The equilibrium constant `(K_c)` for the reaction `N_(2(g)) + O_(2(g)) to 2NO_((g))` at temperature T is `4xx10^(-4)`. The value of `K_c` for the reaction , `NO_((g))+1/2N_(2(g)) + 1/2O_(2(g))` at the same temperature is ….

Options

A. `2.5xx10^2`
B. `4xx10^(-4)`
C. `50.0`
D. `0.02`

Answer

C

Explanation

`N_(2(g)) + O_(2(g)) to 2NO_((g)) , K_c=4xx10^(-4)` …(1)
By multiplying the equation (1) by `1/2`
`1/2N_(2(g)) + 1/2O_(2(g)) to NO_((g))`…(2)
`K._c=sqrtK_c`
`=sqrt(4xx10^(-4))=2xx10^(-2)`
By reversing the equation (2) we get ,
`NO_((g)) to 1/2N_(2(g)) + 1/2O_(2(g))`
`K.._c=1/(K._c)`
`=1/(2xx10^(-2))`= 50.0