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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)]=1.5xx10^(-2) M, [H_(2)]=3.0xx10^(-2)M,` and `[NH_(3)]=1.2xx10^(-2)M`. Calculate the equilibrium constant.

Text Solution

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The equilibrium reaction is :`N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)`
`K_(2) = [NH_(3)]^(2)/ ([N_(2)] [H_(2)]^(3)) = (1* 2 xx 10^(-2))^(2)/((1*5 xx 10^(-2)) (3*0 xx 10^(-2))^(3)) = 3.55 xx 10^(2)`
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