The equilibrium constant for the given reaction is 100.
`N_(2)(g) + 2O_(2)(g) hArr 2NO_(2)(g)` What is the equilibrium constant for the reaction given below?
`NO_(2)(g) hArr 1/2 N_(2)(g) + O_(2)`

Options

A. 10
B. 1
C. 0.1
D. 0.01

Answer

C

Explanation

`(i) N_(2) + 2O_(2) implies 2NO_(2)`
`K_(1) = [NO_(2)]^(2)/([N_(2)][O_(2)]^(2))`………..(1)
`(ii) NO_(2) implies 1/2N_(2) + O_(2)`
`K_(2) = ([N_(2)]^(1/2)[O_(2)])/[NO_(2)]`………..(2)
From the equation (1) and (2), we get,
`(K_(2))^(2) = 1/K_(1) = 1/100 (:' K_(1) = 100)`
`K_(2) = sqrt(1/100) = 0.1`