Consider the following reversible chemical reactions :
`A_(2)(g)+B_(2)(g)overset(K_(1))(hArr)2AB(g)....(1)`
`6AB(g) overset(K_(2))(hArr)3A_(2)(g)+3B_(2)(g).....(2)`
The relation between `K_(1)` and `K_(2)` is :

Options

A. `K_(2) = K_(1)^(3-)`
B. `K_(1)K_(2) = 1/3`
C. `K_(1)K_(2) = 3`
D. `K_(2) = K_(1)^(3)`

Answer

A

Explanation

`A_(2)(g) + B_(2)(g) overset(K_(1))to 2AB(g)` ……(1)
`6AB(g) overset(K_(2))to 3A_(2)(g) + 3B_(2)(g)` ……..(2)
`(1) xx 3`
`3A_(2) + 3B_(2) rarr 6AB K_(1)^(3)`
So, for
`6AB(g) overset(K_(2))to 3A_(2)(g) + 3B_(2)(g)` `1/K_(1)^(3)`
`K_(2) = 1/K^(3)_(1)`