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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

Verified by Experts

`K_(c)=([NH_(3)]^(2))/([N_(2)[H_(2)]^(3)],K_(p))=(P_(NH_(3))^(2))/(P_(N_(2))xxP_(H_(2))^(2))`
Partial pressures of `P_(N_(2)),P_(H_(2))` and `P_(NH_(3))` can be writtenas
`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 these values in `K_(p)` equation
`K_(p)=(C_(NH_(3))^(2))/((C_(N_(2)))(C_(H_(2))^(3))xx[RT]^(-2)`
`:.K_(p)=K_(c)[RT]^(-2)[ :. (C_(NH_(3))^(2))/((C_(N_(2)))(C_(H_(2)))^(3))=K_(C)]`
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