One mole of nitrogen is mixed with three moles of hydrogen in a 4 litre container. If 0.25 per cent of nitrogen is converted into ammonia by the following reaction
` N_2(g) +3H_2 hArr 2NH_3(g)`
calculate the equilibrium constant of the reaction in concentration units. What will be the value of K for the following reaction?
` (1)/(2) N_2 (g) + (3)/(2) H_2 hArr NH_3 (g)`

Explanation

`N_(2) + 3H_(2) rarr 2NH_(3)`
`[N_(2)] = 0.75/4, [H_(2)] = 2.25/4, [NH_(3)] = 0.50/4`
`K_(c) = [NH_(3)]^(2)/([N_(2)][H_(2)]^(3)) = (0.50)^(2)/((0.75)(2.25)^(3)) xx 16 = 0.468 L^(2)mol^(-2)`
Also for : `1/2 N_(2) + 3/2H_(2) rarr NH_(3)`
`K'c = sqrt(Kc) = 0.68`