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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 the air at 25°C is 1.6 ×10–2, what is the concentration of O3?
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Solution
The given reaction is:
\[\ce{3O_{2(g)} ↔ 2O_{3(g)}}\]
Then `"K"_"C" = (["O"_(3("g"))]^2)/["O"_(2("g"))]^3`
It is given that `"K"_"C" = 2.0 xx 10^(-30) and ["O"_(2("g")]] = 1.6 xx 10^(-2)`
Then, we have,
`2.0 xx 10^(-50) = ["O"_(3("g")]]^2/[1.6 xx 10^(-2)]^3`
`=> ["O"_(3("g"))]^2 = 2.0 xx 10^(-50) xx (1.6 xx 10^(-2))^3`
`=> ["O"_(3("g"))]^2 = 8.192 xx 10^(-56)`
`=>["O"_(3("g"))] = 2.86 xx 10^(-28) "M"`
Hence, the concentration of `"O"_3 " is " 2.86 xx 10^(-28)` M
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