A few reactions and their equilbirium constant are given. Find the equilibrium constant of the reaction, `A hArr 3B+E`.
`AhArrB+C,K_(c)=2`

Explanation

(i) `K_(c)=2=([B][C])/([A])`
(ii) `K_(c)=3=([B][D])/([C])`
(iii) `K_(c)=4=([B][E])/([D])`, For the reaction `A hArr 3B+E`,
equilibrium constant, `K_(c)=([B]^(3)[E])/([A])` . . (1)
Multiplying equilibrium constant of reactions (i), (ii) & (iii), we have `2xx3xx4=([B][C])/([A])xx([B][D])/([C])xx([B][E])/([D])`
or, `24=([B]^(3)[E])/([A])` . . [2]
From the equation [1] and [2], we get `K_(c)=24`.
From the equation [1] and [2], we get `K_(c)=24`.
Therefore, the equilibrium constant for the reaction `A hArr 3B+E` is 24.