NEET · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
\(3 O _{2}( g ) \rightleftharpoons 2 O _{3}( g )\) for the above reaction at \(298 K , K _{ c }\) is found to be \(3.0 \times 10^{-59}\). If the concentration of \(O _{2}\) at equilibrium is \(0.040 M\) then concentration of \(O _{3}\) in \(M\) is ...... .
- A \(1.9 \times 10^{-63}\)
- B \(2.4 \times 10^{31}\)
- C \(1.2 \times 10^{21}\)
- D \(4.38 \times 10^{-32}\)
Answer & Solution
Correct Answer
(D) \(4.38 \times 10^{-32}\)
Step-by-step Solution
Detailed explanation
\(3 O _{2}( g ) \rightleftharpoons 2 O _{3}( g )\) \(K _{ c }=\frac{\left[ O _{3}\right]^{2}}{\left[ O _{2}\right]^{3}}\)
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