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The following concentrations were obtain...

The following concentrations were obtained for the formation of `NH_(3)` from `N_(2)` and `H_(2)` at equilibrium at `500 K`. `[N_(2)] = 1.5 xx 10^(-2)M`. `[H_(2)] = 3.0 xx 10^(-2) M` and `[NH_(3)] = 1.2 xx 10^(-2) M`. Calculate equilibrium constant.

Text Solution

Verified by Experts

The equilibrium constant for the reaction,
`aA(g)+bB(g)hArrcC(g)+dD(g)`
can be written as
`K_(c )=([NH_(3)(g)]^(2))/([N_(2)(g)][H_(2)(g)]^(3))=((1.2xx10^(-2))^(2))/((1.5xx10^(-2))(3.0xx10^(-2))^(3))`
`=0.106xx10^(4)=1.06xx10^(3)`
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