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For the hypothetical reactions, the equi...

For the hypothetical reactions, the equilibrium constant `(K)` value are given
`AhArrB,K_(1)=2,BhArrC,K_(2)=4,`
`ChArrD,K_(3)=3`
The equilibrium constant (K) for the reaction
`AhArrD` is

A

`48`

B

`6`

C

`27`

D

`24`

Text Solution

Verified by Experts

The correct Answer is:
D
The reaction `AhArrD` is obtained by adding the three given reactions
`:. K=K_(1)xxK_(2)xxK_(3)=2xx4xx3=24`
or `K_(1)=([B])/([A]), K_(2)([C])/([B]), K_(3)=([D])/([C])`
On multiplying `K_(1),K_(2)` and `K_(3)`, we get
`K=K_(1)xxK_(2)xxK_(3)=([D])/([A])=24`
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