Question

In the CONTACT PROCESS for manufacturing sulphuric acid, the reaction converting \(\text{SO}_2\) to \(\text{SO}_3\) is _______

• A. Exothermic and reversible
• B. Endothermic and reversible
• C. Exothermic and irreversible
• D. Endothermic and irreversible

Hint:

Industrial gas-phase catalytic oxidations, such as \(\text{SO}_2 \to \text{SO}_3\) in the Contact Process and \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\) in the Haber Process, are classic examples of reversible, exothermic reactions.
They both utilize intermediate temperatures and multi-stage designs to maximize equilibrium conversion.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the thermodynamic and chemical equilibrium characteristics of the catalytic oxidation of sulfur dioxide (\(\text{SO}_2\)) to sulfur trioxide (\(\text{SO}_3\)) in the Contact Process, which is the primary industrial method for manufacturing sulfuric acid (\(\text{H}_2\text{SO}_4\)).

Step 2: Key Formula or Approach:
The chemical equation for the oxidation step in the Contact Process is:
\[ 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \quad \Delta H = -197 \text{ kJ/mol} \] We must evaluate whether this reaction is exothermic (\(\Delta H \lt 0\)) or endothermic (\(\Delta H \gt 0\)), and whether it is reversible (\(\rightleftharpoons\)) or irreversible (\(\to\)).

Step 3: Detailed Explanation:
Let us analyze the reaction characteristics:
1. Reversibility: The reaction is highly reversible. In the presence of a catalyst (such as vanadium pentoxide, \(\text{V}_2\text{O}_5\)), a dynamic equilibrium is established between the reactants and the product. This reversibility means that the reaction does not go to complete conversion in a single stage, and Le Chatelier's principle must be applied to optimize the yield of \(\text{SO}_3\).
2. Thermal Effect (Enthalpy Change): The reaction is strongly exothermic, releasing approximately \(197 \text{ kJ}\) of heat energy for every two moles of \(\text{SO}_2\) oxidized. The enthalpy change of the reaction is negative (\(\Delta H = -197 \text{ kJ/mol}\)).
3. Process Optimization: Because the reaction is exothermic and reversible, Le Chatelier's principle indicates that:
- Lower temperatures favor the forward reaction (higher equilibrium conversion), but lower the rate of reaction.
- Therefore, an intermediate optimum temperature of \(400^\circ\text{C}\) to \(450^\circ\text{C}\) is used industrially in a multi-stage converter with interstage cooling to balance thermodynamic conversion and kinetic rate.
Consequently, the reaction is both exothermic and reversible.

Step 4: Final Answer
The correct option is (A).