Question

If equlibrium constnats of reaction, `N_(2)+O_(2)hArr 2NO` is `K_(1) and 1/2N_(2)+1/2O_(2)hArrNO is K_(2),` then
A. `K_(1)=K_(2)`
B. `K_(2)=sqrtK_(1)`
C. `K_(1)=2K_(2)`
D. `K_(1)=1/2K_(2)`

Answer 1

Correct Answer - B
`N_(2)+O_(2)hArr2NO" "...(i)`
`1/2N_(2)+1/2O_(2)hArrNO" "...(ii)`
From equation number (i)
`K_(1)=([NO]^(2))/([N_(2)][O_(2)])" "...(iii)`
For equation number (ii)
`K_(2)=([NO])/([N_(2)]^(1//2)[O_(2)]^(1//2))" "...(iv)`
From equation (iii) & (iv) it is clear that
`K_(2)=(K_(1))^(1//2)=sqrt(K_(1))," ""Hence",K_(2)sqrt(K_(1))`