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?
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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 33: The value of Kc for the reaction 3O2 (g) ↔ 2O3 (g) is 2.0 ×10–50 at 25°C. ....
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