Ncert Exercises & Solutions

7.33 The value of Kc for the reaction 3O2 (g) 2O3 (g) is 2.0 ×10–50 at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6 ×10–2, what is the concentration of O3?

The given reaction is:

$3 O_{2 (g)} \leftrightarrow 2 O_{3 (g)} Then, K_{c} = \frac{{[O_{3 (g)}]}^{2}}{{[O_{2 (g)}]}^{3}}$

It is given that

$K_{c} = 2.0 \times 10^{- 50} and [O_{2 (g)}] = 1.6 \times 10^{- 2} .$

Then we have,

$2.0 \times 10^{- 50} = \frac{{[O_{3 (g)}]}^{2}}{{[1.6 \times 10^{- 2}]}^{3}}$
$\Rightarrow {[O_{3 (g)}]}^{2} = 2.0 \times 10^{- 50} \times {(1.6 \times 10^{- 2})}^{3}$
$\Rightarrow {[O_{3 (g)}]}^{2} = 8.192 \times 10^{- 56}$
$\Rightarrow [O_{3 (g)}] = 2.86 \times 10^{- 28} M$

Hence, the concentration of $O_{3} is 2.86 \times 10^{- 28} M .$

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