Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reaction of sulfur dioxide and oxygen gas. When nitrogen monoxide gas is added to the system, the reaction speeds up significantly because it proceeds through the following steps: First: 2 upper N upper O (g) plus upper O subscript 2 (g) right arrow 2 Upper N upper O subscript 2 (g). Second: 2 upper N upper O subscript 2 (g) plus 2 upper S upper O subscript 2 right arrow 2 Upper N upper o (g) plus 2 upper s upper o subscript 3 (g). Identify the catalyst in this reaction, explain how you know it is the catalyst, and describe how it increases the rate of the reaction.

NO is the catalyst; it provides a reaction pathway with a lower activation energy

Explanation:

1. Identify the catalyst

A catalyst is a species that is present at the beginning of a reaction and reappears at the end.

It does not appear in the overall equation.

Let's apply these concepts to your mechanism:

    First:      2NO(g) + O₂(g) ⟶ 2NO₂(g)

Second: 2NO₂(g) +2SO₂(g) ⟶ 2NO(g) + 2SO₃(g)

Overall:          O₂(g) + 2SO₂ ⟶ 2O₂

We see that NO is present at the beginning of the first step. It reacts and then re-forms in the second step. It does not appear in the final equation.

NO is the catalyst.

2. Explain the action of the catalyst

A catalyst acts by providing an alternate pathway with a lower activation energy.

The direct reaction of SO₂ with oxygen to form SO₃ has a high activation energy.

NO reacts with the oxygen to form an intermediate (NO₂) that then reacts with the SO₂ to form SO₃. Both steps have lower activation energies, so the reaction is faster.

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