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Question
The value of Kc for the reaction 3O2(g)↔2O3(g) is 2.0×10–50at25∘C. If the equilibrium concentration of O2 in air at 25∘C is 1.6×10−2, what is the concentration of O3?
The value of Kc for the reaction 3O2(g)↔2O3(g) is 2.0×10–50at25∘C. If the equilibrium concentration of O2 in air at 25∘C is 1.6×10−2, what is the concentration of O3?
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Solution
The given reaction is:
3O2(g)↔2O3(g)
Then, Kc=−[O3(g)]2[O2(g)]3
It is given that Kc=2.0×10−50and[O2(g)]=1.6×10−2
Then, we have,
2.0×−50=[O3(g)]2[1.6×10−2]3⇒=[O3(g)]2=2.0×10−50×(1.6×10−2)3⇒[O3(g)]2=8.192×10−56⇒[O3(g)]=2.86×10−28M
Hence, the concentration of O3 is 2.86×10−28M.
The given reaction is:
3O2(g)↔2O3(g)
Then, Kc=−[O3(g)]2[O2(g)]3
It is given that Kc=2.0×10−50and[O2(g)]=1.6×10−2
Then, we have,
2.0×−50=[O3(g)]2[1.6×10−2]3⇒=[O3(g)]2=2.0×10−50×(1.6×10−2)3⇒[O3(g)]2=8.192×10−56⇒[O3(g)]=2.86×10−28M
Hence, the concentration of O3 is 2.86×10−28M.
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O3
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