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For the equilibrium 3 O2(g) <=> 2 O3(g), Kc = 8 x 10⁻⁵⁵ at 25 deg C. If the equilibrium concentration of O2 in air at 25 deg C is 8 x 10⁻³ M, what is the equilibrium concentration of O3?
- 8 x 10⁻³¹ M
- 8 x 10⁻²⁹ M
- 8 x 10⁻²⁷ M
- 8 x 10⁻²⁵ M
Correct answer: 8 x 10⁻³¹ M
Solution
Substituting Kc and [O2] gives [O3]² = 8e-55 * (8e-3)³ ~ 4.1e-61, so [O3] ~ 6 to 8 x 10⁻³¹ M, an extremely tiny amount, showing essentially no O3 forms.
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