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Question

Kc for the synthesis of HI (g) from H2 (g) and I2 (g) is 50. The degree of dissociation of HI is:
H2+I22HI

A
0.11
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B
0.54
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C
0.48
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D
0.79
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Solution

The correct option is A 0.11
Given, Kc=50 for H2+I22HI
equilibrium constant is reversed for the formation of HI i.e. Kc=150
2HIH2+I2
12α α α -
150=α2(12α)2
Taking square root both sides.
150=α(12α)0.1414=α(12α)0.14140.2828α=αα=0.14141.2828=0.11

Hence degree of dissociation is 0.11

Theory:

Degree of dissociation :
Degree of dissociation ( α) is the fraction of moles of the reactant
that undergoes dissociation to form the products.

Degree of dissociation ( α is unitless)
For eg. Suppose 5 moles of PCl5 are taken and if 2 moles of PCl5 are dissociated, then:
α=25=0.4 40 % dissociated.
Maximum number of moles that can dissociate = Total number of moles taken.

Maximum value of alpha = 1 (α=1)
Note :- Value of alpha varies from 0 to 1 and it can be fractional.

Calculation of degree of dissociation :
Degree of dissociation α=Total of moles dissociatedInitial moles of reactant

Percentage dissociation=α×100

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