Problem 74 - Chemistry and Chemical Reactivity

Question

A miniature volcano can be made in the laboratory with ammonium dichromate. When ignited, it decomposes in a fiery display. $$ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s}) $$ If 0.95 g of ammonium dichromate is used and the gases from this reaction are trapped in a 15.0 -L flask at \(23^{\circ} \mathrm{C},\) what is the total pressure of the gas in the flask? What are the partial pressures of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\)

Step-by-Step Solution

Answer

Total pressure is 0.03054 atm; \(P_{\mathrm{N}_2} = 0.00611\) atm, \(P_{\mathrm{H}_2\mathrm{O}} = 0.02443\) atm.

Step 1: Molar Mass Calculation

Calculate the molar mass of ammonium dichromate, \((\mathrm{NH}_4)_2 \mathrm{Cr}_2 \mathrm{O}_7\). The molar masses are approximately: N = 14.01 g/mol, H = 1.01 g/mol, Cr = 51.996 g/mol, O = 16.00 g/mol. \((\mathrm{NH}_4)_2 \mathrm{Cr}_2 \mathrm{O}_7=(2 \times (14.01 + 4 \times 1.01)) + (2 \times 51.996) + (7 \times 16.00) = 252.06 \, \text{g/mol}\).

Step 2: Moles of Ammonium Dichromate

Calculate the moles of ammonium dichromate using the mass given: \(\text{moles} = \frac{0.95 \text{ g}}{252.06 \text{ g/mol}} = 0.00377 \text{ mol}\).

Step 3: Stoichiometry

From the balanced equation, 1 mole of \((\mathrm{NH}_4)_2 \mathrm{Cr}_2 \mathrm{O}_7\) decomposes to form 1 mole of \(\mathrm{N}_2\) and 4 moles of \(\mathrm{H}_2\mathrm{O}\). Therefore, 0.00377 mol of \((\mathrm{NH}_4)_2 \mathrm{Cr}_2 \mathrm{O}_7\) produces \(0.00377 \text{ mol of } \mathrm{N}_2\) and \((4 \times 0.00377) \text{ mol of } \mathrm{H}_2\mathrm{O} = 0.01508 \text{ mol}\).

Step 4: Ideal Gas Law for Total Pressure

Use the ideal gas law, \(PV = nRT\), where \(R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\). Convert temperature to Kelvin: \(23^\circ \mathrm{C} = 296 \text{ K}\). Total moles of gas \(= 0.00377 + 0.01508 = 0.01885 \text{ mol}\). Substitute into the equation: \(P \times 15.0 = 0.01885 \times 0.0821 \times 296\), solve for \(P\): \(P = 0.03054 \text{ atm}\).

Step 5: Partial Pressure of \(\mathrm{N}_2\)

Use the formula for partial pressure: \(P_{\mathrm{N}_2} = \frac{n_{\mathrm{N}_2}}{n_{\text{total}}} \times P_{\text{total}}\). \(P_{\mathrm{N}_2} = \frac{0.00377}{0.01885} \times 0.03054 \approx 0.00611 \text{ atm}\).

Step 6: Partial Pressure of \(\mathrm{H}_2\mathrm{O}\)

Similarly, calculate the partial pressure: \(P_{\mathrm{H}_2\mathrm{O}} = \frac{n_{\mathrm{H}_2\mathrm{O}}}{n_{\text{total}}} \times P_{\text{total}}\). \(P_{\mathrm{H}_2\mathrm{O}} = \frac{0.01508}{0.01885} \times 0.03054 \approx 0.02443 \text{ atm}\).

Key Concepts

Ammonium Dichromate Decomposition

Ammonium dichromate, known for its dramatic decomposition reaction, is often used in demonstration experiments. This compound, represented as \((\mathrm{NH}_4)_2 \mathrm{Cr}_2 \mathrm{O}_7\), undergoes a decomposition reaction when ignited: \[ \left( \mathrm{NH}_4 \right)_2 \mathrm{Cr}_2 \mathrm{O}_7 (\mathrm{s}) \rightarrow \mathrm{N}_2 (\mathrm{g}) + 4 \mathrm{H}_2 \mathrm{O} (\mathrm{g}) + \mathrm{Cr}_2 \mathrm{O}_3 (\mathrm{s}) \] This reaction is exothermic and results in a fiery eruption, releasing nitrogen and water vapor while leaving chromium(III) oxide as a solid residue. The decomposition process is favored because it relieves the structural strain in the dichromate, releasing gases that expand dramatically. Here’s a breakdown of what's happening:

• Exothermic Reaction: The heat generated propagates the reaction, causing a visible fiery display.
• Gas Evolution: Nitrogen and water vapor are produced, contributing to the explosive effect.
• Solid Residue: Chromium(III) oxide is created, seen as the green ash left behind.

Understanding this decomposition is critical, as it sets the stage for calculating pressures of the gases produced.

Partial Pressure Calculation

In the context of the decomposition of ammonium dichromate, the calculation of partial pressures of the gases released is crucial. When a gas mixture is confined, such as in a flask, each gas exerts its own pressure. This is called its partial pressure. The total pressure is the sum of these partial pressures. The ideal gas law \(PV = nRT\) allows us to calculate the total pressure exerted by the gas mixture. In the provided scenario:

• Total Pressure: First, the moles of each gas are determined. The total number of moles of gas is then used in the ideal gas equation to find the total pressure.
• Partial Pressure of \(\mathrm{N}_2\): Calculated as \(P_{\mathrm{N}_2} = \frac{n_{\mathrm{N}_2}}{n_{\text{total}}} \times P_{\text{total}}\).
• Partial Pressure of \(\mathrm{H}_2\mathrm{O}\): Similarly calculated using \(P_{\mathrm{H}_2\mathrm{O}} = \frac{n_{\mathrm{H}_2\mathrm{O}}}{n_{\text{total}}} \times P_{\text{total}}\).

These calculations are essential for understanding how much each gas contributes to the total pressure in a closed system.

Stoichiometry

Stoichiometry is a key concept in solving chemical equations and understanding the proportions in which chemicals react. In the decomposition of ammonium dichromate, stoichiometry helps us determine the amounts of nitrogen and water vapor produced. The reaction equation indicates that one mole of ammonium dichromate yields one mole of \(\mathrm{N}_2\) and four moles of \(\mathrm{H}_2\mathrm{O}\). By using stoichiometry, we can scale these relationships to match the quantity of reactant used. Here's a step-by-step application:

• Mole Ratios: From the balanced equation, 1 mole of \((\mathrm{NH}_4)_2 \mathrm{Cr}_2 \mathrm{O}_7\) produces 1 mole of \(\mathrm{N}_2\) and 4 moles of \(\mathrm{H}_2\mathrm{O}\).
• Calculating Moles: The moles of ammonium dichromate available are calculated from its mass, then these moles are used to find the moles of each product gas.
• Predicting Gas Production: Given the moles of reactant, stoichiometry predicts the exact moles of each gaseous product formed.

This approach ensures accuracy in chemical calculations, aiding in predicting reaction outcomes and facilitating pressure calculations in gas mixtures.