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).

In the following system at equilibrium, N_(2)+3H_(2)hArr2NH_(3), the r

1.	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).
Answer» Solution :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)`.