Problem 18 - Chemistry: The Central Science

Question

Hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) and dinitrogentetroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) form a self-igniting mixture that has been used as a rocket propellant. The reaction products are \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Write a balanced chemical equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance serves as the reducing agent, and which as the oxidizing agent?

Step-by-Step Solution

Answer

(a) The balanced chemical equation for this reaction is: \(N_2H_4 + 2 N_2O_4 \rightarrow 3N_2 + 4H_2O\) (b) In this reaction, hydrazine (\(N_2H_4\)) is being oxidized and dinitrogen tetroxide (\(N_2O_4\)) is being reduced. (c) Hydrazine (\(N_2H_4\)) serves as the reducing agent, and dinitrogen tetroxide (\(N_2O_4\)) serves as the oxidizing agent.

Step 1: 1: Writing the balanced chemical equation

First, write the unbalanced chemical equation, which has the reactants and products in their respective sides: \(N_2H_4 + N_2O_4 \rightarrow N_2 + H_2O\) Now we need to balance it. Start by balancing the nitrogen atoms. We have: \(1 \times N_2H_4 + 2 \times N_2O_4 \rightarrow 3 \times N_2 + H_2O\) Next, balance the hydrogen and oxygen atoms. We will need to put a '4' coefficient in front of \(H_2O\): \(1 \times N_2H_4 + 2 \times N_2O_4 \rightarrow 3 \times N_2 + 4 \times H_2O\) Thus the balanced chemical equation is: \( N_2H_4 + 2 N_2O_4 \rightarrow 3N_2 + 4H_2O\)

Step 2: 2: Identifying oxidization and reduction

We need to find out what is being oxidized and what is being reduced. To do this, we should first determine the oxidation number of each element in the reactants and products. - In \(N_2H_4\), N has an oxidation number of -2 and H has an oxidation number of +1. - In \(N_2O_4\), N has an oxidation number of +4 and O has an oxidation number of -2. - In \(N_2\) molecule, N has an oxidation number of 0. - In \(H_2O\), H has an oxidation number of +1 and O has an oxidation number of -2. Now, comparing the oxidation numbers before and after the reaction, we find that: - Nitrogen in Hydrazine is reduced from -2 to 0 (decrease in oxidation number) - Nitrogen in Dinitrogen tetroxide is oxidized from +4 to 0 (increase in oxidation number) So, in this reaction, \(N_2H_4\) is being oxidized, and \(N_2O_4\) is being reduced.

Step 3: 3: Determining the reducing and oxidizing agents

Now that we know what is being oxidized and what is being reduced, we can determine the reducing and oxidizing agents. The substance that is being oxidized (loses electrons) is the reducing agent. In this reaction, the reducing agent is \(N_2H_4\). The substance that is being reduced (gains electrons) is the oxidizing agent. In this reaction, the oxidizing agent is \(N_2O_4\). So, in this reaction, Hydrazine (\( N_2H_4\)) acts as a reducing agent, and Dinitrogen tetroxide (\( N_2O_4\)) acts as an oxidizing agent.

Key Concepts

Balanced Chemical Equation

In a chemical reaction, a balanced equation ensures that the number of atoms of each element is the same on both sides of the equation. This follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

To balance the reaction between hydrazine (N_2H_4) and dinitrogen tetroxide (N_2O_4), we start by writing the unbalanced equation:

• \(N_2H_4 + N_2O_4 \rightarrow N_2 + H_2O\)

Next, we balance the nitrogen atoms by adjusting the coefficients:

• \(1 \times N_2H_4 + 2 \times N_2O_4 \rightarrow 3 \times N_2 + H_2O\)

Finally, balance the hydrogen and oxygen atoms:

• \(1 \times N_2H_4 + 2 \times N_2O_4 \rightarrow 3 \times N_2 + 4 \times H_2O\)

Thus, the balanced chemical equation is:

• \(N_2H_4 + 2N_2O_4 \rightarrow 3N_2 + 4H_2O\)

Oxidation and Reduction

Oxidation and reduction are processes in which electrons are transferred between substances. These processes always occur together in a redox (reduction-oxidation) reaction.

Oxidation involves the loss of electrons, while reduction involves the gain of electrons. To identify these changes, we examine the oxidation numbers of elements in a reaction.

In our reaction:

• Hydrazine ( N_2H_4 ): Nitrogen goes from an oxidation number of -2 to 0, indicating reduction (gain of electrons).
• Dinitrogen tetroxide ( N_2O_4 ): Nitrogen goes from +4 to 0, showing oxidation (loss of electrons).

This illustrates that hydrazine is oxidized, while dinitrogen tetroxide is reduced during the process.

Oxidizing and Reducing Agents

In redox reactions, it's important to identify the substances acting as oxidizing and reducing agents. These agents influence the electron transfers that drive the reaction.

The reducing agent is the substance that donates electrons, thus becoming oxidized itself. For this reaction, hydrazine ( N_2H_4 ) serves as the reducing agent because it loses electrons as its nitrogen's oxidation number increases.

Conversely, the oxidizing agent is the substance that accepts electrons, thus becoming reduced itself. Dinitrogen tetroxide ( N_2O_4 ) is the oxidizing agent here, as it gains electrons with a decrease in nitrogen's oxidation number.

• Reducing Agent: Hydrazine ( N_2H_4 )
• Oxidizing Agent: Dinitrogen tetroxide ( N_2O_4 )

Understanding these roles helps us analyze how redox reactions occur and what changes substances undergo in chemical processes.