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The unit of equilibrium constant of the ...

The unit of equilibrium constant of the reaction, `A+3B hArr nC` is `L^(2)*mol^(-2)`. What is the value of n

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For the given reaction, `K_(c)=([C]^(n))/([A][B]^(3))`
thus, the unit of `K_(c)=((mol*L^(-1))^(n))/((mol*L^(-1))xx(mol*L^(-1))^(3))`
`=(mol*L^(-1))^(n-4)`
Hence, `L^(2)*mol^(-2)=(mol*L^(-1))^(n-4) or, n=2`.
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Knowledge Check

  • H_2+l_2iff2HL , unit of equilibrium constant of the reversible reaction.

    A
    `Mole^(-1) litre`
    B
    `Mole^(-2) litre`
    C
    `Mole litre^(-1)`
    D
    None of these
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