The equilibrium constant for the reaction
`N_(2)(g)+O_(2)(g) hArr 2NO(g)`
at temperature T is `4xx10^(-4)`.
The value of `K_(c)` for the reaction
`NO(g) hArr 1/2 N_(2)(g)+1/2 O_(2)(g)`
at the same temperature is

Options

A. `50*0`
B. `0*02`
C. `2*5 xx 10^(2)`
D. `4 xx 10^(-4)`

Answer

A

Explanation

` N_(2) + O_(2) to NO, K=4 xx 10^(-4)`
For the reserve reaction,
` 2 NO to N_(2) + O_(2), K= 1/(4 xx 10^(-4))`
On dividing by 2 ,`NO to 1/2 N_(2) + 1/2O_(2)`
` K= sqrt(1/(4 xx 10^(-4)))=1/(2 xx10^(-2))=50 `