3 O 2 (   g ) ⇌ 2 O 3 (   g ) for the above reaction at 298   K ,   K c is found to be 3 . 0 × 10 - 59 . If the concentration of O 2 at equilibrium is 0 . 040   M then concentration of O 3 in M is

Step-by-step solution

K c = 3 .0 × 10 - 59 O 2 g = 4 × 10 - 2 . The given reaction 3 O 2 g ⇌ 2 O 3 g K c = p r o d u c t   o f   m o l a r   c o n c e n t r a t i on   o f   p r o d u c ts p r o d u c t   o f   m o l a r   c o n c e n t r a t i o n   o f   r e a c t a n ts = O 3 g 2 O 2 g 3 Put all the given values in the expression of Kc 3 .0 × 10 - 59 = O 3 g 2 4 × 10 - 2 3 ⇒ [ O 3 ( g ) ] 2 = 3 .0 × 10 - 59 × ( 4 × 10 - 2 ) 3 ⇒ O 3