3O_2(g) leftharpoons 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 - Chemistry MCQ NEET 2022 2026 | ExamTest.live

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3 O_{2} (g) ⇌ 2 O_{3} (g) . For the above reaction at 298 K, K_{c} is found to be 3.0 \times 1 0^{- 59} . If the concentration of O_{2} at equilibrium is 0.040 M then concentration of O_{3} in M is:

Solution & Step-by-step Explanation

K_{c} = \frac{[ O _{3} ] ^{2}}{[ O _{2} ] ^{3}} $ 3.0 \times 1 0^{- 59} = \frac{[ O _{3} ] ^{2}}{( 0.040 ) ^{3}} $ [O_{3}]^{2} = (3.0 \times 1 0^{- 59}) \times (4 \times 1 0^{- 2})^{3} [O_{3}]^{2} = (3.0 \times 1 0^{- 59}) \times (64 \times 1 0^{- 6}) = 192 \times 1 0^{- 65} = 1.92 \times 1 0^{- 63} [O_{3}] = 1.92 \times 1 0^{- 63} = 19.2 \times 1 0^{- 64} \approx 4.38 \times 1 0^{- 32} M

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3 O_{2} (g) ⇌ 2 O_{3} (g) . For the above reaction at 298 K, K_{c} is found to be 3.0 \times 1 0^{- 59} . If the concentration of O_{2} at equilibrium is 0.040 M then concentration of O_{3} in M is:

4.38 \times 1 0^{- 32}

1.9 \times 1 0^{- 63}

2.4 \times 1 0^{31}

1.2 \times 1 0^{21}

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