Nitrogen dioxide dimerizes to give dinitrogen tetroxide: 2 NO2(g) ⇌ N2O4(g). At 298 K, 9.66 g of an NO2 and N2O4 mixture exerts a pressure of 0.487 atm in a volume of 6.51 L. What are the mole fractions of the two gases in the mixture?

Verified step by step guidance

Step 1: Calculate the total number of moles of the gas mixture using the ideal gas law, PV = nRT. Use the given pressure (P), volume (V), and temperature (T) to find the total moles (n).

Step 2: Determine the molar masses of NO2 and N2O4. NO2 has a molar mass of approximately 46.01 g/mol, and N2O4 has a molar mass of approximately 92.02 g/mol.

Step 3: Set up a system of equations based on the total mass and total moles. Let x be the moles of NO2 and y be the moles of N2O4. The equations are: 46.01x + 92.02y = 9.66 (total mass) and x + y = total moles (from Step 1).

Step 4: Solve the system of equations to find the values of x and y, which represent the moles of NO2 and N2O4, respectively.

Step 5: Calculate the mole fractions of NO2 and N2O4. The mole fraction of NO2 is x divided by the total moles, and the mole fraction of N2O4 is y divided by the total moles.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. This law is essential for calculating the number of moles of gases in a mixture, which is necessary to determine their mole fractions. In this scenario, the pressure, volume, and temperature are provided, allowing us to find the total number of moles present in the gas mixture.

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture, defined as the ratio of the number of moles of that component to the total number of moles of all components. It is a dimensionless quantity that helps in understanding the composition of gas mixtures. In this question, calculating the mole fractions of NO2 and N2O4 requires knowing the individual moles of each gas after determining the total moles from the Ideal Gas Law.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. In the context of the dimerization of nitrogen dioxide, understanding the equilibrium expression and the relationship between the concentrations of NO2 and N2O4 is crucial for determining their mole fractions. This concept is important for interpreting the behavior of the gases in the mixture and how they relate to each other at equilibrium.

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A 5.00-L vessel contains 25.0 g of PCl3 and 3.00 g of O2 at 15 °C. The vessel is heated to 200.0 °C, and the contents react to give POCl3. What is the final pressure in the vessel, assuming that the reaction goes to completion and that all reactants and products are in the gas phase?

Textbook Question

A steel container with a volume of 500.0 mL is evacuated, and 25.0 g of CaCO₃ is added. The container and contents are then heated to 1500 K, causing the CaCO₃ to decompose completely, according to the equation CaCO₃(s) → CaO(s) + CO₂(g). (a) Using the ideal gas law and ignoring the volume of any solids remaining in the container, calculate the pressure inside the container at 1500 K.

Textbook Question

A steel container with a volume of 500.0 mL is evacuated, and 25.0 g of CaCO₃ is added. The container and contents are then heated to 1500 K, causing the CaCO₃ to decompose completely, according to the equation CaCO₃(s) → CaO(s) + CO₂(g). (b) Now make a more accurate calculation of the pressure inside the container. Take into account the volume of solid CaO (density = 3.34 g/mL) in the container, and use the van der Waals equation to calculate the pressure. The van der Waals constants for CO₂(g) are a = 3.59 (L²-atm)/mol² and b = 0.0427 L/mol.

Textbook Question

An empty 4.00-L steel vessel is filled with 1.00 atm of CH₄(g) and 4.00 atm of O₂(g) at 300 °C. A spark causes the CH₄ to burn completely, according to the equation

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) ΔH° = -802 kJ

(a) What mass of CO₂(g) is produced in the reaction?

Textbook Question

An empty 4.00-L steel vessel is filled with 1.00 atm of CH₄(g) and 4.00 atm of O₂(g) at 300 °C. A spark causes the CH₄ to burn completely, according to the equation

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) ΔH° = -802 kJ

(b) What is the final temperature inside the vessel after combustion, assuming that the steel vessel has a mass of 14.500 kg, the mixture of gases has an average molar heat capacity of 21 J/(mol·°C), and the heat capacity of steel is 0.449 J/(g·°C)?

Textbook Question

An empty 4.00-L steel vessel is filled with 1.00 atm of CH₄(g) and 4.00 atm of O₂(g) at 300 °C. A spark causes the CH₄ to burn completely, according to the equation

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) ΔH° = -802 kJ