InterviewSolution
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InterviewSolution
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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. |
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Answer» `N_(2)(g) + 3H_(2) hArr 2NH_(3)(g)` `K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=([1.2xx10^(-2)M])/([1.5xx10^(-2)M][3.0xx10^(-2)M])=3.55 xx 10^(2)M` |
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