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_(1)=(1//K_(2))^(2)`
B. `K_(1)=K_(2)^(2)`
C. `K_(1)=1//K_(2)`
D. `K_(1)=(K_(2))^(@)`

Answer 1

Correct Answer - A
`N_(2)+O_(2) hArr 2NO` …(i)
`NO hArr 1/2 N_(2)+1/2O_(2)` …(ii)
Equation (ii) is obtained by reversing equation (i) and dividing by `2`.
`:. K_(2)=1/((K_(1))^(1//2))`
`rArr (K_(2))^(2)=1/K_(1)`
`rArr K_(1)=1/((K_(2))^(2))=(1/K_(2))^(2)`