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NEET 2022
Exam
NEET
Year
2022
Subject
Chemistry
Chapter
Chemical Equilibrium

NEET 2022 · Chemistry · Chemical Equilibrium

3 O _ 2 ( g ) ⇌ 2 O _ 3 ( g ) for the above reaction at 298K, K _ c...

3O2(g)⇌2O3(g)3{O}_{2} ( g ) ⇌ 2{O}_{3} ( g )

for the above reaction at 298K,Kc298K, {K}_{c} is found to be 3.0×10−593 . 0 \times {10}^{- 59}. If the concentration of O2{O}_{2} at equilibrium is 0.040M0 . 040M then concentration of O3{O}_{3} in MM is

Options

  1. A1.9×10−631 . 9 \times {10}^{- 63}
  2. B2.4×10312 . 4 \times {10}^{31}
  3. C1.2×10211 . 2 \times {10}^{21}
  4. D4.38×10−324 . 38 \times {10}^{- 32}

Correct answer

D. 4.38×10−324 . 38 \times {10}^{- 32}

Step-by-step solution

Kc=3.0×10−59{K}_{c} = 3 .0 \times {10}^{- 59}
\left[{O}_{2}\left(g\right)\right] = 4 \times {10}^{- 2} .<br/>Thegivenreaction<br />The given reaction3{O}_{2}\left(g\right) ⇌ 2{O}_{3}(g)\left(g\right)
Kc=productofmolarconcentrationofproductsproductofmolarconcentrationofreactants=[O3(g)]2[O2(g)]3{K}_{c} = \dfrac{productofmolarconcentrationofproducts}{productofmolarconcentrationofreactants} = \dfrac{{\left[{O}_{3}\left(g\right)\right]}^{2}}{{\left[{O}_{2}\left(g\right)\right]}^{3}}

Put all the given values in the expression of KcKc

3.0×10−59=[O3(g)]2[4×10−2]33 .0 \times {10}^{- 59} = \dfrac{{\left[{O}_{3}\left(g\right)\right]}^{2}}{{\left[4 \times {10}^{- 2}\right]}^{3}}
⇒[O3(g)]2=3.0×10−59×(4×10−2)3⇒ {[ {O}_{3} ( g ) ]}^{2} = 3 .0 \times {10}^{- 59} \times {( 4 \times {10}^{- 2} )}^{3}
⇒[O3(g)]2=192×10−65⇒ {\left[{O}_{3}\left(g\right)\right]}^{2} = 192 \times {10}^{- 65}
Hence, the concentration of O3=4.38×10−32M{O}_{3}=4 . 38 \times {10}^{- 32}M

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