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Chapter: 11th 12th std standard Class Organic Inorganic Physical Chemistry Higher secondary school College Notes

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Formation of HI from H2 and I2 - Equilibrium constants in terms of degree of dissociation

In the study of dissociation equilibrium, it is easier to derive the equilibrium constant expression in terms of degree of dissociation (x). It is considered as the fraction of total molecules that actually, dissociate into the simpler molecules x has no units.

Degree of dissociation (x)

 

In the study of dissociation equilibrium, it is easier to derive the equilibrium constant expression in terms of degree of dissociation (x). It is considered as the fraction of total molecules that actually, dissociate into the simpler molecules x has no units. If x is the degree of dissociation then for completely dissociating molecules x = 1.0. For all dissociations involving equilibrium state, x is a fractional value. If x is known, Kc or Kp can be calculated and vice-versa.

 

Equilibrium constants in terms of degree of dissociation

 Formation of HI from H2 and I2

 

The formation of HI from H2 and I2 is an example of gaseous homogeneous equilibrium reaction. It can be represented as

H2(g) + I2(g)-- >            < -- 2HI(g)        ∆H=-10.4 kJ

This equilibrium is an exothermic one.

 

Let us consider that one mole of H2 and one mole of I2 are present initially in a vessel of volume V dm3. At equilibrium let us assume that x mole of H2 combines with x mole of I2 to give 2x moles of HI. The concentrations of H2, I2 and HI remaining at equilibrium can be calculated as follows :

 

                                      H2(g) I2(g)  HI(g)

Initial number of moles  I        I        0

Number of moles reacted         x        x        -

Number of moles remaining at equilibrium 1-x    1-x    2x

Equilibrium concentration       1-x/V 1-x/1-x       2x/V

 

According to the law of mass action,

 

 Kc = [HI]2 / [H2] [I2]

Substituting the values of equilibrium concentrations in the above equation, we get

Kc = 1/ (1-x)2

If the initial concentration of H2 and I2 are equal to a and b moles dm-3 respectively, then it can be shown that

Kc  = 4x2  / (a-x)(b-x)

Derivation of Kp in terms of x

 

Let us consider one mole of H2 and one mole of I2 are present initially. At equilibrium, let us assume that x mole of H2 combines with x mole of I2 to give 2x moles of HI. Let the total pressure at equilibrium be P atmosphere. The number of moles of H2, I2 and HI present at equilibrium can be calculated as follows :

                                                H2(g)   I2(g)    HI(g)

Initial number of moles           I        I        O

                            

Number of moles reacted                  x        x        -

Number of moles                     I-x     I-x     2x

remaining at equilibrium                            

                            

The total number of       = 1 - x + 1 - x + 2x = 2

moles at equilibrium                        

 

We know that partial pressure is the product of mole fraction and the total pressure. Mole fraction is the number of moles of that individual component divided by the total number of moles in the mixture. Therefore,

 

PH2 = (1-x / 2) P

PI2 = (1-x / 2) P

PHI = (2x / 2) P

We know that Kp = P2HI / PH2.PI2

Substituting the values of partial pressures in the above equation, we get

Kp= (2xp/2) / ((1-x)/2)P((1-x)/2)P

 Kp = 4x2 / (1-x)2

we see that Kp and Ke are equal in terms of x values. The influence of various factors on the chemical equilibrium can be explained as below:

 

            Influence of pressure : The expressions for the equilibrium constants Kc and Kp involve neither the pressure nor volume term. So the equilibrium constants are independent of pressure and volume. Pressure has therefore no effect on the equilibrium.

 

            Influence of concentration : The addition of either H2 or I2 to the equilibrium mixture well increase the value of the denominator in the equation Ke = [HI]2/[H2][I2] and hence tends to decrease the value of Ke. In order to maintain the constancy of Kc, the increase in the denominator value will be compensated by the corresponding increase in the numerator value. In other words, the forward reaction will be favoured and there will be corresponding increase in the concentration of HI.

 

               Influence of catalyst : A catalyst affects both the forward and reverse reactions to the same extent. So it does not change the relative amounts of reactants and products at equilibrium. The values of Ke and Kp are not affected. However the equilibrium is attained quickly in the presence of a catalyst.


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