Question

How does ozone react with the following:
(i) $PbS$
(ii) $KI$
(iii) ${C_2}{H_4}$
(iv) $NO$

Answer 1

Hint: Ozone or trioxygen is an inorganic molecule with the chemical formula ${O_3}$Ozone is formed from dioxygen by the action of UV light and electrical discharges within the Earth’s atmosphere.
$3{O_2}\xrightarrow[{disch\arg e}]{{Electrical}}2{O_3} + 142KJ/mol$


Complete step by step solution:
(i) Lead sulphide $\left( {PbS} \right)$ is oxidized to lead sulphate $\left( {white} \right)$.
$\mathop {PbS + 4{O_3}}\limits_{Lead{\text{ sulphide}}\left( {Black} \right)} \xrightarrow{{}}\mathop {PbS{O_4} + 4{O_2}}\limits_{Lead{\text{ sulphate}}\left( {White} \right)} $
This reaction proceeds at room temperature and this is a redox reaction.
(ii) Reaction of KI with ozone –
Potassium iodide react with ozone to produce potassium iodate which on further reaction with water gives potassium hydroxide
\[
  KI + 3{O_3}\xrightarrow{{}}KI{O_3} + 3{O_2} \\
  KI{O_3} + {H_2}O\xrightarrow{{}}KOH + {I_2} + {O_2} \\
 \]
(iii) Reaction of ${C_2}{H_4}$with ozone –
Ethylene reacts with ozone to form mole ozonide of ethylene initially then react with $Zn/{H_2}O$ to form formaldehyde.
${C_2}{H_4} + {O_3}\xrightarrow{{Zn/{H_2}O}}\mathop {HCHO}\limits_{Formaldehyde} $
Using excess ozone, the yield of $HCHO$decreases while the yield of $CO$increases with time.
(iv) Reaction of $NO$ with ozone:
Sunlights split nitrogen dioxide $\left( {N{O_2}} \right)$into nitric oxide and an oxygen atom.
$N{O_2} + sunlight\xrightarrow{{}}\mathop {NO + \left[ O \right]}\limits_{Nitric{\text{ oxide}}} $
Ozone reacts with nitric oxide to yield nitrogen dioxide and oxygen.
$\mathop {NO + {O_3}}\limits_{Nitric{\text{ oxide ozone}}} \xrightarrow{{}}N{O_2} + {O_2}$


Note: Ozone reacts continuously with sunlight and a wide variety of natural and human produced chemicals in the stratosphere. In each reaction, an ozone molecule is lost and other chemical compounds are produced.