Question

A silent electric discharge is passed through pure oxygen. The volume of the gas is reduced by $\dfrac{100}{3}%$. The percentage conversion of oxygen to ozone is:
A. $\dfrac{200}{3}$
B. 75
C. 80
D. 100

Answer 1

Hint: Silent electric discharge is a method used to obtain ozone, without the oxygen being dissociated. This method uses low heat energy, and electricity without sparks.

Complete step by step solution: We have been given pure oxygen, through which a silent electrical discharge is passed. This reduces the volume of oxygen by $\dfrac{100}{3}%$. As the conversion of oxygen to ozone happens in this silent discharge, we have to find the percent of ozone after the reduction of oxygen.
The reaction that happens in the silent electrical discharge is,
\[3{{O}_{2}}\to 2{{O}_{3}}\]
So, assume that the initial volume of oxygen, ${{O}_{2}}$(V) = 100 mL
So, the percentage of ozone converted from this is assumed to be x mL. Therefore, at equilibrium the volume will become, ${{V}_{equi}}$ = $100-x+\dfrac{2x}{3}$
The final volume will be, $100-x+\dfrac{2x}{3}=100-\dfrac{100}{3}$,
Solving this for x, we will get the percent of ozone to be 100 mL.
Hence, the percentage conversion of oxygen to ozone is 100.

Note: When the equilibrium is established, the reaction happens, so, the volume will be 100 – x, as volume of oxygen is assumed to be 100, also it has an added factor of $\dfrac{2x}{3}$ which shows that 2 moles of x mL, ozone and 3 moles of oxygen are employed in the reaction. Then solving for x we will take 100 and x common, so the answer will be 100.