If `K_(c)` is the equlibrium constant for the formation of `KH_(3).` the dissociation constant of ammonia under the same temperaturee will be

Options

A. `K_c`
B. `sqrt(K_c)`
C. `K_c^2`
D. `1//K_c`

Answer

D

Explanation

`NH_3 hArr 1/2 N_2 +3/2 H_2`
`K_c = ([N_2]^(1//2)[H_2]^(3//2))/(NH_3)and 1/2N_2+3/2H_2 hArrNH_3`
`K_c =([NH_3])/([N_2]^(1//2)[H_2]^(3//2))`
So for dissociation `=1/K_c`