Question
Describe the manufacturing of $H_2SO_4$ by the contact process.

Answer

Sulfuric acid is manufactured by the contact process, which involves the following three steps.

1) Roasting in air:
Sulfur or sulfide ore (iron pyrites) on burning or roasting in air produces sulfur dioxide.
i) $\underset{\text{Sulfur}}{S _{( s )}}+ O _{2( g )} \stackrel{\Delta}{\longrightarrow} \underset{\text{Sulfur dioxide}}{SO _{2( g )}}$
ii) $\underset{\text{Iron sulfide}}{4 FeS _{2( s )}}+11 O _{2( g )} \stackrel{\Delta}{\longrightarrow} 2 Fe _2 O _{3( s )}+\underset{\text{Sulfur dioxide}}{8 SO _{2( g )}}$

2) Catalytic oxidation of sulfur dioxide:

i) Sulfur dioxide is oxidised catalytically with oxygen to sulfur trioxide, in the presence of $V _2 O _5$ catalyst.
$\underset{\text { Sulfuir dioxide }}{2 SO_{2(g)}}+O_{2(g)} \xrightarrow{V_2 O_5} \underset{\text { Sulfur trioxide }}{2 SO_{3(g)}}$
ii) The reaction is exothermic and reversible and the forward reaction leads to decrease in volume. Therefore, low temperature $\left(720 K\right.$ ) and high pressure ( $2$ bar ) are favourable conditions for maximum yield of $SO _3$.

3) Absorption, followed by dilution of sulfur trioxide gas:

i) Sulfur trioxide gas (from the catalytic converter) is absorbed in concentrated $H _2 SO _4$ to produce oleum.
$\underset{\text { Sulfur trioxide }}{SO_3}+H_2 SO_4 \longrightarrow \underset{\text { Oleum }}{H_2 S_2 O_7}$
ii) Dilution of oleum with water gives sulfuric acid of desired concentration.
$\underset{\text { Oleum }}{H_2 S_2 O_7}+H_2 O \longrightarrow \underset{\text { Sulfuric acid }}{2 H_2 SO_4}$
iii) The sulfuric acid obtained by contact process is $96-98 \%$ pure.

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