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The equilibrium constant for the reversi...

The equilibrium constant for the reversible reaction `N_(2)+3H_(2) hArr 2NH_(3)` is K and for the reaction `(1)/(2) N_(2)+(3)/(2) H_(2) hArr NH_(3)` , the equilibrium constant is `K',.K` and `K'` will be related as

A

`K=K`

B

`K = sqrt(k)`

C

`K=sqrt(K)`

D

`Kxx K=1`

Text Solution

Verified by Experts

The correct Answer is:
B
`K=sqrt(K)`
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Knowledge Check

  • The equlibrium constant for the reversible reaction, N_(2)+3H_(2)2NH_(3) is K and for the reaction 1/2N_(2)+3/2H_(2)hArrNH_(3) the equlibrium constant is K' K and K' will be related as

    A
    `K=K'`
    B
    `K'=sqrtK`
    C
    `K=sqrt(K')`
    D
    `KxxK'=1`

  • K_(C ) for the reaction (1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) hArr NH_(3(g)) is

    A
    `K_(c )=([NH_(3)])/([N_(2)][H_(2)])`
    B
    `K_(c )=([N_(2)][H_(2)])/([NH_(3)])`
    C
    `K_(c )=([NH_(3)])/([N_(2)]^(1/2)[H_(2)]^(3/2))`
    D
    `K_(c )=([N_(2)]^(1/2)[H_(2)]^(3/2))/([NH_(3)])`

  • The equlibrium constant for the reaction N_(2)+3H_(2)hArr2NH_(3) is K, then the equlibrium constant for the equlibrium NH_(3) 1/2N_(2)+3/2H_(2) is

    A
    `1//K`
    B
    `1//K^(2)`
    C
    `sqrtK`
    D
    `(1)/(sqrtK)`

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