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Derive the relation between K(p) and K(c...

Derive the relation between `K_(p)` and `K_(c)` for the equilibrium reaction.
`N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`

Text Solution

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A. `K_( c) = ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3)), K_(p) = (P_(NH_(3))^(2))/(P_(N_(2)) xx P_(H_(2))^(3))`
Partial pressures of `P_(N_(2)), P_(H_(2))` and `P_(NH_(3))` can be written as:
`P_(N_(2)) = (n_(N_(2))RT)/V = C_(N_(2))RT`
`P_(H_(2)) = (n_(H_(2))RT)/V = C_(H_(2))RT`
`P_(NH_(3)) = (n_(NH_(3))RT)/V = C_(NH_(3))RT`
Substituting the values in `K_(p)` equation:
`K_(p) = C_(NH_(3))^(2)/((C_(N_(2))(C_(H_(2))) xx [RT]^(-2)`
`therefore K_(p) = K_( c)[RT]^(-2) [ therefore C_(NH_(3))^(2)/((C_(N_(2))).(C_(H_(2)))^(3)) = K_( c)]`
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