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Hint: Lead-chamber process is a method of producing sulfur dioxide with moist air, using gaseous nitrogen oxide. The reaction takes place primarily in a series of large, boxlike chambers of sheet lead. Here, the role of NO is to accelerate the rate of reaction.
Complete step by step solution:
It is considered as the most common manufacturing technique. Most common reactions of > > Lead Chamber Process is given below-
\[2S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}}\]
The $SO_3$ reacts with water to get $H_2SO_4$ which is shown below-
\[S{{O}_{3}}+{{H}_{2}}O\to {{H}_{2}}S{{O}_{4}}\]
> We should know that classic lead chamber process consist of 3 stages: Glover tower, lead chambers and Gay-Lussac Tower,therefore, we are proceeding into the depth of the process.
The gases issuing from the chambers consist mainly of nitrogen dioxide, nitric oxide and atmospheric nitrogen. The two former are dissolved by the sulphuric acis in the Gay-Lussac tower with formation of a solution of nitrosylsuphuric acid in excess of sulfuric acid:
\[\text{NO+N}{{\text{O}}_{2}}+2{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\to 2\text{N}{{\text{O}}_{2}}.\text{S}{{\text{O}}_{2}}.\text{OH+}{{\text{H}}_{2}}\text{O}\]
> As the formation of nitrosyl sulphuric acid is a reversible process, dilution of the sulphuric acid solution in the Glover tower tends to cause hydrolysis, which is aided by the high temperature; the oxides of nitrogen resulting from the decomposition pass on with the sulphur dioxide and excess of air to the chambers, whilst the sulphuric acid descends and issues at the bottom of the tower.
> In addition to the forgoing process, it should be noted that the Glover tower actually produces sulphuric acid. This is brought about by interaction of the sulphur dioxide in the burner gases with nitrosyl sulphuric acid, as follows
$2\text{N}{{\text{O}}_{2}}.\text{S}{{\text{O}}_{2}}.\text{OH+S}{{\text{O}}_{2}}+2{{\text{H}}_{2}}\text{O}\to \text{3}{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}+2\text{NO}$
This suggests that NO acts as the catalyst in the above exothermic reaction.Therefore, the correct option is (c) Catalyst
Note: It should be noted that the catalyst used in the manufacture of sulphuric acid is gaseous nitrogen oxides.Here, both product and catalyst are both gaseous in nature. Therefore, it can be termed as a homogeneous catalyst. It results in the production of B grade sulfuric acid of 50-60% purity.
Complete step by step solution:
It is considered as the most common manufacturing technique. Most common reactions of > > Lead Chamber Process is given below-
\[2S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}}\]
The $SO_3$ reacts with water to get $H_2SO_4$ which is shown below-
\[S{{O}_{3}}+{{H}_{2}}O\to {{H}_{2}}S{{O}_{4}}\]
> We should know that classic lead chamber process consist of 3 stages: Glover tower, lead chambers and Gay-Lussac Tower,therefore, we are proceeding into the depth of the process.
The gases issuing from the chambers consist mainly of nitrogen dioxide, nitric oxide and atmospheric nitrogen. The two former are dissolved by the sulphuric acis in the Gay-Lussac tower with formation of a solution of nitrosylsuphuric acid in excess of sulfuric acid:
\[\text{NO+N}{{\text{O}}_{2}}+2{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\to 2\text{N}{{\text{O}}_{2}}.\text{S}{{\text{O}}_{2}}.\text{OH+}{{\text{H}}_{2}}\text{O}\]
> As the formation of nitrosyl sulphuric acid is a reversible process, dilution of the sulphuric acid solution in the Glover tower tends to cause hydrolysis, which is aided by the high temperature; the oxides of nitrogen resulting from the decomposition pass on with the sulphur dioxide and excess of air to the chambers, whilst the sulphuric acid descends and issues at the bottom of the tower.
> In addition to the forgoing process, it should be noted that the Glover tower actually produces sulphuric acid. This is brought about by interaction of the sulphur dioxide in the burner gases with nitrosyl sulphuric acid, as follows
$2\text{N}{{\text{O}}_{2}}.\text{S}{{\text{O}}_{2}}.\text{OH+S}{{\text{O}}_{2}}+2{{\text{H}}_{2}}\text{O}\to \text{3}{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}+2\text{NO}$
This suggests that NO acts as the catalyst in the above exothermic reaction.Therefore, the correct option is (c) Catalyst
Note: It should be noted that the catalyst used in the manufacture of sulphuric acid is gaseous nitrogen oxides.Here, both product and catalyst are both gaseous in nature. Therefore, it can be termed as a homogeneous catalyst. It results in the production of B grade sulfuric acid of 50-60% purity.
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