Question

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 is
A. `9//16`
B. `1//9`
C. `16//9`
D. 9

Answer 1

Correct Answer - D
`AB +CD hArr AD +CB`
Original concentration of `[AB]=[CD]=(1)/(V)` mol `L^(-1)`
where V is the volume of the vessel in L.
`K_(c )=([AD][CB])/([AB][CD])`
At equilibrium
`[AB]=[CD]=(1-3//4)/(V)=(1)/(4V)"mol" L^(-1)`
`[AD]=[CB] =(3)/(4V) "mol" L^(-1)`
`K_(c )=(((3)/(4V))((3)/(4V)))/(((1)/(4V))((1)/(4V)))=9`