In the following system at equilibrium, `N_(2)+3H_(2)hArr2NH_(3)`, the reaction mixture contains 0.005 mol of `N_(2)`, 0.012 mol of `H_(2)` and 0.002 mol of `NH_(3)` in a 2 litre vessel. Calculate `K_(c)`.

Explanation

Volume of the vessel is 2 litre
`[NH_(3)]=0.002/2=0.001"mol"L^(-1),[N_(2)]=0.005/2=0.0025"mol"L^(-1)`
`[H_(2)]=0.012/2=0.006"mol"L^(-1)`
`K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=((0.001)^(2))/((0.0025)(0.006)^(3))=1.852xx10^(3)`.