The equilibrium constant `K_(c)` for the following reaction at `842^(@)`C is `7.90xx10^(-3)`. What is `K_(p)`at same temperature ?
` (1)/(2)F_(2(g)) hArr F_(g)`

Options

A. `8.64xx10^(5)`
B. `8.26xx10^(-4)`
C. `7.90xx10^(-2)`
D. `7.56xx10^(-2)`

Answer

D

Explanation

`K_(P)=(7.9xx10^(-3))xx(0.0821xx115)^(0.5)`
`K_(P)=7.56xx10^(-2)`