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Column -1( Reaction Equilibrium ) Co...

Column -1( Reaction Equilibrium ) Column -II ( Degree of dissociation of X in terms of their equilibrium constants assuming degree of dissociation of Y is same that of X whenever applicable )
( P ) `X ( g) rarr Z (g) + Y ( s)` ( 1) `( 2 sqrt( K))/( 1+ 2 sqrt(K))`
( Q ) `2X(g) rarr Y (g) + Z ( g)` ( 2) `( 2 sqrt( K))/( 1+ sqrt(K))`
( R ) ` X ( g) +Y(g) rarr Z(g) + W(g) ` ( 3) `( K )/( K +1)`
( S ) `X ( g) rarr ( Y )/( 2)(s) + ( Z)/( 2) (g)` ( 4) `( 2K)/( 1+ 2K)`
Codes `:`

A

`{:(P,Q,R,S),(3,2,2,4):}`

B

`{:(P,Q,R,S),(3,1,2,4):}`

C

`{:(P,Q,R,S),(1,3,4,2):}`

D

`{:(P,Q,R,S),(1,4,2,3):}`

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • For the reaction, N_2(g) +O_2(g) hArr 2NO(g) Equilibrium constant k_c=2 Degree of association is

    A
    `1/(1-sqrt2)`
    B
    `1/(1+sqrt2)`
    C
    `2/(1+sqrt2)`
    D
    `2/(1-sqrt2)`

  • At equilibrium degree of dissociation of SO_(3) is 50% for the reaction 2SO_(3)(g)hArr 2SO_(2)(g)+O_(2)(g) . The equilibrium constant for the reaction is

    A
    0.15
    B
    0.25
    C
    2.5
    D
    0.025

  • For the reaction 2AB(g)hArrA_(2)(g)+B_(2)(g) The degree of dissociation (alpha) of HI(g) is related to equilibrium constant K_(p) by the expression

    A
    `(1+2sqrt(K_(p)))/(2sqrt(K_(p)))`
    B
    `sqrt((1+2K_(p))/2)`
    C
    `sqrt((2K_(p))/(1+2K_(p)))`
    D
    `(2sqrt(K_(p)))/(1+2sqrt(K_(p)))`