Question

For the reactions,
`{:(AhArrB, K_(c)=2), (BhArrC, K_(c)=4), (ChArrD, K_(c)=6):}`
`K_(c)` for the reaction, `AhArrD` is:
A. `(2+4+6)`
B. `(2xx4)//6`
C. `(4xx6)//2`
D. `2xx4xx6`

Answer 1

`2=([B])/([A])`, `4=([C])/([B])` and `6=([D])/([C])`
Thus on multiplying, `2xx4xx6=([D])/([A])`, i.e., `K_(c)` for the reaction.