InterviewSolution
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InterviewSolution
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following equilibrium is studied by taking 1 mole of `N_(2)` and 3 moles of `H_(2)` in a 1L flask at a given temperature? `N_(2)(g) +3H_(2) (g) hArr 2NH_(3)(g)` `NH_(3)(g)` formed at equilibrium is neutralised by 200 mL of 1M HCl. Calculate equilibrium constant. |
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Answer» `200 mL of 1M HCl = (1)/(1000) XX 200 = 0.2 M` `NH_(3) ` and HCl react in equimolar Proporation to form `NH_(4)Cl` `:. NH_(3)` evolved n the reaction = 0.2 M `N_(2)(g) + 3H_(2) (g) hArr 2NH_(3)(g)` No. of moles of `NH_(3)(g)` formed =0.2M No. of moles of `N_(2)(g) " reacted " =(0.2)/(2) = 0.1M` No. of moles of `H_(2)(g) " reacted " =0.2xx(3)/(2) =0.3M` The molar concentration of various species at the equilibrium point is : `{:(,N_(2)(g),+,3YH_(2)(g),hArr,2NH_(3)(g)),("Initial moles/litre",1,,3,,0),("Moles/litre at eqm. point",1-0.1,,3-0.3,,0.2),(,=0.9,,2.7,,):}` Applying Law of chemical equilibrium : `K_(c) = [[NH_(3)(g)]^(2))/[[N_(2)(g)][H_(2)(g)]^(3))= (0.2)^(2)/(0.9xx(2.7)^(3))=(0.04)/(17.71)=2..26xx 10^(-3)` |
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