1 mole of `N_(2)` and 3 mole of `H_(2)` are mixed in a closed vessel of 1 `dm^(3)` capacity. At equilibrium if the vessel contains a total of 2.4 moles, calculate equilibrium constant `K_(c)" for "N_(2)+3H_(2)hArr2NH_(3)`.

Explanation

Initial no. of moles: `underset(1)(N_(2))+underset(3)(3H_(2))hArrunderset(0)(2NH_(3))`
`{:("No. of moles reacting"),("at equilibrium"):}}x" "3x" "-`
`{:("No. of moles present"),("at equilibrium"):}}1-x" "3-3x" "2x`
`therefore` Total no. of moles present at equilibrium = (1 - x) + (3 - 3x) + 2x = 4 - 2x
Given 4 - 2x = 2.4 `" "therefore2x=4-2.4orx=1.6/2=0.8`
Since the volume of the vessel is 1 `dm^(3)`,
`[N_(2)]=(1-x)/1=(1-0.8)/1=0.2" mol"//dm^(3),[H_(2)]=(3-3x)/1=(3-2.4)/1=0.6"mol"//dm^(3)`
`[NH_(3)]=(2x)/1=1.6/1=1.6"mol"//dm^(3)`
`K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=((1.6)^(2))/((0.2)xx(0.6)^(3))=59.26("mol"//dm^(3))^(-2)`.