The value of `K_(c)` for the reaction, `N_(2)(g)+harr2NO_(2)(g)` at a certain temperature is `400`. Calculate the value of equilibrium constant for.
(i) `2NO_(2)(g)harrN_(2)(g)+2O_(2)(g)`, (ii) `1//2N_(2)(g)+O_(2)(g)harrNO_(2)(g)`

Explanation

Given: Equilibrium constant `(K_(c))` for the reaction `N_(2)(g)+20_(2)(g)harr2NO_(2)(g) " is " K_(c)=400`
Formulae: `K_(c)=([NO_(2)])^(2)=400`
Asked: (i) For the reaction `2NO_(2)(g)harrN_(2)+20_(2)(g),K_(c)=([N_(2)][O_(2)]^(2))/([NO_(2)]^(2))=(1)/K_(c))`
Substitution & Calculation: `K_(c)=(1)/(400)=0.0025`
Asked: For the reaction `1//2N_(2)(g)+O_(2)(g)harrNO_(2)(g)`
Substitution & Calculation: `K"_(c)=([NO_(2)])/([N_(2)]^(1//2)[O_(2)])=sqrtK_(c)`
`rArr=K'_(c)=sqrt400=20`