Question

`K_(1)` and `K_(2)` are equilibrium constants for reaction (i) and (ii)
`N_(2)(g)+O_(2)(g) hArr 2NO(g)` …(i)
`NO(g) hArr 1//2 N_(2)(g)+1//2O_(2)(g)` …(ii)
then,
A. `K_(2) = (1)/(K_(1))`
B. `K_(2) = K_(1)^(2)`
C. `K_(2) = (K_(1))/(2)`
D. `K_(2) = (1)/(K_(1)^(2))`

Answer 1

Correct Answer - D
`K_(1) = ([NO_(2)])/([NO][O_(2)]^(1//2)), K_(2) = ([NO]^(2) [O_(2)])/([NO_(2)]^(2))`
`implies ([NO_(2)]^(2))/([NO][O_(2)]) = (1)/(K_(2))`
`implies ([NO_(2)])/([NO][O_(2)]^(1//2)) = (1)/(sqrt(K_(2)))`
`implies K_(1) = (1)/(sqrt(K_(2))), K_(2) = (1)/(K_(1)^(2))`