At T K, 100 L of dry oxygen is present in a | vedclass

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(D) The chemical reaction for the formation of ozone is: $3O_2(g) \rightarrow 2O_3(g)$.Let the initial volume of $O_2$ be $100 \ L$.Let $x \ L$ of $O_

Vedclass · Accepted Answer

$40$

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(D) The chemical reaction for the formation of ozone is: $3O_2(g) ightarrow 2O_3(g)$.Let the initial volume of $O_2$ be $100 \ L$.Let $x \ L$ of $O_2$ be converted into ozone.According to the stoichiometry, $3 \ L$ of $O_2$ produces $2 \ L$ of $O_3$.Therefore, $x \ L$ of $O_2$ will produce $\frac{2}{3}x \ L$ of $O_3$.The remaining volume of $O_2$ is $(100 - x) \ L$.The volume of $O_3$ formed is $\frac{2}{3}x \ L$.Given that the volumes of $O_2$ and $O_3$ become equal:$100 - x = \frac{2}{3}x$$100 = x + \frac{2}{3}x = \frac{5}{3}x$$x = \frac{100 	imes 3}{5} = 60 \ L$.The volume of ozone formed is $\frac{2}{3}x = \frac{2}{3} 	imes 60 = 40 \ L$.

At $T \ K$, $100 \ L$ of dry oxygen is present in a sealed container. It is subjected to silent electric discharge, until the volumes of oxygen and ozone become equal. What is the volume (in litres) of ozone formed at $T \ K$?

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