The value of Kc for the reaction 3O2 (g) ⇌ 2O3 (g) is $$2.0\times 10^{-50}$$ at 25°C. If the equilibrium concentration of O2 in air at 25°C is $$1.6 \times 10^{-2}$$ , what is the concentration of O3?

Asked by Pragya Singh | 11 months ago |  86

##### Solution :-

Given,

3O2 (g) ⇌ 2O3 (g)

Then, Kc = $$\dfrac{ [O_{3}(g)]^{2}}{[O_{2}(g)]^{3}}$$

Given that Kc = $$2.0 \times 10^{-50}\;and \;[O_2(g)]$$

$$1.6 \times 10^{-2}$$

Then,

$$2.0 \times 10^{-50}$$

$$= \dfrac { [O_{3}(g)]^{2} }{ [1.6 \times 10^{ -2 }]^{3}}\\ \\ \Rightarrow [O_{3} (g)]^{2}$$

$$= 2.0 \times 10^{-50} \times (1.6 \times 10^{-2})^{3}\\ \\ \Rightarrow [O_{3} (g)]^{2}$$

$$= 8.192 \times 10^{-56}\\ \\$$

$$\Rightarrow [O_{3} (g)]^{2}$$

$$= 2.86 \times 10^{-28} M$$

So, the conc. of O3 is $$2.86 \times 10^{-28}M$$.

Answered by Abhisek | 11 months ago

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