Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: P Block Elements

Question:

The manufacture of sulphuric acid involves the following steps. Identify \(X_1\), \(X_2\). \(X_3\) and \(X_4\) in the reaction involved:

(i) \(X_1 + O_2 (g) \longrightarrow SO_2 (g)\)

(ii) \(2SO_2 (g) + O_2 \overset{X_2}{\longrightarrow} 2X_3\)

(iii) \(X_3 + H_2SO_4 \longrightarrow X_4\)

Options:
\(X_1\) \(X_2\) \(X_3\) \(X_4\)
\(S\) \(V_2O_5\) \(SO_3\) \(H_2S_2O_7\)
\(X_1\) \(X_2\) \(X_3\) \(X_4\)
\(SO_3\) \(V_2O_3\) \(S\) \(H_2SO_5\)
\(X_1\) \(X_2\) \(X_3\) \(X_4\)
\(H_2S\) \(VO_2\) \(SO_3\) \(H_2SO_5\)
\(X_1\) \(X_2\) \(X_3\) \(X_4\)
\(S\) \(V_2O_3\) \(SO_3\) \(H_2S_2O_7\)
Correct Answer:
\(X_1\) \(X_2\) \(X_3\) \(X_4\)
\(S\) \(V_2O_5\) \(SO_3\) \(H_2S_2O_7\)
Explanation:

The correct answer is option 1.

\(X_1\) \(X_2\) \(X_3\) \(X_4\)
\(S\) \(V_2O_5\) \(SO_3\) \(H_2S_2O_7\)

The manufacture of sulfuric acid involves a series of well-defined chemical reactions. Here is a detailed explanation of each step:

Step 1: Production of Sulfur Dioxide

Reaction: \(S + O_2 \rightarrow SO_2 \)

Reactant \(X_1\): Sulfur (\(S\))

Process: Sulfur is burned in air (or pure oxygen) to produce sulfur dioxide (\(SO_2\)). This reaction is straightforward and involves the combustion of sulfur.

Chemical Equation: When sulfur (\(S\)) reacts with oxygen (\(O_2\)), it forms sulfur dioxide (\(SO_2\)).

Step 2: Oxidation of Sulfur Dioxide to Sulfur Trioxide

Reaction: \(2SO_2 + O_2 \overset{X_2}{\longrightarrow} 2SO_3 \)

Catalyst \(X_2\): Vanadium pentoxide (\(V_2O_5\))

Process: Sulfur dioxide (\(SO_2\)) is further oxidized to sulfur trioxide (\(SO_3\)) in the presence of a catalyst. This step is crucial and typically takes place in a reactor called a contact chamber.

Catalyst Function: Vanadium pentoxide (\(V_2O_5\)) is used as a catalyst to speed up the reaction without being consumed in the process. This oxidation reaction is performed at high temperatures (450-600°C) and under high pressure.

Step 3: Formation of Oleum

Reaction: \(SO_3 + H_2SO_4 \rightarrow H_2S_2O_7 \)

Reactant \(X_3\): Sulfur trioxide (\(SO_3\))

Product \(X_4\): Oleum (\(H_2S_2O_7\))

Process: Sulfur trioxide (\(SO_3\)) is absorbed into existing sulfuric acid (\(H_2SO_4\)) to form oleum, also known as pyrosulfuric acid. Oleum is a solution of sulfur trioxide in sulfuric acid and is an intermediate compound in the production of sulfuric acid.

Chemical Equation: Sulfur trioxide reacts with sulfuric acid to form oleum. This reaction occurs because sulfur trioxide is highly reactive and forms a mixture with sulfuric acid, producing oleum.

Summary of Steps

Production of Sulfur Dioxide: Combustion of sulfur to produce \(SO_2\).

Oxidation of Sulfur Dioxide: Catalyzed by \(V_2O_5\) to produce \(SO_3\)

Formation of Oleum: \(SO_3\) reacts with \(H_2SO_4\) to form \(H_2S_2O_7\) (oleum).

So, the final values for \(X_1\), \(X_2\), \(X_3\), and \(X_4\) are:

\(X_1\) = \(S\) (Sulfur)

\(X_2\) = \(V_2O_5\) (Vanadium pentoxide)

\(X_3\) = \(SO_3\) (Sulfur trioxide)

\(X_4\) = \(H_2S_2O_7\) (Oleum)

The correct choice is 1. \(S\) \(V_2O_5\) \(SO_3\) \(H_2S_2O_7\).