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For a reaction at equilibrium                                                                                      

A ( g ) ? B ( g ) + 1 2 C ( g )

The relation between dissociation constant (K), degree of dissociation (a) and equilibrium pressure (p) is given by

Option 1 -

K = α 1 2 P 3 2 ( 1 + 3 2 α ) 1 2 ( 1 − α )

Option 2 -

K = α 3 2 p 1 2 ( 2 + α ) 1 2 ( 1 − α )

Option 3 -

K = ( α p ) 3 2 ( 1 + 3 2 α ) 1 2 ( 1 − α )

Option 4 -

K = ( α p ) 3 2 ( 1 + α ) ( 1 − α ) 1 2

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 2


    Detailed Solution:

                                A ( g ) ?                   B ( g ) +                     1 2 C ( g ) t = 0       a     m o l e s                           0                                         0                   − a α     m o l e s         + a α     m o l e s       + a α 2 m o l e s _

    Eq.   a(1 - α)          aα           (aα/2)

    Moles           moles       moles

    Total no. of moles at equilibrium

    = nA + nB + nC

    = a ( 1 − α ) + a α + a α 2 = a [ 1 + α 2 ]

    ( P A ) e q = ( x A ) e q × P e q = a ( 1 − α ) ( 1 + α / 2 ) P = 1 − α ( 1 + α / 2 ) P

    ( P B ) e q = ( x B ) e q × P e q = a α a ( 1 + α / 2 ) P = α ( 1 + α / 2 ) P

    ( P C ) e q = ( x C ) e q × P e q = a α / 2 a ( 1 + α / 2 ) P = α / 2 ( 1 + α / 2 ) P

    K P = ( P B ) e q × ( P C ) e q 1 / 2 ( P A ) e q = ( α ) 3 / 2 P 1 / 2 ( 1 − α ) ( 2 + α ) 1 / 2

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