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

Options

A. `K_1 = K_2`
B. `K_2 = sqrt(K_1)`
C. `K_1 = 2K_2`
D. `K_(1) =1/(2)K_2`

Answer

B

Explanation

`N_2 +O_2 hArr 2NO " "....(i)`
`1/2 N_(2) + 1/2 O_2 hArr NO " "....(ii)`
For 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)`