) Find out the heterogeneous equilibrium from the following reaction

(A) CaCO3 (s)  CaO(s) + CO2 (g)

(B) H2(g)    + I2(g)     2HI (g)

Exercise 5.

What is the value of equilibrium constant for the following reaction

If  A    D

A   B     K1 = 2

 B   C     K2 = 5

 C   D  K3 = 3   

 

Exercise 6.

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

 

SIGNIFICANCE OF THE MAGNITUDE OF EQUILIBRIUM CONSTANT

(i) A very large value of KC or KP signifies that the forward reaction goes to completion or very nearly so.

(ii) A very small value of KC or KP signifies that the forward reaction does not occur to any significant extent.

(iii) A reaction is most likely to reach a state of equilibrium in which both reactants and products are present if the numerical value of Kc or KP is neither very large nor very small.

Units of Equilibrium Constant

As we have learned that numerical value of equilibrium constant is a function of stoichiometric coefficient used for any balanced chemical equation.

But, what will happen in the size of units of equilibrium constant when the stoichiometric coefficient of the reaction changes? (If sum of the exponent for product is not equal to reactant side sum)

So for the sake of simplicity in the units of Kp for Kc the relative molarity or pressure of reactants and products are used with respect to standard condition. (For solution standard state 1 mole/litre and for gas standard pressure = 1 atm)

Now the resulting equilibrium constant becomes unitless by using relative molarity and pressure. 

 

Illustration 7. The value of Kp for the reaction 2H2O(g) + 2Cl2(g) 4HCl(g) + O2(g) is
0.035 atm at 400oC, when the partial pressures are expressed in atmosphere. Calculate Kc for the reaction,

Cl2(g) + H2O(g)

Solution: KP = KC (RT)Δn

Δn = moles of product − moles of reactants = 5 − 4 = 1

R = 0.082 L atm/mol K, 

T = 400 + 273 = 673 K

∴ 0.035 = KC (0.082 × 673)

KC = 6.342 × 10-4 mol l-1

∴ for the reverse reaction would be

∴ = = 1576.8 (mol l-1)-1

When a reaction is multiplied by any number n (integer or a fraction) the or becomes (KC)n or (KP)n of the original reaction.

∴ KC for O2(g) + 2HCl(g) Cl2(g) + H2O(g)

is 39.7 (mol.l-1)-½

 

Illustration 8. Kp for the equilibrium, FeO(s) + CO(g) Fe(s) + CO2(g) at 1000oC is 0.4. If CO(g) at a pressure of 1 atm and excess FeO(s) are placed in a container at 1000oC, what are the pressures of CO(g) and CO2(g) when equilibrium is attained?

Solution: Assuming ideal gas behaviour, partial pressures are proportional to the no. of moles present. Since moles of CO2 formed equals moles of CO consumed, the drop in partial pressure of CO will equal the partial pressure of CO2 produced. Let the partial pressure of CO2 at equilibrium be ‘x’ atm. Then, partial pressure of CO will be (1 – x) atm.

Since   ⇒   x = 0.286

Hence PCO = 1-x 

            = 0.714 atm.

 

Illustration 9. At 800 K a reaction mixture contained 0.5 mole of SO2, 0.12 mole of O2 and 5 mole of SO3 at equilibrium. Kc for the equilibrium 2SO2 + O2 2SO3 is 833 lit/ mole. If the volume of the container is 1 litre, calculate how much O2 is to be added at this equilibrium in order to get 5.2 moles of SO3 at the same temperature.

Solution: Suppose x mole of O2 is added by which equilibrium shifts to right hand side and y mole of O2 changes to SO3. 

The new equilibrium concentration may be 

2 SO2 + O2 2SO3

Moles at I equilibrium 0.5 0.12 5

      II equilibrium   (0.5-2y) (0.12+x-y) (5+2y) 

5 + 2y = 5.2

∴ y = 0.1

Kc =

    = 833

Substituting y = 0.1 (V = 1 lit) 

We get x = 0.34 mole

Exercise 7.

Ammonium hydrogen sulphide dissociates as follows 

NH4HS(s) NH3(g) + H2S(g)

If solid NH4HS is placed in an evacuated flask at certain temperature it will dissociate until the total pressure is 600 torr. 

(a) Calculate the value of equilibrium constant for the dissociation reaction 

(b) Additional NH3 is introduced into the equilibrium mixture without changing the temperature until partial pressure of NH3 is 750 torr, what is the partial pressure of H2S under these conditions? What is the total pressure in the flask?

 

Exercise 8.

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.

 

THE REACTION QUOTIENT ‘Q’  

Consider the equilibrium

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

At equilibrium   =  KC. When the reaction is not at equilibrium this ratio  is called ‘QC’ i.e., QC is the general term used for the above given ratio at any instant of time. And at equilibrium QC becomes KC. 

Similarly, is called QP and at equilibrium it becomes KP.

• If the reaction is at equilibrium, Q = Kc
• A net reaction proceeds from left to right (forward direction) if Q < KC.
• A net reaction proceeds from right to left (the reverse direction) if Q >Kc

Illustration 10. For the reaction,

A(g) + B(g) 2C(g) at 25°C, in a 2 litre vessel contains 1, 2, 3 moles of respectively. Predict the direction of the reaction if

(a) Kc for the reaction is 3

(b) Kc for the reaction is 6

(c) Kc for the reaction is 4.5

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

Reaction quotient Q =

(a) therefore backward reaction will be followed 

(b)

The forward reaction is followed

(c) Q = Kc

∴ The reaction is at equilibrium

Illustration 11. For the reaction : A(aq)+ B(aq) C(aq) +D(aq) , the net rate of consumption of B at 25°C and at any time ‘t’ is as given below 

= {4×10-4[A] [B] – 1.33×10-5 [C] [D]} mol L-1 min-1 

Predict whether the reaction will be spontaneous in the direction as written in reaction mixture in which each A, B, C and D is having a concentration of
1 mol L-1? 

 

Solution: K =

 

Q =

    = 1 < K

Since Q < K, so the above reaction is spontaneous in the forward direction.

 

Exercise 9.

The equilibrium constant is 0.403 at 1000K for the process 

FeO(s) + CO(g) Fe(s)+ CO2)g)

A steam of pure CO is passed over powdered FeO at 1000K until equilibrium is reached. What is the mole fraction of CO in the gas stream leaving the reaction zone?

 

LE CHATELIER’S PRINCIPLE

“When an equilibrium is subjected to either a change in concentration, temperature or, in external pressure, the equilibrium will shift in that direction where the effects caused by these changes are nullified”.

This can be understood by the following example. Overall we can also predict the direction of equilibrium by keeping in mind following theoretical assumption.

PCl5 PCl3 + Cl2

Let us assume that we have this reaction at equilibrium and the moles of Cl2, PCl3 and PCl5 at equilibrium are a, b and c respectively, and the total pressure be PT.

∴ =

Since PT =

∴ KP =

Now if d moles of PCl3 is added to the system, the value of Q would be,

We can see that this is more than KP. So the system would move reverse to attain equilibrium. 

(i) If we increase the volume of the system, the Q becomes where V′ > V.  

∴ Q becomes less, and the system would move forward to attain equilibrium.

(ii) If we add a noble gas at constant pressure, it amounts to increasing the volume of the system and therefore the reaction moves forward.

(iii) If we add the noble gas at constant volume, the expression of Q remains as Q = and the system continues to be in equilibrium.  

∴ nothing happens.

(iv) Therefore for using Le-Chatlier’s principle, convert the expression of KP and KC into basic terms and then see the effect of various changes.

 

Effect of Concentration

Let us have a general reaction,

aA + bB cC + dD

at a given temperature, the equilibrium constant,

again if α, β, γ and δ are the number of mole of A, B, C and D are at equilibrium

then,

If any of product will be added, to keep the Kc constant, concentration of reactants will increase i.e. the reaction will move in reverse direction. Similarly if any change or disturbance in reactant side will be done, change in product’s concentration will take place to minimise the effect.

Temperature Effect

The effect of change in temperature on an equilibrium cannot be immediately seen because on changing temperature the equilibrium constant itself changes. So first we must find out as to how the equilibrium constant changes with temperature.

For the forward reaction, according to the Arrhenius equation,

And for the reverse reaction, ∴ ∴  It can be seen that or For any reaction, ΔH = Eaf − Ear =or =  From the equation, it is clear that

R = Reactant, P = Product

(a)  If ΔHo is +ve (endothermic), an increase in temperature (T2 > T1) will make K2 > K1, i.e., the reaction goes more towards the forward direction and vice-versa.

(b) If ΔHo is -ve (exothermic), an increase in temperature (T2 > T1), will make K2 < K1 i.e., the reaction goes in the reverse direction.

(i) Increase in temperature will shift the reaction towards left in case of exothermic reactions and right in endothermic reactions.

(ii) Increase of pressure (decrease in volume) will shift the reaction to the side having fewer moles of the gas; while decreases of pressure (increase in volume) will shift the reaction to the side having more moles of the gas.

(iii) If no gases are involved in the reaction higher pressure favours the reaction to shift towards higher density solid or liquid.

 

Illustration 12. Under what conditions will the following reactions go in the forward direction?

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

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

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

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

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

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

(vii) N2O4(g) 2NO2(g) – 14 k cal.

 

Solution: (i)  Low T, High P, excess of N2 and H2.

(ii)  Low T, High P, excess of SO2 and O2.

(iii) High T, any P, excess of N2 and O2

(iv)Low T, High P, excess of NO and O2
(v) Low T, Low P, excess of C and H2O

(vi) High T, Low P, excess of PCl5
(vii) High T, Low P, excess of N2O4.

 

Illustration 13. The equilibrium constant KP for the reaction N2(g) + 3H2(g) 2NH3(g) is 1.6 × 10-4 atm-2 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: Equilibrium constants at different temperature and heat of the reaction are related by the equation,   =
2.303 log = 

= + log 1.64 × 10-4
log
∴

Effect of change of Pressure

The effect of change of pressure on chemical equilibrium can be done by the formula.

Case A: When Δn = 0

Kp = Qp

And equilibria is independent of pressure.

Case B: When Δn = – ve

Here with increase in external pressure, will shift the equilibrium towards forward direction to maintain Kp. Similarly, decrease in pressure will shift the equilibrium in the reverse direction.

Case C: When Δn = +ve

and, effect will be opposite to that of Case ‘B’

 

Exercise 10.

N2 + 3H2   2NH3;           Δ H = negative

What are the conditions of temperature and pressure favorable for this reaction?

 

EFFECT OF CATALYST ON EQUILIBRIUM

Since the catalyst is associated with forward as well as reverse direction reaction. So, at equilibrium, rate of forward reaction will be equal to rate of reverse reaction and hence catalyst effect will be same on both forward as well as reverse. Hence catalyst never effect the point of equilibrium but it reduces the time to attain the equilibrium.

Effect on equilibrium due to the addition of inert gases

Following are the cases, where this effect in different manner.

Case A: When Δn = 0 and total volume change at equilibrium remain constant.

So, addition of inert gases will increase the number of moles in the mixture but partial pressure of each component remains constant. Hence equilibrium remains unaffected.

Case B: When Δn = +ve and ΔVT ≠ 0

The total volume will increase with addition of inert gases. This will shift the equilibrium in forward direction.

Case C: When Δn = –ve and ΔVT ≠ 0

In this case, the addition of inert gases will increase the total volume and the reaction will shift in reverse direction.

 

Dependence of KP or Kc on Temperature

 

With the increase of temperature, equilibrium favours forward reaction in case of an endothermic (ΔH > 0) reaction while it favours backward reaction in the case of an exothermic (ΔH < 0) reaction.

Illustration 14. For the equilibrium

NH4I(g) NH3(g) + HI(g)     

What will be the effect on the equilibrium constant on increasing the temperature.

Solution: Since the forward reaction is endothermic, so increasing the temperature, the forward reaction is favoured. Thereby the equilibrium constant will increase.

Illustration 15. Kp for the reaction 2BaO2(s) 2BaO(s) + O2(g) is 1.6 × 10–4 atm, at 400°C. Heat of reaction is – 25.14 kcal. What will be the no. of moles of O2 gas produced at 500°C temperature, if it is carried in 2 litre reaction vessel?

Solution: We know that

Since, PV = nRT

= 4.60 × 10−7

Exercise 11.

Equilibrium constant Kp for the reaction H2(g) + N2(g) NH3(g) are 0.0266 and 0.0129 at 310°C and 400°C respectively. Calculate the heat of formation of gaseous ammonia.

Standard Free Energy Change of a Reaction and its Equilibrium Constant

Let ΔG0 be the difference in free energy of the reaction when all the reactants and products are in the standard state and Kc or, Kp be the thermodynamic equilibrium constant of the reaction. Both related to each other at temperature T by the following relation.

ΔG0 = – 2.303 RT logKc  and ΔG0 = – 2.303 RT log Kp (in case of ideal gases)

Now, we know that thermodynamically,

ΔG0 = ΔH0 – TΔS0

here ΔH0 is standard enthalpy of reaction, and ΔS0 is standard entropy change

(i) When ΔG0 = 0, then, Kc = 1

(ii) When, ΔG0 > 0, i.e. +ve, then Kc < 1, in this case reverse reaction is feasible showing thereby a less concentration of products at equilibrium rate.

(iii) When ΔG0 < 0, i.e. −ve, then, Kc > 1; In this case, forward reaction is feasible showing thereby a large concentrations of product at equilibrium state.

Illustration 16. A system is in equilibrium as

PCl5 + Heat PCl3 + Cl2

Why does the temperature of the system decreases, when PCl3 are being removed from the equilibrium mixture at constant volume?

Solution: When PCl3 are being removed from the system, the reaction moved to the right. This consumed heat and therefore, temperature is decreased.

 

Illustration 17. NO and Br2 at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at 300K. At equilibrium the total pressure was 110.5 torr. 

Calculate the value of the equilibrium constant and the standard free energy change at 300K for the reaction 2NO(g) + Br2(g) 2NOBr(g).

 

Solution: 2NO(g) + Br2(g) 2NOBr(g)

Initial pressure  98.4          41.3 0

At eqlbm. 98.4-x    41.3 – x

The total pressure at equilibrium is 110.5 torr.

∴ 98.4-x + 41.3-+ x = 110.5

x = 58.4 torr

Now,  1 atm = 760.4 torr,     

∴ x = 7.68 × 10-2 atm.

pNOBr = 7.68 × 10-2 atm  ;     pNO = 98.4-x = 40 torr = 5.26 × 10-2 atm.

pBr2 = 41.3 – = 12.1 torr = 1.59 × 10-2 atm.

= atm-1

ΔGo = -2.303 RT log K = -2.303 (1.99) (300) (log 134)  

= −2.92 k cal = −12.2 kJ.

[If R is used as 1.99 cal/mol K, then ΔGo will be in cal. If R is used as 8.314 J/mol K, then ΔGo will be in Joules. But KP must always be in (atm)Δn.]

 

Exercise 12.

What is the value of R if value of ΔGO is in?

(i) Joules (ii) Calories

 

Exercise 13.

N2O4 is 25% dissociated at 27°C and 1 atm pressure. 

Calculate (i)  Kp and (ii) the percentage dissociation at 0.1 atm and 27°C.

 

Dependence of Degree of Dissociation on Density Measurements

The degree of dissociation of a substance is defined as the fraction of its molecules dissociating at a given time. 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 KP 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 equilibrium V is a constant and ρ 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−α) Cα     Cα

∴ = 

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

 

Illustration 18. When PCl5 is heated it dissociates into PCl3 and Cl2. The density of the gas mixture at 200oC and at 250oC is 70.2 and 57.9 respectively. Find the degree of dissociation at 200oC and 250oC.

 

Solution: PCl5(g) PCl3(g) + Cl2(g)
We are given the vapour densities at equilibrium at 200oC and 250oC.
The initial vapour density will be the same at both the temperatures as it would be
∴  Initial vapour density =

Vapour density at equilibrium at 200oC = 70.2
∴

=

∴ α = 0.485
At 250oC, 1 + α =

∴ α = 0.8

Illustration 19. The degree of dissociation of PCl5 is 60%, then find out the observed molar mass of the mixture.

 

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

Initially moles 1 0         0

At eqm. 1 – α               α         α

Where α = degree of dissociation = 0.6

Total moles at equilibrium = 1 + α = observed mole

= 129.06

 

Exercise 14.

Vapour density of N2O4 which dissociates according to the equilibrium
N2O4(g) 2NO2(g) is 25.67 at 100°C and a pressure of 1 atm. Calculate the degree of dissociation and Kp for the reaction.

 

Exercise 15.

How much PCl5 must be added to a one litre vessel at 250oC in order to obtain a concentration of 0.1 mole of Cl2? Kc for is 0.0414 mol/litre.

 

ANSWERS TO EXERCISES

Exercise 1:

A 4 atm 

B 3 atm 

 

Exercise 2:

 

Exercise 3:

 

Exercise 4: (i) Kc = 

Kp =

(ii) (A) is a heterogeneous equilibrium.

 

Exercise 5:

K = 30

 

Exercise 6:

1.8

 

Exercise 7:

(a) 0.155 atm2

(b) 119.32 torr, 869.32 torr 

 

Exercise 8:

0.4, 0, 0.4

 

Exercise 9:

0.713

 

Exercise 10:

Low temperature, high pressure and high concentration of N2 and H2.

 

Exercise 11.

 

Exercise 12:

(i) 8.314 J mole-1K-1

(ii) 2 cal mole-1K-1

 

Exercise 13:

(i) 0.266 atm

(ii) 93.27%

 

Exercise 14:

 

Exercise 15: 0.3415 mole

MISCELLANEOUS EXERCISES

 

Exercise 1: For an equilibrium, the rate constant for the forward and the backward reaction are 2.38 × 10−4 and 8.15 × 10−5 respectively. Calculate the equilibrium constant for the reaction. 

 

Exercise 2: In an experiment starting with 1 mole C2H5OH, 1 mole CH3COOH and 1 mole of water, the equilibrium mixture on analysis shows that 54.3% of the acid is esterified. Calculated Kc. 

 

Exercise 3: From the values of the equilibrium constants, indicate in which case, does the reaction go farthest to completion: K1 = 10−10, K2 = 1010, K3 = 105. 

 

Exercise 4: The value of the equilibrium constant is less than zero. What does it indicate? 

 

Exercise 5: Would you expect equilibrium constant for the reaction to increase or decrease as temperature increases? Assign reason. 

 

Exercise 6: Acetic acid is highly soluble in water but still a weak electrolyte. Why? 

 

Exercise 7: Which of the following reactions involve homogenous and which involve heterogenous equilibria? 

(a)

(b)

(c)

(d)

(e)

Exercise 8: For the reaction, at 1173 K, the magnitude of Kc is 3.6. For each of the following composition, decide whether the reaction mixture is at equilibrium. If not, decide to which direction the reaction should go? 

(a)

(b)

 

Exercise 9: Kp for the reaction at 460°C is 49. If the partial pressure of H2 and I2 is 0.5 atm respectively, determine the partial pressure of each gas in the equilibrium mixture. 

 

Exercise 10: 0.16 g N2H4 (hydrazine) are dissolved in water and the total volume is made upto 500 mL. Calculate the percentage of N2H4 that has reacted with water in solution. The Kb for N2H4 is 4.0 × 10−6 M. 

 

ANSWERS TO MISCELLANEOUS EXERCISES

 

Exercise 1: 2.92 

 

Exercise 2: Kc = 4.012 

 

Exercise 3: High value of equilibrium constant (K) means that the molar concentration of the products is quite large. In other words, the reaction has proceeded to a large extent in the forward direction before attaining the equilibrium. Thus for K2 = 1010, the forward reaction has proceeded maximum before attaining the equilibrium. 

 

Exercise 4: This indicates that the forward reaction has proceeded only to a small extent before the equilibrium is attained. 

 

Exercise 5: In the forward reaction, energy is needed to bring about the dissociation of I2(g) molecules. This means that the forward reaction is endothermic in nature. The increase in temperature will favour the forward reaction. Therefore, the equilibrium constant will increase with rise in temperature. 

 

Exercise 6: The strength of the electrolyte does not depend upon its amount present in solution but on its ionization in solution. Since acetic acid is a weak acid (organic acid), it is ionized only to small extent. Therefore, it is a weak electrolyte. 

 

Exercise 7: Homogenous equilibria : (a), (b) (d) 

Heterogenous equilibria : (c) and (e)

 

Exercise 8: (a) Backward direction 

(b) Backward direction 

 

Exercise 9: Partial pressure of H2 (g) = 0.112 atm 

Partial pressure of I2(g) = 0.112 atm 

Partial pressure of HI (g) = 0.776 atm

 

Exercise 10: 2.0%

 

SOLVED PROBLEMS

 

Subjective:

 

Board Type Questions

 

Prob 1. At 444°C, 15g mole of hydrogen are mixed with 5.2g mole of I2 vapour. When equilibrium was established, 10g mole of HI was formed. Calculate the equilibrium constant for the reaction.

 

Sol.

a          b                0

a – x    b – x          2x

So,

= = 50

 

Prob 2. The degree of dissociation is 0.4 at 400K and 1 atm for the gaseous reaction

Assuming ideal behaviour of all the gases, calculate the density of equilibrium mixture at 400K and 1 atm pressure.

(Atomic mass of P = 31 and Cl = 35.5)

 

Sol.

at t = 0 1       0         0

at eqm. 1 – 0.4       0.4      0.4

Total no. of moles after dissociation = 0.6 + 0.4 + 0.4

Now,

So, density =

 

Prob 3. When 3.06g of solid NH4HS is introduced into a 2 litre evacuated flask at 27°C, 30% of the solid decomposed into gaseous ammonia and hydrogen sulphide.

(i) Calculate Kc for reaction at 27°C.

(ii) What would happen to the equilibrium when more solid NH4HS is introduced into the flask?

 

Sol. Moles of NH4HS introduced into flask =

0.06 (1 – x)        0.06x           0.06x

as x = 30%, so, x = 0.3

So, 0.06 × 0.7    0.06×0.3 0    0.6 × 0.3

So,

Addition of NH4HS(s) does not change position of equilibrium.

Prob 4. Determine Kc for the reaction ½N2(g) + ½O2(g) + ½Br2(g) NOBr(g) from the
following information (at 298oK)

Kc = 2.4 × 1030  for  2NO(g) N2(g) + O2(g) ;  

= 1.4 for NO(g) + ½Br2(g)  NOBr(g)

 

Sol. N2(g) + O2(g)    2NO (g)

Kc/  =

(i) ½N2(g) + ½O2(g )  NO(g)

Kc// =

(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

 

Prob 5. The heat of reaction at constant volume for an endothermic reaction in equilibrium is 1200 cal more than at constant pressure at 300K. Calculate the ratio of equilibrium constant Kp and Kc.

 

Sol. Given that

Also we have

∴

or

Now,

 

IIT Level Question 

 

Prob 6. Anhydrous CaCl2 is often used as dessicant. In presence of excess of CaCl2, the amount of water taken up is governed by Kp = 1.28 × 1085 for the following reaction at room temperature, CaCl2(s) + 6H2O(g) CaCl2.6H2O(s)

What is the equilibrium pressure of water in a closed vessel that contains CaCl2(s)?

 

Sol.

Here,

Prob 7. NO2(g) NO(g) + 1/2O2(g)

Consider the above equilibrium where one mole of NO2 is placed in a vessel and allowed to come to equilibrium at a total pressure of 1 atm. Experimental analysis shows that 

Temperature	700K	800K
	0.872	2.50

(a) Calculate Kp at 700K and 800K

(b) Calculate ΔG0

 

Sol. (a) NO2(g) NO(g) +   O2(g)

at t = 0 1 mole 0       0

t eq. 1 – 2x 2x       x

partial  

at equilibrium pressure.

Given that

at 700K

⇒ x = 0.2329

∴

=   = 0.372 atm1/2

Similarly can be calculated

(b)

= 5.65kJ/mol

 

Prob 8. 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 80g of N2O4 confined to a 7 litre vessel.

(iii) What volume of above mixture will diffuse if 20 ml of pure O2 diffuses in 10 minutes at same temperature and pressure?

 

Sol. N2O4(g) 2NO2(g)

t equilibrium 1 – α 2α

Partial pressure at equilibrium

= = 0.56 atm

(i)

Solving, α = 0.64

% dissociation = 64%

(ii) Since

Here Δn = 1

= 0.0228 mole/litre

where x = no. of moles of N2O4(g) = 80/92

Neglecting α with respect to 1

i.e. 1 – α ≈ 1

0.0228 =

(iii)

Let V(ml) volume of mixture diffused in Now, from Graham’s law of diffusion.

⇒  V = 12.998 ml

 

Prob 9. 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 remains 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

 

Sol. P = = 0.3618 atm

= 0.181 atm

Kp = (0.181)2 = 0.033 atm2

NH4I(s) NH3(g) +   HI(g)

          (0.181 + p)       (0.181  + p – p′) Kp = 3.3 × 10–2 atm2

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

(0.181 + p – p′) Kp = 0.015

0.033 = (0.018 – p) (0.181 + p – p′)

0.015 = ⇒ p′ = 0.04 atm

p = 0.026 atm

∴ total pressure = 2(0.181 + p) = 0.414 = 314.64 mm Hg

 

Prob 10. For the gaseous reaction: C2H2 + D2O C2D2 + H2O, ΔH is 530 cal. At 25°C,
KP = 0.82. Calculate how much C2D2 will be formed if 1 mole of C2H2 and 2 moles of D2O are put together at a total pressure of 1 atm at 100°C?

 

Sol. Firstly calculate KP at 100°C by using equation 

log 

On solving we get, (KP)100°C = 0.98 

C2H2 + D2O C2D2 + H2O

Initial mole 1 2 0 0

At eq. mole 1–α 2–α α α

∴ 0.98 =

0.02α2 + 2.94α – 1.96 = 0

On solving we get,  α = 0.75

 

Prob 11. An equilibrium mixture at 300K contains N2O4 and NO2 at 0.28 and 1.1 atm respectively. If the volume of container is doubled, calculate the new equilibrium pressure of two gases.

 

Sol.

0.28                 1.1

When volume is doubled pressure will reduced to half

     

   

Prob 12. At 540K, 0.10 mole of PCl5 are heated in a 8 litre flask. The pressure of the equilibrium mixture is found to be 1 atm. Calculate the Kp for the reaction.

Sol.

0.1 – α       α         α

Total no. of moles =

Now,

So,

atm

 

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

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

 

Sol. Let the initial moles of N2 and H2 be 1 and 3 respectively (this assumption is valid as KP 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 NH3 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

∴ KP =

          = 7.31 × 10-4 atm–2

 

Prob 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).

 

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

    N2O4(g)    2NO2(g)

Initial           C             0

At equilibrium C(1−α)           2Cα

Where α 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 + α =

∴  α = 0.752

∴  Kp =

  = 5.2 atm

 

Prob 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, Kp and the total pressure P.

 

Sol. 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, KP =

 

Objective:

 

Prob 1. Reaction between iron and steam is reversible if it is carried out

(A) at constant T (B) at constant P

(C) in an open vessel (D) in a closed vessel

 

Sol. In open vessel H2 gas will escape.

∴ (D)

 

Prob 2. For the reaction

The value of Kc at 250°C is 26. The value of Kp at this temperature will be

(A) 0.61 (B) 0.57

(C) 0.83 (D) 0.46

 

Sol.

∴ (A)

 

Prob 3. 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) becomes (original)/4

(C) doubled

(D) constant

 

Sol. Kcor Kp do not depend on concentration, but only on temperature.

∴ (D)

 

Prob 4. The equilibrium constant for the reaction,

In presence of a catalyst, equilibrium is attained ten times faster. Therefore, the equilibrium constant, in presence of the catalyst at 2000K is

(A) (B)

(C) (D) difficulty to compute

 

Sol. Equilibrium is constant at constant temperature for a reaction

∴ (B)

 

Prob 5. 64g of HI are present in a 2 litre vessel. The active mass of HI is:

(A) 0.5 (B) 0.25

(C) 1 (D) none

 

Sol. Molecular mass of HI = 128

HI] =

∴ (B)

 

Prob 6. For which of the following Kp may be equal to 0.5 atm 

(A) 2HI H2 + I2 (B) PCl5 PCl3 + Cl2

(C) N2 + 3H2 2NH3 (D) 2NO2 N2O4

Sol. For Kp = 0.5 atm

Δn = 1 (since the unit is atm)

and PCl5 PCl3 + Cl2

Δn = 1

∴ (B)

 

Prob 7. The vapour density of undecomposed N2O4 is 46. When heated, vapour density decreases to 24.5 due to its dissociation to NO2. The % dissociation of N2O4 at the final temperature is

(A) 80 (B) 60

(C) 40 (D) 70

 

Sol. N2O4 2NO2

1             0 at initial

1 – α           2α at equilibrium

∴

1.8 = 1 + α ⇒ α = 0.8   or 80%

∴ (A)

 

Prob 8. If pressure is applied to the following equilibrium, liquid vapours the boiling point of liquid

(A) will increase (B) will decrease

(C) may increase or decrease (D) will not change

 

Sol. Boiling point of a liquid is the temperature at which vapour pressure became equal to atm pressure. If the pressure is applied to the above equilibrium the reaction will go to the backward direction, i.e. vapour pressure decrease hence the boiling point increase.

∴ (A)

 

Prob 9. For the reaction

A(g) + B(g) 3C(g) at 250°C, a 3 litre vessel contains 1, 2, 4 mole of A, B and C respectively. If KC for the reaction is 10, the reaction will proceed in 

(A) Forward direction (B) Backward direction

(C) In either direction (D) In equilibrium

 

Sol. = 10.66

Q [C] =

[A] =   ⇒  [B] = Q KC = 10, and Q > KC

∴ reaction will proceed in backward direction

∴ (B)

 

Prob 10. If CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(l) Kp = 1.086 × 10–4 atm2 at 25°C. The efflorescent  nature of CuSO4.5H2O can be noticed when vapour pressure of H2O in atmosphere is

(A) > 7.29 mm (B) < 7.92 mm

(C) ≥ 7.92 mm (D) None

Sol. An efflorescent salt is one that loses H2O to atmosphere.

For the reaction

CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(l)

Kp = ()2 = 1.086 × 10–4

= 1.042 × 10–2 atm = 7.92 mm

Q If at 25°C < 7.92 mm only then, reaction will proceed in forward direction.

∴ (B)

 

Prob 11. The enthalpies of two reaction are ΔH1 and ΔH2 (both positive) with
ΔH2 > ΔH1. If the temperature of reacting system is increased from T1 to T2, predict which of the following alternatives is correct? 

(A) (B)

(C) (D) None

 

Sol. As the temperature of reacting system is increased the equilibrium constant of reaction is also increased for endothermic reactions so for two reactions on increasing the temperature by equal amounts 

Hence, (C) is correct.

 

Prob 12. At a certain temperature 2 moles of carbonmonoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction CO + Cl2 COCl2 in a 5 lit vessel. At equilibrium if one mole of CO is present then equilibrium constant for the reaction is:

(A) 2 (B) 2.5

(C) 3.0 (D) 4

 

Sol. CO + Cl2    COCl2

      At t = 0 2   3 0

At equilibrium (2–1) (3–1)       1

Concentrations (V = 5 lit)      

Kc = = = or Kc = 2.5

Hence, (B) is correct.

 

Prob 13. The reaction: 3O2 2O3, ΔH = + 69,000 calories is favoured by:

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

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

 

Sol. According to Le Chatalier principle formation of ozone is favoured by high temperature (endothermic reaction) and high pressure.

Hence, (B) is correct.

Prob 14. The equilibrium constant at 323°C is 1000. What would be its value in the presence of a catalyst in the forward reaction?

A + B C + D + 38 Kcal

(A) 1000 × concentration of catalyst (B) 1000

(C) (D) impossible to tell

 

Sol. The equilibrium constant varies only with temperature. At constant temperature it will not vary.

Hence, (B) is correct.

 

Prob 15. Which of the following reactions represent a heterogenous equilibrium? 

(A)

(B)

(C)

(D)

 

Sol. In heterogenous equilibrium, physical state of all the reactants and products are not same. 

Hence, (B) is correct. 

 

True/False

 

Prob 16. For a reversible system at constant temperature the value of Kc increase if the concentration are changed at equilibrium.

 

Sol. False

 

Prob 17. The apparent molecular weight if PCl5 on dissociation shows lower value.

 

Sol. True

 

Prob 18. The value of Kc for the reaction will increase if the volume of reaction vessel is increased.

 

Sol. False

 

Fill in the Blanks

 

Prob 19. Free energy change at equilibrium is …………..

 

Sol. Zero

 

Prob 20. The equilibrium constant is ………. of velocity constant of forward and backward reactions.

Sol. Ratio

ASSIGNMENT PROBLEMS

 

Subjective:

 

Level – O

 

1. What happens to rate of forward and rate of backward reaction when equilibrium is attained?

 

2. What happens to an equilibrium in a reversible reaction, if a catalyst is added to it?

 

3. The equilibrium constant of a reaction at 27°C and 127°C are 1.6 × 10–3 and 7.6 × 10–2 respectively. Is the reaction exothermic or, endothermic?

 

4. Ice melts slowly at higher altitudes. Explain why?

 

5. Predict the effect of compression on the following equilibrium reaction.

 

6. List any four important characteristics of a chemical equilibrium

 

7. The equilibrium constant for the reaction

N2 + 3H2  2NH3 is K. What will be the equilibrium constant for the reaction?

 

8. What are reversible and irreversible reactions? Explain with the help of some examples?

 

9. Why gas fizzes out when soda water bottle is opened?

 

10. What will happen when in the following equilibrium:

H2 + I2 2HI + Q

Temperature and pressure be increased?

 

11. State and explain Le−Chatelier’s principle. What effect will be there by addition of inert gas in an equilibrium mixture?

 

12. For the reaction

CaCO3(s) CaO(s) + CO2(g), Kp = 1.16 atm at 800°C. If 20g of CaCO3 were kept in a 10 litre container and heated upto 800°C; what percentage by weight of CaCO3 would remain unreacted at equilibrium?

 

13. For the reaction

Do not use the formula. Start finding Kc and then final Kp.

 

14. A cylinder fitted with an air tight piston contains a small amount of a liquid at a fixed temperature. The piston is moved out so that the volume increases.

(a) What will be the effect on the change of vapour pressure initially?

(b) How will the rates of evaporation and condensation change initially?

(c) What will happen when equilibrium is restored finally and what will be the final vapour pressure?

15. What happens to the equilibrium?

if nitrogen gas is added to it (i) at constant volume (ii) at constant pressure? Give reasons.

Level – I

 

1. An air sample containing 21:79 of O2 and N2 (molar ratio) is heated to 2400°C. If the mole percent of NO at equilibrium is 1.8%. Calculate the Kp of the reaction
   N2 + O2 2NO

 

2. For the reaction F2 2F. Calculate the degree of dissociation of fluorine at 4 atm and 1000K, when Kp = 1.4 × 10–2 atm.

 

3. The equilibrium constant of ester formation of propionic acid with ethyl alcohol is 7.36 at 50°C. Calculate the weight of ethyl propionate in gram existing in an equilibrium mixture when 0.5 mole of propionic acid is heated with 0.5 mole of ethyl alcohol at 50°C.

 

4. 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

 

5. 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 KC for the reaction.

 

6. 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.

 

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

 

8. Kp for the reaction PCl5 PCl3 + Cl2 at 250°C is 0.82. Calculate 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?

 

9. (a) Calculate the equilibrium constant for the process

N2O4 2NO2(g)

At a temperature 350 K, given that K = 0.14 at 298 K and ΔH = 57.2 kJ

(b) 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?

 

10. Ammonium carbamate dissociates as . In a closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that partial pressure of NH3 now equals to original total pressure, calculate the ratio of total pressure now to the original pressure.

 

Level – II

 

1. KC for PCl5(g) PCl3(g) + Cl2(g) is 0.04 at 250°C. How many mole of PCl5 must be added to a 3 litre flask to obtain a Cl2 concentration of 0.15 M?

 

2. KC for N2O4(g) 2NO2(g) is 0.00466 at 298 K. If a one litre container initially contained 0.8 mole of N2O4, what would be the concentrations of N2O4 and NO2 at equilibrium? Also calculate equilibrium concentration of N2O4 and NO2 if the volume is halved at the same temperature.

 

3. A mixture of SO2, SO3 and O2 gas is maintained in a 10 litre flask at a temperature at which KC= 100 for the reaction 

2SO2 + O2 2SO3

At equilibrium:

(a) If the number of moles of SO2 and SO3 in the flask are equal, how much O2 is present?

(b) If the number of moles of SO3 in the flask is twice the number of moles of SO2. How much O2 is present?

 

4. At 800K a reaction mixture contained 0.5 mole of SO2, 0.12 mole of O2 and 5 mole of SO3 at equilibrium. KC for the equilibrium 2SO2 + O2 2SO3 is 833 lit/mol. If the volume of the container is 1 litre. Calculate how much O2 is to be added at this equilibrium in order to get 5.2 mole of SO3 at the same temperature.

 

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

 

6. Show that KP for the reaction 

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

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

 

7. When PCl5 is heated it dissociates into PCl3 and Cl2. The vapour density of the gas mixture at 200oC and at 250oC is 70.2 and 57.9 respectively. Find the degree of dissociation at 200oC and 250oC.

 

8. 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 is introduced into the flask?

 

9. When 1-pentyne (A) is treated with 4N 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 ΔG0 for the following equilibria.

B A ΔG10 =? B C ΔG20 = ?

From the calculated value of ΔG10 and ΔG20 indicate the order of stability of A, B and C. Write a reaction mechanism showing all intermediates leading to A, B and C. 

 

10. The density of an equilibrium mixture of N2O4 and NO2 at 1atm. and 348 K is 1.84 g dm-3. Calculate the equilibrium constant of the reaction, N2O4(g) 2NO2(g).

Objective:

 

Level – I

 

1. In which of the following cases, does the reaction go furthest to completion?

(A) (B)

(C) (D) K = 1

 

2. The vapour density of undecomposed N2O4 is 46. When heated, the vapour density decreases to 24.5 due to its dissociation to NO2. The % dissociation of N2O4 at the final temperature is 

(A) 87% (B) 60% 

(C) 40% (D) 70%

 

3. In a reaction

the equilibrium constant can be expressed in units of

(A) (B)

(C) (D)

 

4. In a chemical reaction

The concentration A, B, C and D are 0.5, 0.8, 0.4 and 1.0 at equilibrium respectively.
The equilibrium constant is:

(A) 0.1 (B) 1.0

(C) 10 (D)

 

5. One mole of was placed in a litre reaction vessel at a certain temperature. The following equilibrium was established.. At equilibrium, 0.6 moles of SO2 were formed. The equilibrium constant of the reaction will be

(A) 0.36 (B) 0.45

(C) 0.54 (D) 0.675

 

6. HI was heated in a sealed tube at 440°C till the equilibrium was reached. HI was found to be 22% decomposed. The equilibrium constant for dissociation is

(A) 0.282 (B) 0.0769

(C) 0.0155 (D) 0.0199

 

7. For the reaction: , the partial pressure of CO2 and CO are 4 and 8 atm respectively. Hence Kp for the reaction is 

(A) 16 (B) 2 

(C) 0.5 (D) 4   

 

8. When KOH is dissolved in water, heat is evolved. If the temperature is raised, the solubility of KOH

(A) Increases (B) Decreases

(C) Remains the same (D) Cannot be predicted

 

9. A 1 litre vessel contains 2 moles each of gases A, B, C and D at equilibrium. If 1 mole each of A and B are removed, Kc for A + B C + D; will be 

(A) 4 (B) 1 

(C) 1/4 (D) 2     

10. Consider the water gas equilibrium reaction

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

Which of the following statement is true at equilibrium?

(A) If the amount of C(s) is increased, less water would be formed

(B) If the amount of C(s) is increased, more CO and H2 would be formed

(C) If the pressure on the system is increased by halving the volume, more water would be formed.

(D) If the pressure on the system is increased by halving the volume, more CO and H2 would be formed.

 

11. A reaction takes place in two steps with equilibrium constants 10–2 for slow step and 102 for fast step. The equilibrium constant of the overall reaction will be

(A) 104 (B) 10—4

(C) 1 (D) 10–2

 

12. For the reaction PCl3(g) + Cl2(g) PCl5(g), the value of KC at 250°C is 26 litre/mole. The value of Kp at this temperature will be

(A) 0.61 atm–1 (B) 0.57 atm–1

(C) 0.85 atm–1 (D) 0.46 atm–1

 

13. According to Le Chatlier’s principle adding heat to a solid and liquid in equilibrium with endothermic nature will cause the

(A) Amount of solid to decrease (B) Amount of liquid to decrease

(C) Temperature to rise (D) Temperature to fall

 

14. An aqueous solution of hydrogen sulphide shows the equilibrium

H2S H+ + HS–

If dilute hydrochloric acid is added to an aqueous solution of H2S, without any change in temperature, then

(A) The equilibrium constant will change

(B) The concentration of HS– will increase

(C) The concentration of undissociated hydrogen sulphide will decrease

(D) The concentration of HS– will decrease

 

15. What would happen to a reversible reaction at equilibrium when temperature is raised, given that its ΔH is positive 

(A) More of the products are formed (B) Less of the products are formed 

(C) More of the reactants are formed (D) It remains in equilibrium 

 

16. 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 moles 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 

 

17. One mole of ethanol is treated with one mole of ethanoic acid at 25°C.
    One–fourth of the acid changes into ester at equilibrium. The equilibrium constant for the  reaction will be

(A) 1/9 (B) 4/9

(C) 9 (D) 9/4

18. The equilibrium, PCl5(g) PCl3(g) + Cl2(g) is attained at 25°C in a closed container and an inert gas He is introduced. Which of the following statements are correct?

(A) concentration of PCl5, PCl3 and Cl2 are changed 

(B) more Cl2 is formed 

(C) concentration of PCl3 is reduced 

(D) Nothing happens 

 

19. In which of the following equilibrium, the value of KP is less than KC?

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

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

 

20. In the reaction A2(g) + 4B2 (g) 2AB4(g) , ΔH > 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 

Level – II

 

1. The equilibrium constant for a reaction is 10+22 at 300K, the standard free energy change for this reaction is

(A) – 115 kJ (B) +115kJ

(C) 166kJ (D) – 166 kJ

 

2. The normal vapour density of PCl5 is 104.25. At 250°C, its vapour density is 57.9. The degree of dissociation of PCl5 at this temperature will be

(A) 40% (B) 60%

(C) 80% (D) 90%

 

3. For the reactions,

,  , 

Kc for the reaction A D is

(A) 2 + 4 + 6 (B)

(C) (D)

 

4. The equilibrium constant for the reaction. The reaction is at 500K and 700K are 10–10 and 10–5 respectively. Hence the reaction is 

(A) Endothermic (B) Exothermic

(C) Fast (D) Slow

 

5. If pressure is applied to the liquid vapour equilibria

the boiling point of the liquid

(A) will decrease (B) will increase

(C) many increase or decrease (D) will not change

 

6. of the reaction

(A) (B)  

(C) (D) 1

 

7. For a system

eqm. conc.     0.06   0.12          0.216

The value of Kc is

(A) 250 (B) 416

(C) 30 (D) 125

 

8. On applying pressure to the equilibrium,   Ice water

Which phenomenon will happen

(A) More ice will be formed (B) More water will be formed

(C) Equilibrium will not be disturbed (D) Water will evaporate

 

9. Densities of diamond and graphite are 3.5 and 2.3 grams respectively. Increase of pressure on the equilibrium Cdiamond Cgraphite

(A) Favours backward reaction (B) Favours forward reaction

(C) Have no effect (D) Increase the reaction rate

10. The decomposition of N2O4 into NO2 is carried out at 280 K in chloroform. When equilibrium has been established, 0.2 moles of N2O4 and 2 × 10−3 moles of NO2 are present in 2 litre of the solution. The equilibrium constant for the reaction, , is 

(A) 1 × 10−2 (B) 2 × 10−3 

(C) 1 × 10−5 (D) 2 × 10−5

 

11. One mole of N2O4(g) at 300K is kept in a closed container under one atm. It is heated to 600K 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

 

12. At 30°C, Kp for the dissociation reaction 

SO2Cl2(g) SO2(g) + Cl2(g)  is 2.9 × 10–2 atm. If the total pressure is 1 atm, the degree of dissociation of SO2Cl2 is

(A) 87% (B) 13%

(C) 17% (D) 29%

 

13. A vessel at 1000K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. The value of K if the total pressure at equilibrium is 0.8 atm is

(A) 1.8 atm (B) 3 atm

(C) 0.3 atm (D) 0.18 atm

 

14. Vapour density of PCl5 is 104.16 but when heated at 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%

15. For the reaction SO2(g) + O2(g) SO3(g) Kp = 1.7 × 1012 at 20°C and 1 atm pressure. Calculate Kc.

(A) 1.7 × 1012  (B) 0.7 × 1012 

(C) 8.40 × 1012 (D) 1.2 × 1012

 

16. For a gaseous equilibrium 

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

(A) ≈ 0.031 (B) ≈ 32

(C) ≈ 44.4 (D) ≈ 1.3  × 10–3

 

17. For the reaction N2O4(g) 2NO2(g) the reaction connecting the degree of dissociation (α) of N2O4(g) with its equilibrium constant KP is 

(A) α = (B) α =

(C) α = (D) α =

 

18. 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 a 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 partial pressure of O2 (g) at equilibrium will be 

    (A) 0.66 atm (B) 0.493 atm

(C) 0.33 atm (D)  0.20 atm

19. KP 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 KC 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

 

20. The energy profile for the reaction:  is shown as:  The equilibrium constant for the said equilibrium (A) increases with the increase in temperature (B) decrease with the increase in temperature  (C) does not change with the change in temperature (C) is equal to the rate constant of the forward reaction

 

 

ANSWERS TO ASSIGNMENT PROBLEMS

 

Subjective:

 

Level – O

 

1. The rate of forward reaction becomes equal to the rate of backward reaction at equilibrium.

 

2. When a catalyst is added, the state of equilibrium is not disturbed but equilibrium is attained quickly.

 

3. As the equilibrium constant has increased with increase in temperature, the reaction is endothermic in the forward direction.

 

4.

The melting point of ice is favoured at high pressure because there is decrease in volume in the forward reaction. Since at high altitudes, atmospheric pressure is low and therefore, ice melts slowly.

 

5. The increase of pressure i.e. compression will favour the formation of reactants.

 

7.

 

12. 35% 

 

Level – I

 

1. 5.131 × 10–4 2. 2.96%

 

3. 37.25 gms 4. 0.03114

 

5. 3.142 6. 67.57%

 

7. (i)  0.267 atm,   (ii)   40% 8. 37.53% and 27.52%

 

9. (a)  4.32           (b)  1.462 × 10–4 atm 10. 31:27

 

Level – II

 

1. 2.1375 moles                                                    2. (i)  [N2O4] = 0.77M; [NO2] = 0.06 M 

(ii) [N2O4]=1.557 M; [NO2] = 0.086 M 

 

3. (a) 0.1  (b) 0.4

 

4. 0.34                                                                   5.
5. 2.334 × 10–6 7. 0.48 and 0.79

 

8. Kp = 0.049 atm2 and Kc = 8.1 × 10−5, No effect

 

9. , B > C > A 10. 5.14 atm

 

Objective: 

Level – I

1. C 2. A 3. B
2. B 5. D 6. C
3. A 8. B 9. B
4. C 11. C 12. A 13. A 14. D 15. A
5. A 17. A 18. D
6. C & D 20. D

Level – II

1. A 2. C 3. D
2. A 5. B 6. B
3. C 8. B 9. A
4. C 11. B 12. C 13. A 14. B 15. C
5. B 17. C 18. D
6. C 20. A