For the reaction N 2 O 4 ( g ) ⇌ 2 NO 2 ( g ) , K p = 0 . 492 atm at 300 K . K c for the reaction at same temperature is _ _ _ _ _ _ × 10 - 2 . (Given : R = 0 . 082 L atm mol - 1 K - 1 )

Step-by-step solution

The given reaction is, N 2 O 4 ( g ) ⇌ 2 NO 2 ( g ) K p = 0 . 492 atm T = 300 K K P = K C · ( RT ) Δn g Δn g = number of gaseous molecules of products – number of gaseous molecules of reactants. Δn g = 1 By putting the values in the above equation, ⇒ K c = K p RT = 0 . 492 0 . 082 × 300 = 2 × 10 - 2