3O2(g)⇌2O3(g) for the above reaction at 298 K, Kc is found to be 3.0 x 10-59. If the concentration of O2at equilibrium is 0.040 M then concentration of O3 in M is

Kc=O32O23O3=KcO23=4.38×10−32

3O2(g)⇌2O3(g) for the above reaction at 298 K, Kc is found to be 3.0 x 10-59. If the concentration of O2at equilibrium is 0.040 M then concentration of O3 in M is

Kc=O32O23O3=KcO23=4.38×10−32

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$3 O_{2} (g) ⇌ 2 O_{3} (g)$ for the above reaction at 298 K, K_cis found to be 3.0 x 10^-59. If the concentration of O₂
at equilibrium is 0.040 M then concentration of O₃ in M is

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1.9 x 10^-63

2.4 x 10³¹

1.2 x 10²¹

4.38 x 10^-32

answer is D.

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$\begin{array}{l} K_{c} = \frac{{[O_{3}]}^{2}}{{[O_{2}]}^{3}} \\ [O_{3}] = \sqrt{K_{c} {[O_{2}]}^{3}} = 4 .38 \times 10^{- 32} \end{array}$

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3O2(g)⇌2O3(g) for the above reaction at 298 K, Kc is found to be 3.0 x 10-59. If the concentration of O2at equilibrium is 0.040 M then concentration of O3 in M is

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$3 O_{2} (g) ⇌ 2 O_{3} (g)$ for the above reaction at 298 K, K_cis found to be 3.0 x 10^-59. If the concentration of O₂
at equilibrium is 0.040 M then concentration of O₃ in M is