For the reaction N _ 2 O _ 4 ( g ) ⇌ 2 NO _ 2 ( g ) , K _ p = 0 ....

The given reaction is,

 ${N}_{2}{O}_{4} ( g ) ⇌ 2{NO}_{2} ( g )$

${K}_{p} = 0 . 492atm$

$T = 300K$

${K}_{P} = {K}_{C} \cdot ( RT{)}^{{\Delta n}_{g}}$

${\Delta n}_{g} =$ number of gaseous molecules of products – number of gaseous molecules of reactants.

${\Delta n}_{g} = 1$

By putting the values in the above equation,

$⇒ {K}_{c} = \dfrac{{K}_{p}}{RT} = \dfrac{0 . 492}{0 . 082 \times 300} = 2 \times {10}^{- 2}$