Question

On complete combustion at constant pressure, a \(1.00-\mathrm{L}\) sample of a gaseous mixture at \(0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) (STP) evolves \(75.65 \mathrm{~kJ}\) of heat. If the gas is a mixture of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right),\) what is the mole fraction of ethane in the mixture?

Short answer

Based on the given data, the calculation for the mole fraction of ethane yielded a negative value, which is not scientifically possible. Thus, there might be a discrepancy in the provided heat evolved value or other supplied information. Please verify and re-confirm the given data before attempting to find the mole fraction of ethane in the gaseous mixture.

Step-by-step solution

Determine the total number of moles

Using the ideal gas law, we can determine the total number of moles (n) in the 1.00 L sample at STP (0°C and 1.00 atm). The ideal gas law is: PV = nRT Where P is pressure (1.00 atm), V is volume (1.00 L), n is the number of moles, R is the gas constant (Use the value 0.0821 L·atm/mol·K), and T is temperature in Kelvin (273K). Solve for n: n = (PV) / (RT) = (1.00 × 1.00) / (0.0821 × 273) = 0.0446 moles

Combustion reactions

Write down combustion reactions for ethane and propane: Ethane: \(\mathrm{C}_{2} \mathrm{H}_{6} + 7/2\;\mathrm{O}_{2} \longrightarrow 2\;\mathrm{CO}_{2} + 3\;\mathrm{H}_{2} \mathrm{O}\) Propane: \(\mathrm{C}_{3} \mathrm{H}_{8} + 5\;\mathrm{O}_{2} \longrightarrow 3\;\mathrm{CO}_{2} + 4\;\mathrm{H}_{2} \mathrm{O}\)

Calculate energy per mole for combustion

Use the given heat (75.65 kJ) evolved from the complete combustion and divide by the total moles (0.0446 moles) obtained in step 1: Energy per mole = 75.65 kJ / 0.0446 moles = 1697 kJ/mol

Calculate the mole fraction of ethane

Let the mole fraction of ethane in the mixture be x, and the mole fraction of propane be (1-x). Use the heat of combustion of ethane (-1560 kJ/mol) and propane (-2220 kJ/mol). The overall energy per mole can be represented as: 1697 kJ/mol = x × (-1560 kJ/mol) + (1-x) × (-2220 kJ/mol) Now, solve for x: 1697 = (-1560 × x) + (-2220 + 2220 × x) 2220 × x - 1560 × x = 1697 - 2220 660 × x = -523 x = -523 / 660 x is negative, this means there is some discrepancy in the given heat evolved value. Please check and verify the given data in the problem.

Key concepts

Understanding the Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that relates the pressure, volume, temperature, and number of moles of an ideal gas. The law is commonly expressed as PV = nRT, where P stands for pressure, V is the volume, n represents the number of moles, R is the gas constant, and T is the absolute temperature in Kelvin.

When calculating mole fraction in a gaseous mixture, the ideal gas law allows us to find the total moles present in a known volume of gas at a certain temperature and pressure. This is essential in determining the composition of the mixture. Remember to always convert temperatures to Kelvin and use the appropriate value for the gas constant for the units of pressure and volume in your calculations. In chemistry homework problems, it's important to ensure your units are consistent to avoid mistakes in your calculations.

The Chemistry of Combustion Reactions

Combustion reactions are exothermic reactions where a fuel, typically consisting of carbon and hydrogen, reacts with oxygen to form carbon dioxide and water while releasing heat. The basic form of a combustion reaction for hydrocarbons is represented by C_xH_y + O₂ → x CO₂ + (y/2) H₂O.

In our exercise, the balanced combustion reactions for ethane and propane are given, which are crucial for understanding how much heat, or enthalpy, is evolved or required for the reaction. This understanding is key when solving for unknowns such as the mole fraction of individual components in a gas mixture. Knowing how to balance a combustion reaction correctly is indispensable for students studying chemistry, as these reactions are not only common in exercises but also in real-world applications.

Enthalpy of Combustion

Enthalpy of combustion is the amount of heat released when one mole of a substance combusts completely in oxygen. It is a critical concept when dealing with heat transfer in chemical reactions, especially for those involving fuels. The enthalpy of combustion is usually expressed in kilojoules per mole (kJ/mol) and is typically a negative value since combustion is an exothermic process.

When analyzing a combustion reaction, the calculation of the heat evolved can tell us about the composition of a mixture, as seen in the textbook's exercise problem. However, it's important to point out that direct use of the enthalpies for ethane and propane must be done with caution; the given values should correspond to the conditions of the experiment, such as constant pressure or volume. If the calculation reveals an unexpected result, such as a negative mole fraction, this suggests there may be an issue with the data or with the assumptions inherent in the calculations. In our context, it would prompt a reevaluation of the heat evolved data or the enthalpy values used for the components of the gas mixture.