SELECT * FROM question_mgmt as q WHERE id=3067 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=48 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 11th chemistry Equilibrium the value of kc for the reaction 3o2 g harr

Question:

The value of Kc for the reaction

3O2 (g) ↔ 2O3 (g) is 2.0 ×10–50 at 25°C.

If the equilibrium concentration of O2 in air at 25°C is 1.6 ×10–2, what is the concentration of O3?

Answer:

The given reaction is:

3O2 (g) ↔ 2O3 (g)

Then, Kc =  [O3 (g)]2  /  [O2 (g)]3

It is given that Kc =  2.0 ×10–50  and O2 (g) = 1.6 ×10–2

Then we have,

2.0 ×10–50  =  [O3 (g)]2   / [1.6 ×10–2 ]3

⇒  [O3 (g)]2 = [2.0 ×10–50] x [1.6 ×10–2 ]3

⇒ [O3 (g)]2   = 8.192 x 10-56

⇒ O3 (g)     =   2.86x10-28 M

Hence, the concentration of O2 (g)  = 2.86x10-28 M


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