Chapter 14: Problem 102
The rate of the reaction between \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) to form \(\mathrm{HI}\) increases with the intensity of visible light. (a) Explain why this fact supports a two-step mechanism. \(\left(\mathrm{I}_{2}\right.\) vapor is purple.) (b) Explain why the visible light has no effect on the formation of \(\mathrm{H}\) atoms.
Short Answer
Step by step solution
Understand the Reaction
Analyze the Impact of Light on Rate
Consider a Two-Step Mechanism
Explain Lack of Effect on \(\mathrm{H}\) Formation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rate of Reaction
Visible Light Effect
- The color of a substance is often linked to the wavelengths of light it can absorb. In this case, the purple color of iodine suggests it absorbs visible light effectively.
- When enough energy is absorbed, it can cause molecule dissociation, creating reactive radicals.
Photolytic Step
Two-Step Mechanism
- The first step is photolytic: \(\text{I}_2 \rightarrow 2\text{I}\). This step is driven by visible light, which breaks the \(\text{I}_2\) molecules into iodine radicals.
- The second step involves these radicals: \(\text{H}_2 + \text{I} \rightarrow 2\text{HI}\), where the highly reactive iodine radicals react with hydrogen to form the product.