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

Initial moles of `N_(2)=1`
Moles of `N_(2)` converted into Ammonia= `(1 xx 0.25)/(100)= 0.0025`
According to the given equation, one mole of Nitrogen combines with 3 moles of Hydrogen, to give 2 moles of Ammonia. Therefore, 0.0025 mole of Nitrogen combines with `3 xx 0.0025=0.0075` mole of Hydrogen, to give `2 xx 0.0025= 0.0050` mole of Ammonia.
`{:(,N_(2)(g) ,+,3H_(2)(g),hArr,2NH_(3)(g)),("Initial moles",1,,3,,0),("Moles at equilibrium",1-0.0025,,3-0.0075,,0.0050):}`
Hence, `[N_(2)]= (1- 0.0025)/(4)= 0.25`
`[H_(2)]= (3-0.0075)/(4)= 0.75`
`[NH_(3)]= (0.0050)/(4) = 1.25 xx 10^(-3)`
(since total volume= 4 litres)
Therefore, `K= ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
`=((1.25 xx 10^(-3))^(2))/(0..25 xx (0.75)^(3))= 1.48 xx 10^(-5)`
For the reaction,
`(1)/(2) N_(2)(g) + (3)/(2) H_(2)(g) hArr NH_(3)(g)`
`K.= ([NH_(3)])/([N_(2)]^(½)[H_(2)]^(""^(3)//_(2)))`
Hence, `K.= (K)^(½) = (1.48 xx 10^(-5))^(½)= 3.85 xx 10^(-3)`