1. 1.2 × 10²¹ 2. 4.38 × 10⁻³² 3. 1.9 × 10⁻⁶³ 4. 2.4 × 10³¹

Answer: 2. 4.38 × 10⁻³² Key idea: K_c=[O₃]²/[O₂]³ . [O₃]²=3.0×10⁻⁵⁹×(0.040)³=3.0×10⁻⁵⁹×6.4×10⁻⁵=1.92×10⁻⁶³ . [O₃]=sqrt1.92×10⁻⁶³=4.38×10⁻³² M. Option 3 forgets to take the square root.

Topic: Law of chemical equilibrium, Kp and Kc· Equilibrium

3O_2(g) ⇌ 2O_3(g)for the above reaction at 298 K, K_c is found to be 3.0 × 10^-59. If the concentration of O_2 at equilibrium is 0.040 M then concentration of O_3 in M is

1. \(1.2 \times 10^{21}\)

2. \(4.38 \times 10^{-32}\)

3. \(1.9 \times 10^{-63}\)

4. \(2.4 \times 10^{31}\)

Doubtly Asked 30 Oct 2026 NEET UG 2022 Practice test neet Save WhatsApp Report

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Answer: 2. \(4.38 \times 10^{-32}\)

Key idea: \(K_c=\dfrac{[\mathrm{O_3}]^2}{[\mathrm{O_2}]^3}\).

\([\mathrm{O_3}]^2=3.0\times10^{-59}\times(0.040)^3=3.0\times10^{-59}\times6.4\times10^{-5}=1.92\times10^{-63}\).

\([\mathrm{O_3}]=\sqrt{1.92\times10^{-63}}=4.38\times10^{-32}\) M.

Option 3 forgets to take the square root.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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3O_2(g) ⇌ 2O_3(g)for the above reaction at 298 K, K_c is found to be 3.0 × 10^-59. If the concentration of O_2 at equilibrium is 0.040 M then concentration of O_3 in M is

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