`K_(c)` for the reaction `N_(2)O(g)hArr2NO_(2)(g)` is `4.63xx10^(-3)` at `25^(2)C`.
a. What is the value of `K_(p)` at this temperature?
b. At `25^(@)C`, if the partial pressure of `N_(2)O_(4)(g)` at equilibrium is 0.2 atm, calculate equilibrium pressure of `NO_(2)(g)`

Explanation

`N_(2)O_(4)(g)hArr2NO_(2)(g) K_(c)=4.63xx10^(-3)`
a. `K_(p)=K_(c)[RT]^(Deltan)`
`Deltan` for the above reaction `=2-1=1`
`:.K_(p)=4.63xx10^(-3)xx0.0821xx298`
`=0.1132atm`
b. `K_(p)=(P_(NO_(2))^(2))/(P_(N_(2)O_(4)))`
`0.1132=(P_(NO_(2))^(2))/0.2=0.15atm`
`:.` Equilibrium pressure of `NO_(2)=0.15` atm