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Chemical Equilibrium Class 11, JEE, NEET & AIIMS

Chemical Equilibrium

Dynamic equilibrium:

Types of Chemical Reactions in Chemical equilibrium:

Reversible Irreversible Reactions
Reactions which do not proceed to competition in forward direction & also proceed in backward directions under suitable condition.e.g     N2(g)+ 3H2(g) ⇄ 2NH3 Reactions only proceed to competition in a forward direction. AgNO3 + NaCl → AgCl + NaNO3
These reactions attain equilibrium. These reactions never attain equilibrium

Equilibrium Classification:

Physical Equilibrium:
Equilibrium set up in physical processes like the evaporation of water, melting of solids, dissolution of solutes, etc., is called physical equilibrium, e.g.,    Ice ⇔ Water   At equilibrium,  Rate of melting of ice = Rate of freezing of water.
Chemical Equilibrium

Is a reversible reaction is carried out in a closed vessel, a stage is attained where the speed of the forward reaction equals the speed of the backward reaction. It corresponds to chemical equilibrium. At equilibrium,
Rate of forward reaction = Rate of backward reaction.

Characteristics of Chemical Equilibrium: 

Types of Equilibrium:

  1. Homogeneous Equilibrium:  In homogeneous equilibrium, the reactants and products are present in the same phase or physical state (gaseous or liquid).

 2SO2 (g) + O2 (g) ⇔ 2SO3 (g)

2. Heterogeneous Equilibrium: In heterogeneous equilibrium, the reactants and products are present in two or more physical states or phases.

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

Law of Mass Action in Chemical equilibrium:

Guldberg and Waage state that the rate of a chemical reaction is directly proportional to the product of the active masses of the reacting substances. Hence For a general reaction,

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

Rate of backward reaction  ∝ [C]c [D]d = kb[C]c [D]d
Where kf & kb are rate constant.

Characteristics of  Partial pressure or Equilibrium constant :

Use of Partial Pressures Instead of Concentration:

For gaseous reactions, partial pressures are conveniently used since at any fixed temperature partial pressure is directly proportional to concentration. For a general gaseous reaction.

Relation between Kc and Partial pressure:

Where, Δng = moles of products – moles of reactants (gaseous only).

Relation between Equilibrium constant  and Partial pressure for different types of reactions:

Δng   0  +ve,  -ve
KP or Kc KP = Kc 
  Kp > Kc   Kp < Kc 

Units of Kp and Equilibrium constant:

Reaction Quotient:

For a general reaction:       aA     +     bB      ⇄    cC    +     dD  

Qc  =  [C]c  [D]d / [A]a [B]b

Le – Chatelier’s Principle:

Effect of Change of Concentration:

If at equilibrium the concentration of one of the reactants is increased. The equilibrium will shift in the forward direction and vice-versa.

Effect of Change in Pressure:

      e.g. N2 + O2⇔ 2NO.

      e.g., N2 + 3H2 ⇔ 2NH3   [High p. high yield of NH3]

Effect of Temperature

When the process is exothermic, low temperature favors the forward reaction. Whereas a process is endothermic.  high temperature favors the formation of products.

Effect of Addition of Inert Gas

If inert gas is added. It will increase the volume of the system. Therefore. The equilibrium will shift in a direction in which there is an increase in the number of moles of gases.

If keeping volume of the system constant, an inert gas is added. The relative molar concentration of the substance will not change. Hence. The equilibrium position of the reaction remains unaffected.

The presence of catalyst does not change the position of equilibrium. Hence it simply fastens the attainment of equilibrium.

Le-Chatelier’s Principle Applicable to Physical Equilibrium:

The increased pressure will increase the solubility of gas and vice-versa.

Some substances dissolve with the absorption of heat. So the solubility of such substances will increase with the increase of temperature and vice-versa, e.g., dissolution of NH4CI, KCI, KNO3 , etc. The dissolution of calcium acetate and calcium hydroxide is exothermic, so their solubility is lowered at a higher temperature.

 Ice ⇔ liquid water. The ice occupy the more volume than liquid water, so increased pressure will result in       melting of ice according to Le-Chatelier’s Principle

Favorable conditions for some chemical equilibria to get a higher yield of product:

Calculation of the Degree of Dissociation (α) from Density Measurements in Chemical equilibrium:

α = D – d /d (y-1)
Where,

D = theoretical vapour density
d = observed vapour density
Now,
Molecular mass = 2 x VD
∴                                                            α = Mc – Mo / Mo
Where,
Mc = calculated molecular weight.
Mo = observed molecular weight.

Important Question & Answers in Chemical equilibrium:

  1.  Find out Kc for the given reaction in equilibrium state: [SO2]= 0.6 M, [O2] = 0.82 M and [SO3] = 1.9 M ?

ANS: As per the question,

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

Kc = [SO3]2/ [SO2]2 [O2]

= (1.9)2 M2/(0.6)2 (0.82) M3

= 12.229 M-1 

2.  For the following equilibrium, K =   6.3×1014 at 1000K
    NO (g) +O3 (g) NO2 (g) +O2 (g)
Both the reverse and forward reactions in the equilibrium are elementary bimolecular reactions. Calculate Kc, for the reverse reaction?

ANS:

For the reverse reaction,               Kc = 1/Kc
= 1/6.3×1014
= 1.5×10-15

3. At equilibrium, the concentrations of N2= 3.0 × 10–3M, O2 = 4.2 ×10-3M & NO= 2.8 ×10-3M in a sealed vessel at 800K. What will be Kc for the reaction N2 (g) + O2 (g) ⇄ 2NO (g)

ANS:       

For the reaction equilibrium constant Kc is written as :

Kc = [NO]2 / [N2] [O2]

= (2.8 10-3 M)2 / (3.0 x 10-3 M) (4.2 x 10-3 M)

= 0.622

4. PCl5 ,PCl3 & Cl2 are at equilibrium at 500 K and having concentration 1.59 M PCl3, 1.59 MCl2 &   1.41 M PCl5. Calculate Kc for the reaction,
                                                              PCl5 ⇄ PCl3 + Cl2

ANS: 

Equilibrium constant Kc for above reaction is:
 PCl5 ⇄ PCl3 + Cl2

Kc = [PCl3] [Cl2] / [PCl5]

=  (1.59)2 / (1.41)

= 1.79.

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