Problem 92 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

(c) KI + H_2SO_4 does not effectively prepare pure HI.

Step 1: Evaluate Reaction (a)

The reaction (a) involves the reaction between iodine (I_2) and hydrogen sulfide (H_2S) to produce hydrogen iodide (HI) and sulfur. This reaction is feasible and used to prepare HI.

Step 2: Evaluate Reaction (b)

Reaction (b) is between phosphorus triiodide (PI_3) and water (H_2O), leading to the production of hydrogen iodide (HI) and phosphorous acid. This is a known method for preparing HI.

Step 3: Evaluate Reaction (d)

In reaction (d), hydrogen gas (H_2) and iodine (I_2) react in the presence of a platinum catalyst to form hydrogen iodide (HI). This is a feasible method for preparing HI.

Step 4: Evaluate Reaction (c)

Reaction (c) involves reacting potassium iodide (KI) with sulfuric acid (H_2SO_4). This reaction produces HI temporarily; however, it does not yield pure HI due to HI decomposing further into iodine (I_2) and other byproducts. This method is not suitable for producing HI that's pure.

Step 5: Conclusion

After analyzing all reactions, reaction (c) does not effectively prepare pure HI and stands out as the exception among the given methods.

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into new substances called products. These processes involve breaking and forming of chemical bonds, leading to new molecular arrangements. Understanding chemical reactions is crucial because it helps us predict how substances will interact under various conditions.

There are several types of reactions, such as synthesis, decomposition, single replacement, and double replacement. In the context of preparing hydrogen iodide (HI), different chemical reactions have different efficiencies and purities when producing this compound. For instance, some reactions might lead to unwanted byproducts, affecting the purity of HI. By analyzing the specific reactions used in HI preparation, we can deduce which method is most effective and why others fail or succeed differently.

Iodine and Hydrogen Sulfide Reaction

The reaction between iodine (\( \text{I}_2 \)) and hydrogen sulfide (\( \text{H}_2 \text{S} \)) is a noteworthy method for producing hydrogen iodide. In this reaction:

• Iodine acts as an oxidizing agent.
• Hydrogen sulfide serves as a reducing agent.

This results in the formation of hydrogen iodide and sulfur. The balanced equation is as follows:

\[ \text{I}_2 + \text{H}_2\text{S} \rightarrow \text{2HI} + \text{S} \]

This reaction is favorable for HI preparation because it results in a fairly direct conversion, making it viable for acquiring HI. Importantly, the sulfur produced is a byproduct that does not interfere with the purity of the HI in a significant way, allowing this method to be feasible in practical applications.

Phosphorus Triiodide and Water Reaction

When reacting phosphorus triiodide (\( \text{PI}_3 \)) with water (\( \text{H}_2\text{O} \)), hydrogen iodide is obtained along with phosphorous acid. This reaction is an example of hydrolysis and is presented by the equation:

\[ \text{PI}_3 + 3\text{H}_2\text{O} \rightarrow \text{3HI} + \text{H}_3\text{PO}_3 \]

Key points in this reaction:

• Phosphorus triiodide reacts swiftly with water.
• This is a reliable method for laboratory synthesis of HI due to the controlled environment, minimizing byproducts.

Despite the reaction forming phosphorous acid alongside HI, the process effectively yields hydrogen iodide due to its clean reaction pathway. This makes it an excellent choice for scenarios demanding relatively pure HI.

Potassium Iodide and Sulfuric Acid Reaction

The reaction between potassium iodide (\( \text{KI} \)) and sulfuric acid (\( \text{H}_2\text{SO}_4 \)) is unique because it doesn’t succeed in producing pure hydrogen iodide. Here’s the general reaction:

\[ \text{2KI} + \text{H}_2\text{SO}_4 \rightarrow \text{2HI} + \text{K}_2\text{SO}_4 \]

However, the challenge is that hydrogen iodide rapidly decomposes under these conditions:

• HI decomposes to iodine and byproducts, reducing purity.
• The presence of excess reactants shifts the reaction backwards, further complicating pure HI production.

This reaction isn't efficient for obtaining pure HI due to its tendency to degrade and the formation of impurities. While it can create HI in situ, it’s not ideal for applications requiring concentrated or unmixed HI.