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One mole of a compound AB reacts with 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 is

A

`(9)/(16)`

B

`(1)/(9)`

C

`(16)/(9)`

D

9

Text Solution

Verified by Experts

The correct Answer is:
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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Knowledge Check

  • One mole of a compound AB reacts with one mole of a compound CD accoeding to the equation AB+CDhArrAD+CB. When equilibrium had been estblished it was found that 3/4 mole each of rectant AB and CD had been converted to AD and CB. There is no change in volume. The equlibrium constant for the reaction is

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  • One mole of a compound AB reacts with one mole of a compound CD according to the equation, AB_((g)) + CD_((g)) Leftrightarrow AD_((g))+CB_((g)) . When equilibrium had been established it was found that 3/4 mole each of reactants AB and CD had been converted to AD and CB. There is no change in volume. The equilibrium constant for the reaction is

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