Problem 97 - General Chemistry: Principles and Modern Applications

Question

A 3.05 g sample of \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s})\) is introduced into an evacuated 2.18 L flask and then heated to \(250^{\circ} \mathrm{C}\).What is the total gas pressure, in atmospheres, in the flask at \(250^{\circ} \mathrm{C}\) when the \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) has completely decomposed? $$\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s}) \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$

Step-by-Step Solution

Answer

The total gas pressure, in atmospheres, in the flask at \(250^{\circ} \mathrm{C}\) when the \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) has completely decomposed is \(1.25 atm\).

Step 1: Finding the Number of Moles of \(\mathrm{NH}_{4} \(\mathrm{NO}_{3}\)

Begin by calculating the number of moles of the compound \(\mathrm{NH}_{4} \(\mathrm{NO}_{3}\). Molar mass of \(\mathrm{NH}_{4} \(\mathrm{NO}_{3}\) is approximately 80.052 g/mol. So, the number of moles can be calculated using the formula \(\text{Number of moles = Mass / Molar Mass}\). Upon substituting the given values, we get \(\text{Number of moles = }3.05 g /\ 80.052 g/mol = 0.0381 mol\).

Step 2: Applying the Stoichiometry

Next, apply the stoichiometry of the reaction. The balanced chemical equation shows that 1 mole of the reactant produces 1 mole of \(\mathrm{N}_{2}\mathrm{O}(\mathrm{g})\) and 2 moles of \(\mathrm{H}_{2}\mathrm{O}(\mathrm{g})\) upon decomposition. Therefore, the total moles of gas produced = 0.0381 mol \( + 2 * 0.0381 mol = 0.1143 mol\).

Step 3: Ideal Gas Law Calculation

Finally, use the ideal gas law equation \(P V = nRT\) to solve for the pressure. In this equation, P represents the pressure we want to determine, V is the volume which is 2.18 L, n is the number of moles calculated in the previous step, R is the gas constant which is about 0.0821 L.atm/mol.K, and T is the temperature in Kelvin, which is 250^0 C + 273 = 523 K. Substituting these values, we get \(P = nRT / V\). Thus, \(P = 0.1143 mol * 0.0821 L.atm/(mol.K) * 523 K / 2.18 L\), which gives the pressure = 1.25 atm.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a crucial equation in chemistry, widely used to calculate the behavior of gases under different conditions. It is expressed as \( PV = nRT \), where:

• \( P \) represents the pressure of the gas,
• \( V \) is the volume the gas occupies,
• \( n \) stands for the number of moles of gas present,
• \( R \) is the ideal gas constant, approximately equal to 0.0821 L.atm/mol.K, and
• \( T \) is the temperature in Kelvin.

When using the ideal gas law, it's important to convert temperature from Celsius to Kelvin by adding 273. This formula helps predict the pressure, volume, or temperature of a gas given the other parameters. It's a theoretical model that assumes gases behave ideally, which means no interactions between molecules and that they occupy no volume.

Chemical Stoichiometry

Chemical stoichiometry deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to use balanced chemical equations to quantify the amounts of substances involved. For the decomposition of ammonium nitrate \( (\mathrm{NH}_4\mathrm{NO}_3) \), the balanced chemical equation shows that it decomposes to produce nitrous oxide \( (\mathrm{N}_2\mathrm{O}) \) and water \( (\mathrm{H}_2\mathrm{O}) \) in specific mole ratios:

• 1 mole of \( \mathrm{NH}_4\mathrm{NO}_3 \) produces 1 mole of \( \mathrm{N}_2\mathrm{O} \).
• 1 mole of \( \mathrm{NH}_4\mathrm{NO}_3 \) produces 2 moles of \( \mathrm{H}_2\mathrm{O} \).

These ratios allow us to determine how much product forms from a given amount of reactant. Here, 0.0381 mol of ammonium nitrate yields 0.1143 mol of gaseous products when decomposed completely.

Thermal Decomposition

Thermal decomposition is a chemical process in which a compound breaks down into two or more products when heated. Ammonium nitrate \( (\mathrm{NH}_4\mathrm{NO}_3) \) thermally decomposes when heated to produce nitrous oxide \( (\mathrm{N}_2\mathrm{O}) \) and water \( (\mathrm{H}_2\mathrm{O}) \) in gaseous form. This type of reaction is endothermic, meaning it absorbs heat, triggering the breakdown of the compound. By permanently altering its structure, it forms simpler substances. Understanding thermal decomposition is essential for safely conducting reactions and predicting the products formed at high temperatures.

Mole Calculation

Mole calculation is fundamental in chemistry and involves determining the number of moles in a given substance. This is useful for converting masses of reactants and products using their molar masses. The molar mass, a substance's weight for one mole, is crucial for these calculations.In the exercise, to find the number of moles of ammonium nitrate \( (\mathrm{NH}_4\mathrm{NO}_3) \), we use its molar mass, about 80.052 g/mol:\[\text{Number of moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.05 \text{ g}}{80.052 \text{ g/mol}} = 0.0381 \text{ mol}\]This calculation forms the basis for stoichiometric calculations, linking masses to moles and allowing for further analysis of chemical reactions.