One mole of a compound AB reacts with 1 mole of a compound CD according to the equation `AB + CD hArr AD + CB`. When equilibrium had been established it was found that `(3)/(4)` mole each of reactant AB and CD has been converted to AD and CB. There is no change in volume. The equilibrium constant for the reaction isA. `(9)/(16)`B. `(1)/(9)`C. `(16)/(9)`D. 9

One mole of a compound AB reacts with 1 mole of a compound CD accordin

1.	One mole of a compound AB reacts with 1 mole of a compound CD according to the equation `AB + CD hArr AD + CB`. When equilibrium had been established it was found that `(3)/(4)` mole each of reactant AB and CD has been converted to AD and CB. There is no change in volume. The equilibrium constant for the reaction isA. `(9)/(16)`B. `(1)/(9)`C. `(16)/(9)`D. 9
Answer» Correct Answer - D `AB + CDhArr AD + CD` mole at t = 0 ,1,1,0,0 Mole at equilibrium `(1-3/4)(1-3/4)hArr(3/4)(3/4)` `0.25` `0.25` `0.75` `0.75` `K_(c) = (0.75xx0.75)/(0.25xx0.25) = (0.5625)/(0.0625) = 9`

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