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) \(\overline{\mathrm{KI}}+\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{H}_{2}+\mathrm{I}_{2} \stackrel{\mathrm{Pt}}{\mathrm{Pt}}\)

Step-by-Step Solution

Answer

Methods (a), (b), and (d) produce HI; method (c) does not.

Step 1: Analyze Option (a)

Consider the reaction \( \mathrm{I}_{2} + \mathrm{H}_{2} \mathrm{~S} \). In this method, iodine reacts with hydrogen sulfide to form hydrogen iodide and sulfur. The balanced chemical equation is \( \mathrm{I}_{2} + \mathrm{H}_{2} \mathrm{S} \rightarrow 2\mathrm{HI} + \mathrm{S} \), indicating that HI is produced.

Step 2: Analyze Option (b)

Consider the reaction \( \mathrm{PI}_{3} + \mathrm{H}_{2} \mathrm{O} \). This reaction involves phosphorus triiodide reacting with water to produce hydrogen iodide and phosphorous acid. The reaction is \( \mathrm{PI}_{3} + 3 \mathrm{H}_{2} \mathrm{O} \rightarrow 3\mathrm{HI} + \mathrm{H}_{3}\mathrm{PO}_{3} \), showing that HI is generated.

Step 3: Analyze Option (c)

Consider the reaction \( \overline{\mathrm{KI}} + \mathrm{H}_{2} \mathrm{SO}_{4} \). When potassium iodide is reacted with sulfuric acid, it typically results in the formation of iodine, sulfur dioxide, potassium bisulfate, and water, rather than HI. This is due to the oxidizing nature of \( \mathrm{H}_{2} \mathrm{SO}_{4} \). Thus, HI is not formed here.

Step 4: Analyze Option (d)

Consider the reaction \( \mathrm{H}_{2} + \mathrm{I}_{2} \stackrel{\mathrm{Pt}}{\rightarrow} 2\mathrm{HI} \). In this catalytic reaction using platinum, hydrogen gas reacts with iodine gas to form hydrogen iodide. This indicates that HI is produced in this method.

Step 5: Conclusion

Based on the analysis, options (a), (b), and (d) are suitable methods for preparing HI, whereas option (c) does not produce HI.

Key Concepts

Chemical Reactions

Understanding chemical reactions is crucial for the study of chemistry. They involve breaking bonds in reactants and forming new bonds to create products. For example, in the reaction:

• Iodine (I_2) reacts with hydrogen sulfide (H_2S) to form hydrogen iodide (HI) and sulfur (S).
• Similarly, phosphorus triiodide (PI_3) reacts with water (H_2O) to produce HI and phosphorous acid (H_3PO_3).
• Hydrogen (H_2) and iodine (I_2) gases can react to form HI when catalyzed by platinum.

These reactions are about rearranging atoms to form new substances. Balance is key; the same number of each type of atom must appear on both sides of the equation. In each of these reactions, hydrogen iodide is a crucial product.

Iodine Chemistry

Iodine is a versatile element in chemistry, known for its reactivity with various elements and compounds. It's utilized in multiple synthesis reactions to produce valuable compounds such as hydrogen iodide (HI).

• In the formation of HI using iodine, it's combined with H_2S or directly with H_2 in the presence of a catalyst.
• Phosphorus triiodide (PI_3) can also release iodine atoms in reaction with water, forming HI.
• Iodine can oxidize other substances, which is why when combined with potassium iodide (KI) and sulfuric acid (H_2SO_4), elemental iodine rather than HI is produced.

The chemistry of iodine showcases its importance in forming both simple and more complex compounds. It plays a significant role in chemical synthesis and analytical chemistry.

Catalysis

Catalysis is pivotal to speeding up chemical reactions without being consumed in the process. Catalysts provide an alternative pathway with a lower activation energy, making reactions faster. A prime example from this exercise is the use of platinum

• Platinum is used as a catalyst in the reaction of hydrogen and iodine gases.
• This pathway allows the transformation into HI, which might otherwise take a lot longer without the catalyst.

Catalysts are incredibly important in the chemical industry and scientific research because they increase efficiency and reduce the energy required for reactions. Platinum is especially effective thanks to its unique properties and ability to facilitate the exchange of electrons.

Sulfuric Acid Reactions

Sulfuric acid (H_2SO_4) is one of the most widely used chemicals in various industrial reactions. It is known for its strong oxidizing properties which can influence different chemical pathways.

• For instance, when H_2SO_4 is combined with potassium iodide (KI), it typically results in the formation of iodine rather than hydrogen iodide due to sulfuric acid's ability to oxidize iodide ions.
• The outcome is different products: iodine, potassium bisulfate, sulfur dioxide, and water, avoiding the formation of HI.

Understanding the interactions of sulfuric acid is crucial, as it often dictates the end product in a reaction involving oxidizing or reducing conditions. It is a key reagent in many processes due to its reactivity and versatility.