Problem 163 - A Complete Resource Book in Chemistry for JEE Main

Question

HI can be prepared by which of the following methods (a) \(\mathrm{I}_{2}+\mathrm{H}_{2} \mathrm{~S}\) (b) \(\mathrm{PI}_{3}+\mathrm{H}_{2} \mathrm{O}\) (c) \(\hat{\mathrm{KI}}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{H}_{2}+\mathrm{I}_{2} \stackrel{\mathrm{Pt}}{\mathrm{Pt}}\)

Step-by-Step Solution

Answer

HI can be prepared by options (a), (b), and (d).

Step 1: Analyze Option (a)

The reaction is between iodine, \( \mathrm{I}_2 \), and hydrogen sulfide, \( \mathrm{H}_2 \mathrm{~S} \). This reaction produces hydrogen iodide, \( \mathrm{HI} \), and sulfur, \( \mathrm{S} \), according to the equation: \[\text{I}_2 + \text{H}_2\text{S} \rightarrow 2\text{HI} + \text{S}\]This option is a valid method to prepare HI.

Step 2: Analyze Option (b)

Reaction with phosphorus triiodide, \( \mathrm{PI}_3 \), and water, \( \mathrm{H}_2 \mathrm{O} \), typically yields a phosphoric acid and hydrogen iodide. The reaction is given by:\[\text{PI}_3 + 3\text{H}_2\text{O} \rightarrow 3\text{HI} + \text{H}_3\text{PO}_3\]Thus, this method can also be used to prepare HI.

Step 3: Analyze Option (c)

The reaction involves potassium iodide, \( \mathrm{KI} \), and sulfuric acid, \( \mathrm{H}_2 \mathrm{SO}_4 \). This reaction does not produce \( \mathrm{HI} \). Instead, it typically forms iodine and other by-products:\[2\text{KI} + 2\text{H}_2\text{SO}_4 \rightarrow \text{I}_2 + \text{SO}_2 + \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}\]Thus, this method is not suitable for preparing HI.

Step 4: Analyze Option (d)

The reaction involves hydrogen gas, \( \mathrm{H}_2 \), and iodine, \( \mathrm{I}_2 \), in the presence of a platinum catalyst. This reaction produces hydrogen iodide as shown by:\[\text{H}_2 + \text{I}_2 \xrightarrow{\text{Pt}} 2\text{HI}\]Therefore, this option is also a valid method to prepare HI.

Key Concepts

Iodine and Hydrogen Sulfide Reaction

The reaction between iodine (\( \mathrm{I}_2 \)) and hydrogen sulfide (\( \mathrm{H}_2 \mathrm{~S} \)) is an efficient method to produce hydrogen iodide (\( \mathrm{HI} \)). This reaction occurs as follows: \[\mathrm{I}_2 + \mathrm{H}_2\mathrm{S} \rightarrow 2\mathrm{HI} + \mathrm{S}\]. This equation suggests that iodine reacts directly with hydrogen sulfide to form hydrogen iodide and sulfur.

There are several points to note about this reaction:

• The reaction is a straightforward synthesis. It involves the combination of two reactants to form products directly.
• The formation of sulfur (\( \mathrm{S} \)) as a solid precipitate can be observed, making it easier to identify that the reaction has occurred.
• This reaction is generally conducted under controlled conditions to maximize yield and ensure safety, as hydrogen sulfide is a toxic gas.

Understanding the basic nature of this reaction helps in recognizing it as a suitable method for the production of hydrogen iodide.

Phosphorus Triiodide and Water Reaction

The reaction between phosphorus triiodide (\( \mathrm{PI}_3 \)) and water (\( \mathrm{H}_2 \mathrm{O} \)) is another method to prepare hydrogen iodide. This reaction is represented by the following equation:\[\mathrm{PI}_3 + 3\mathrm{H}_2\mathrm{O} \rightarrow 3\mathrm{HI} + \mathrm{H}_3\mathrm{PO}_3\].

In this reaction, phosphorus triiodide reacts with water to produce hydrogen iodide and phosphorous acid (\( \mathrm{H}_3\mathrm{PO}_3 \)). Here are some key points:

• The hydrolysis of phosphorus triiodide is an exothermic reaction, which means it releases heat.
• This reaction is quite practical because it typically occurs at a relatively low temperature, making it manageable in a laboratory setting.
• The by-product, phosphorous acid, can be used separately, depending on application needs.

This reaction not only provides hydrogen iodide efficiently but also demonstrates the utility of phosphorus compounds in inorganic chemistry.

Platinum Catalyzed Reaction

The platinum catalyzed reaction involves the interaction of hydrogen gas (\( \mathrm{H}_2 \)) and iodine (\( \mathrm{I}_2 \)) in the presence of a platinum catalyst to produce hydrogen iodide. The equation for the reaction is given as:\[\mathrm{H}_2 + \mathrm{I}_2 \xrightarrow{\mathrm{Pt}} 2\mathrm{HI}\].

In this approach, several essential points about the process can be highlighted:

• Platinum acts as a catalyst and speeds up the reaction without being consumed in the process.
• This method is particularly useful due to the high efficiency and relatively simple setup required.
• The reaction typically requires a controlled environment to achieve optimal results, ensuring that the reactants and catalyst are pure, and conditions are just right.

Utilizing a platinum catalyst is a smart way to facilitate reactions, providing insights into both catalytic action and industrial processes for hydrogen iodide preparation.