The value of Kc for the reaction 3O2(g)⇔2O3(g) is 2.0 ×10-50at 250 c. If the Equilibrium concentration of O2 in air at 250 c is 1.6 ×10-2,what is the concentration of O3

Kc=[O3]2[O2]3 2.0 ×10-50=[O3]2(1.6×10-2)3 [O3]2 =(1.6×10-2)3×2.0 ×10-50 [O3]2 =8.192×10-56 O3 =2.86×10-28M

The value of Kc for the reaction 3O2(g)⇔2O3(g) is 2.0 ×10-50at 250 c. If the Equilibrium concentration of O2 in air at 250 c is 1.6 ×10-2,what is the concentration of O3

Kc=[O3]2[O2]3 2.0 ×10-50=[O3]2(1.6×10-2)3 [O3]2 =(1.6×10-2)3×2.0 ×10-50 [O3]2 =8.192×10-56 O3 =2.86×10-28M

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The value of K_c for the reaction $3 O_{2} (g) \Leftrightarrow 2 O_{3} (g) is 2.0 \times 10^{- 50}$ at 25⁰ c. If the Equilibrium concentration of O₂ in air at 25⁰ c is
$1.6 \times 10^{- 2}$ ,what is the concentration of O₃

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8.192 $\times$ 10^-56

2.86 $\times$ 10^-28

8.192 $\times$ 10^-28

2.86 $\times$ 10^-56

answer is B.

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Detailed Solution

$K_{c} = \frac{{[O_{3}]}^{2}}{{[O_{2}]}^{3}} 2.0 \times 10^{- 50} = \frac{{[O_{3}]}^{2}}{{(1.6 \times 10^{- 2})}^{3}} {[O_{3}]}^{2} = {(1.6 \times 10^{- 2})}^{3} \times 2.0 \times 10^{- 50} {[O_{3}]}^{2} = 8.192 \times 10^{- 56} O_{3} = 2.86 \times 10^{- 28} M$