Problem 4 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Predict the products and write balanced equations for the following reactions (Section 25.2): (a) \(\quad \mathrm{CsH}+\) water (b) \(\quad B_{2} H_{6}+\) pyridine (c) \(\quad \mathrm{S}_{2} \mathrm{H}_{6}+\) oxygen (d) \(\quad \mathrm{N}_{2} \mathrm{H}_{4}+\mathrm{HNO}_{3}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) Products: \(\text{CsOH} + \text{H}_2\). (b) Products: \(\text{B}_2\text{H}_6\cdot 2 \text{C}_5\text{H}_5\text{N}\). (c) Products: \(\text{SO}_2 + \text{H}_2\text{O}\). (d) Products: \(\text{N}_2 + \text{H}_2\text{O}\).

Step 1: Determine Reaction Products for CsH + Water

When cesium hydride \(\text{CsH}\) reacts with water \(\text{H}_2\text{O}\), it forms cesium hydroxide \(\text{CsOH}\) and hydrogen gas \(\text{H}_2\). The balanced equation for this reaction is: \[ \text{CsH} + \text{H}_2\text{O} \rightarrow \text{CsOH} + \text{H}_2} \]

Step 2: Determine Reaction Products for \(B_2H_6\) + Pyridine

Diborane \(\text{B}_2\text{H}_6\) is known to form a complex with pyridine. The likely product is a compound where one borane is associated with pyridine: \(\text{B}_2\text{H}_6\cdot 2 \text{C}_5\text{H}_5\text{N}\). The balanced equation is: \[\text{B}_2\text{H}_6 + 2\text{C}_5\text{H}_5\text{N} \rightarrow \text{B}_2\text{H}_6\cdot 2\text{C}_5\text{H}_5\text{N} \]

Step 3: Determine Reaction Products for \(S_2H_6\) + Oxygen

Disulfur hexahydride \(\text{S}_2\text{H}_6\) will react with oxygen \(\text{O}_2\) to produce sulfur dioxide \(\text{SO}_2\) and water \(\text{H}_2\text{O}\). The balanced chemical equation is: \[ \text{S}_2\text{H}_6 + 5\text{O}_2 \rightarrow 2\text{SO}_2 + 3\text{H}_2\text{O} \]

Step 4: Determine Reaction Products for \(N_2H_4\) + \(HNO_3 (aq)\)

The hydrazine \(\text{N}_2\text{H}_4\) and nitric acid \(\text{HNO}_3\) reaction is a redox reaction forming nitrogen \(\text{N}_2\), nitrous oxide \(\text{N}_2\text{O}\), and water. The balanced equation is: \[ 3\text{N}_2\text{H}_4 + 4\text{HNO}_3 \rightarrow 4\text{N}_2 + 4\text{H}_2\text{O} \]

Key Concepts

Balancing Chemical Equations

Understanding how to balance chemical equations is a crucial skill in chemistry. This process ensures that the number of atoms of each element is the same on both sides of the equation. It reflects the principle of the conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
Let's dive into the approach using cesium hydride's reaction with water as an example:

• Recognize the reactants and products. For \(\text{CsH} + \text{H}_2\text{O} \rightarrow \text{CsOH} + \text{H}_2\), the reactants are cesium hydride and water, and the products are cesium hydroxide and hydrogen gas.
• Count the number of each type of atom in the reactants and products. Initially, you might find different counts on both sides.
• Add coefficients to balance the equation. Here, each compound is already balanced with one molecule of reactant yielding one molecule of product.

The balanced equation should read: \[\text{CsH} + \text{H}_2\text{O} \rightarrow \text{CsOH} + \text{H}_2}\]
Balancing ensures that reactions are represented accurately and helps predict the amounts of substances involved in reactions, which is fundamental for lab work and industrial applications.

Inorganic Chemistry

Inorganic chemistry focuses on the behavior and synthesis of inorganic compounds, which aren't based on carbon-hydrogen bonds. These include metals, minerals, and organometallic compounds.
An interesting reaction in this realm is between diborane \(\text{B}_2\text{H}_6\) and pyridine. Diborane is a boron compound with hydrogen, and pyridine is a nitrogen-containing heterocyclic compound. Here, they form a coordination complex:

• Coordination complexes are structures where molecules like pyridine can donate a lone pair of electrons to central atoms like boron.
• The balanced reaction is \(\text{B}_2\text{H}_6 + 2\text{C}_5\text{H}_5\text{N} \rightarrow \text{B}_2\text{H}_6\cdot 2\text{C}_5\text{H}_5\text{N}\), showing how two pyridine molecules coordinate with diborane.

Inorganic chemistry often deals with unique bonding situations and diverse reaction types, making it a vibrant area of study. Whether involving acids or bases, metals or nonmetals, its scope is vast and essential for understanding material properties and reactivity.

Reaction Mechanism

A reaction mechanism provides a step-by-step description of how a chemical reaction occurs. It includes all the intermediates and states that particles go through during the overall transformation from reactants to products.
Consider the reaction between disulfur hexahydride \(\text{S}_2\text{H}_6\) and oxygen \(\text{O}_2\), resulting in sulfur dioxide and water:

• Initially, \(\text{S}_2\text{H}_6\) interacts with \(\text{O}_2\) to form sulfur intermediate compounds and hydrogen.
• Eventually, it forms stable products: \(\text{SO}_2\) and \(\text{H}_2\text{O}\).
• The balanced equation \(\text{S}_2\text{H}_6 + 5\text{O}_2 \rightarrow 2\text{SO}_2 + 3\text{H}_2\text{O}\) represents the end result of this stepwise process.

This process sheds light on not just what products are formed, but how bonds are broken and new ones formed. Understanding reaction mechanisms is fundamental for predicting product ratios, reaction conditions, and for synthesizing desired compounds effectively.