${NO}_{2}$ 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?

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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.

Watch 3-min video & get full concept clarity