Question

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) can be obtained by the thermal decomposition of ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\). (a) Write a balanced equation for the reaction. (b) In a certain experiment, a student obtains \(0.340 \mathrm{~L}\) of the gas at \(718 \mathrm{mmHg}\) and \(24^{\circ} \mathrm{C}\). If the gas weighs \(0.580 \mathrm{~g}\), calculate the value of the gas constant.

Short answer

(a) \( \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2\text{H}_2\text{O} \). (b) \( R \approx 0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \).

Step-by-step solution

Write the Unbalanced Equation

Start by writing the chemical formula for the decomposition reaction of ammonium nitrate. The reaction is \( \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O} \).

Balance the Chemical Equation

Balance the equation by ensuring the number of atoms for each element is the same on both sides: \( \text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2\text{H}_2\text{O} \).

Use Ideal Gas Law

Apply the ideal gas law \( PV = nRT \) to find \( R \), where \( P = 718 \, \text{mmHg} \), \( V = 0.340 \, \text{L} \), \( T = 24^{\circ}\text{C} \) (converted to Kelvin).

Convert Pressure and Temperature

Convert pressure from mmHg to atm: \( 718 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 0.9447 \, \text{atm} \). Convert temperature: \( 24^{\circ}\text{C} = 24 + 273.15 = 297.15 \, \text{K} \).

Calculate Moles of Gas

Calculate moles of \( \text{N}_2\text{O} \) using its molar mass. The molar mass of \( \text{N}_2\text{O} \approx 44.01 \, \text{g/mol} \). Moles \( n = \frac{0.580 \, \text{g}}{44.01 \, \text{g/mol}} = 0.01318 \, \text{mol} \).

Solve for Gas Constant \( R \)

Use the equation \( R = \frac{PV}{nT} \). Substituting the values: \( R = \frac{(0.9447 \, \text{atm})(0.340 \, \text{L})}{(0.01318 \, \text{mol})(297.15 \, \text{K})} \approx 0.0821 \, \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}} \).

Key concepts

Balanced Chemical Equation

A balanced chemical equation is crucial for understanding and predicting the products of a chemical reaction. When dealing with reactions, it’s important to ensure that the number of atoms for each element is equal on both sides of the equation. This represents the principle of conservation of mass, meaning that matter cannot be created or destroyed.
In our exercise, we’re looking at the thermal decomposition of ammonium nitrate (\(\text{NH}_4\text{NO}_3\)). The unbalanced equation can be written as: \[\text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}.\] Balancing this equation involves ensuring that the number of nitrogen, hydrogen, and oxygen atoms are the same on each side. Here is how it is balanced: \[\text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2\text{H}_2\text{O}.\] In this balanced equation, the nitrogen and oxygen atoms are equal on both sides, meeting the conservation of mass requirements.

Gas Constant Calculation

The gas constant, represented by \( R \), is vital when working with gases, particularly when using the ideal gas law. The ideal gas law is given by: \[PV = nRT,\]where:

• \( P \) is the pressure,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( T \) is the temperature.

In this exercise, the pressure is initially given in mmHg, which you convert to atm to work with standard units (1 atm = 760 mmHg). Thus, \[718 \, \text{mmHg} \times \frac{1 \, \text{atm}}{760 \, \text{mmHg}} = 0.9447 \, \text{atm}.\]Similarly, temperature in Celsius is converted to Kelvin by adding 273.15, so \[24^{\circ}C = 297.15 \, K.\]Once you have these values, you can solve for \( R \) using \[R = \frac{PV}{nT}.\]Substituting the values gives \[R \approx 0.0821 \, \frac{L\cdot atm}{mol\cdot K},\]matching the typical value for \( R \) when using these units.

Decomposition Reaction

Decomposition reactions are a type of chemical reaction where one compound breaks down into two or more simpler substances. This process is crucial in chemistry for understanding how complex molecules can form simpler products.
In the exercise, ammonium nitrate (\(\text{NH}_4\text{NO}_3\)) undergoes thermal decomposition. This means it decomposes when heated. The equation for this reaction is:\[\text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2\text{H}_2\text{O}.\]The decomposition reaction here produces nitrous oxide (\(\text{N}_2\text{O}\)) and water (\(\text{H}_2\text{O}\)). Decomposition reactions are important for many industrial processes, such as the production of agricultural chemicals, and in laboratory settings for understanding chemical behaviors.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, and it's typically expressed in grams per mole (\(\text{g/mol}\)). Calculating the molar mass correctly is essential for converting mass into moles, which is a key step in many equations and reactions.
For nitrous oxide, \(\text{N}_2\text{O}\), you calculate the molar mass by adding the atomic masses of all the atoms in the molecule:

• Nitrogen \((\text{N})\) has an atomic mass of about \(14.01 \, \text{g/mol}\), and there are two nitrogen atoms in \(\text{N}_2\text{O}\), contributing \(28.02 \, \text{g/mol}\).
• Oxygen \((\text{O})\) has an atomic mass of \(16.00 \, \text{g/mol}\), adding up to the total.

Thus, the molar mass of \(\text{N}_2\text{O}\) is approximately:\[28.02 \, \text{g/mol} + 16.00 \, \text{g/mol} = 44.01 \, \text{g/mol}.\]This value is essential for calculating the number of moles from a given mass of a gas, which is a frequent requirement in stoichiometry and gas law calculations.