${NO}_{2}$ is involved in the formation of smog and acid rain. A reaction that is important in the formation of $N O_{2}$ is $O_{9} (g) + NO (g) ⇌ O_{2} (g) + {NO}_{2} (g); K_{c} = 6 \times 10^{94}$ , if the air over KOTA contained $1 \times 10^{- 6} M O_{3}$ , $1 \times 10^{- 5} M N O_{2} . 5 \times 10^{- 4} M N O_{2}$ and $8.2 \times 10^{- 3} {MO}_{2}$ , what can we conclude?

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There will be a tendency to form more $NO$ and $O_{3}$

There will be a tendency to form more ${NO}_{2}$ and $O_{2}$

There will be a tendency to form more ${NO}_{2}$ and
$O_{3}$

There will be no tendency for change because the reaction is at equilibrium

answer is B.

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Detailed Solution

For the given reaction,

$O_{3} (g) + NO (g) ⇌ O_{2} (g) + {NO}_{2} (g)$

Given, The value of $K_{C} = 6 * 10^{34}$

For the above reaction equilibrium quotient is,

$Q = \frac{[O_{2}] [{NO}_{2}]}{[O_{3}] [NO]}$

$Q = \frac{[8.2 \times 10^{- 3}] [2.5 \times 10^{- 4}]}{[1.0 \times 10^{- 6}] [1.0 \times 10^{- 5}]}$

$Q = 8.2 \times 2.5 \times 10^{4}$

$Q = 20.5 \times 10^{4}$

As $Q < K_{C}$ the reaction will hive a tendency to move forward.

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