Question

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right),\) the "laughing gas" used as an anesthetic by dentists, is made by thermal decomposition of solid \(\mathrm{NH}_{4} \mathrm{NO}_{3}\). Write a balanced equation for this reaction. What are the oxidation states of \(\mathrm{N}\) in \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) and in \(\mathrm{N}_{2} \mathrm{O} ?\)

Short answer

The balanced equation is \(\text{NH}_{4}\text{NO}_{3} \rightarrow \text{N}_{2}\text{O} + 2\text{H}_{2}\text{O}\). The oxidation states of N in \(\text{NH}_{4}\text{NO}_{3}\) are -3 and +5, and in \(\text{N}_{2}\text{O}\) it is +1.

Step-by-step solution

Write the Unbalanced Equation

The thermal decomposition of \(\text{NH}_{4}\text{NO}_{3}\) produces nitrous oxide (laughing gas) and water (H_2O). The unbalanced chemical equation is: \[\text{NH}_{4}\text{NO}_{3} \rightarrow \text{N}_{2}\text{O} + \text{H}_{2}\text{O}\]

Balance the Chemical Equation

To balance the equation, ensure equal numbers of each type of atom on both sides of the equation. Start with nitrogen (N) atoms: \[\text{NH}_{4}\text{NO}_{3} \rightarrow \text{N}_{2}\text{O} + \text{H}_{2}\text{O}\] There are 2 nitrogen atoms on the product side, so we'll use the coefficient 2 for H_2O: \[\text{NH}_{4}\text{NO}_{3} \rightarrow \text{N}_{2}\text{O} + 2\text{H}_{2}\text{O}\]

Verify the Balance

Double-check that the number of each type of atom matches on both sides: \[\text{NH}_{4}\text{NO}_{3} \rightarrow \text{N}_{2}\text{O} + 2\text{H}_{2}\text{O}\] Nitrogen atoms: 2 on each side \ Oxygen atoms: 3 on each side \ Hydrogen atoms: 4 on each side \ The equation is balanced.

Determine Oxidation States in \(\text{NH}_{4}\text{NO}_{3}\)

In \(\text{NH}_{4}\text{NO}_{3}\), nitrogen is present in two different environments: \(\text{NH}_4^+\) and \(\text{NO}_3^-\). \(\text{NH}_4^+\): the oxidation state of hydrogen (H) is +1. Since the total charge is +1, let the oxidation state of N be \(x\): \[\text{N} + 4(\text{H}) = +1 \rightarrow x + 4(+1) = +1 \rightarrow x = -3\] So, the oxidation state of nitrogen in \(\text{NH}_4^+\) is -3. \(\text{NO}_3^-\): the oxidation state of oxygen (O) is -2. Since the total charge is -1, let the oxidation state of N be \(y\): \[\text{N} + 3(\text{O}) = -1 \rightarrow y + 3(-2) = -1 \rightarrow y - 6 = -1 \rightarrow y = +5\] So, the oxidation state of nitrogen in \(\text{NO}_3^-\) is +5.

Determine Oxidation States in \(\text{N}_{2}\text{O}\)

For \(\text{N}_{2}\text{O}\), let the oxidation state of nitrogen be \(z\). The oxidation state of oxygen is -2: \[\text{N}_2 + \text{O} = 0 \rightarrow 2z + (-2) = 0 \rightarrow 2z - 2 = 0 \rightarrow 2z = 2 \rightarrow z = +1\] So, the oxidation state of nitrogen in \(\text{N}_{2}\text{O}\) is +1.

Key concepts

balanced chemical equation

A balanced chemical equation shows the exact ratio of different atoms involved in a chemical reaction. This ensures that the Law of Conservation of Mass is followed, meaning the mass of the reactants equals the mass of the products. In the case of nitrous oxide decomposition, the unbalanced chemical equation is given by:
\[ \text{NH}_{4}\text{NO}_{3} \rightarrow \text{N}_{2}\text{O} + \text{H}_{2}\text{O} \]
To balance this equation, we have to ensure there are equal numbers of each type of atom on both sides. After adding the necessary coefficients, we obtain the balanced equation:
\[ \text{NH}_{4}\text{NO}_{3} \rightarrow \text{N}_{2}\text{O} + 2\text{H}_{2}\text{O} \]
Each side has the same number of nitrogen, oxygen, and hydrogen atoms, confirming that the equation respects the conservation of mass.

oxidation states

Oxidation states (or oxidation numbers) are an important concept in chemistry. They indicate the degree of oxidation (loss of electrons) or reduction (gain of electrons) of an element in a molecule. The oxidation state is basically the hypothetical charge an atom would have if all the bonds to different atoms were completely ionic. For example, the oxidation state of oxygen is usually -2, while for hydrogen, it is usually +1. Understanding oxidation states helps us keep track of electron transfer and balance redox reactions.

thermal decomposition

Thermal decomposition is a type of chemical reaction where a compound breaks down into simpler compounds or elements when heated. In this process, heat energy is used to break the chemical bonds in the compound. For instance, nitrous oxide (\(\text{N}_{2}\text{O}\)) is produced when ammonium nitrate (\(\text{NH}_{4}\text{NO}_{3}\)) is heated. The reaction is:
\[ \text{NH}_{4}\text{NO}_{3} \rightarrow \text{N}_{2}\text{O} + 2\text{H}_{2}\text{O} \]
In this reaction, ammonium nitrate decomposes into water and nitrous oxide, allowing us to understand the changes that take place at the molecular level due to the addition of heat.

chemical reaction balance

Balancing a chemical reaction is necessary for accurately representing the reaction. It ensures that the number of atoms of each element is conserved in both the reactants and products.
Steps to balance a reaction include:

• Write the unbalanced equation.
• Count the number of atoms of each element on both sides.
• Add coefficients to balance the number of atoms of each element.
• Check the balance by counting atoms again.

In the nitrous oxide decomposition reaction, starting with the unbalanced equation:
\[ \text{NH}_{4}\text{NO}_{3} \rightarrow \text{N}_{2}\text{O} + \text{H}_{2}\text{O} \]
we see that nitrogen is already balanced but we need to balance hydrogen and oxygen by adding the coefficient 2 before \(\text{H}_{2}\text{O}\) resulting in:
\[ \text{NH}_{4}\text{NO}_{3} \rightarrow \text{N}_{2}\text{O} + 2\text{H}_{2}\text{O} \]
This verifies a balanced equation with equal atoms on both sides.

oxidation state of nitrogen

Determining the oxidation state of nitrogen in different compounds helps us understand how electrons are distributed. In ammonium nitrate (\(\text{NH}_{4}\text{NO}_{3}\)), nitrogen exists in two different environments: in \(\text{NH}_4^+\) and in \(\text{NO}_3^-\).
For \(\text{NH}_4^+\), the oxidation state of nitrogen is calculated by solving: \[ \text{N} + 4\text{H} = +1 \rightarrow x + 4(+1) = +1 \rightarrow x = -3 \] So, nitrogen has an oxidation state of -3 in \(\text{NH}_4^+\).
For \(\text{NO}_3^-\), we solve as follows: \[ \text{N} + 3\text{O} = -1 \rightarrow y + 3(-2) = -1 \rightarrow y - 6 = -1 \rightarrow y = +5 \] Here, nitrogen has an oxidation state of +5 in \(\text{NO}_3^-\). For \(\text{N}_{2}\text{O}\), let’s assume the oxidation state of nitrogen is z: \[ \text{N}_2 + \text{O} = 0 \rightarrow 2z + (-2) = 0 \rightarrow 2z - 2 = 0 \rightarrow 2z = 2 \rightarrow z = +1 \] Thus, nitrogen has an oxidation state of +1 in \(\text{N}_{2}\text{O}\). Calculation of these states is essential for understanding redox processes and reaction mechanisms.