Problem 62 - Chemistry and Chemical Reactivity

Question

Ethane burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) $$2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ (a) Four gases are involved in this reaction. Place them in order of increasing rms speed. (Assume all are at the same temperature.) (b) A \(3.26-\) - flask contains \(C_{2} H_{6}\) at a pressure of \(256 \mathrm{mm} \mathrm{Hg}\) and a temperature of \(25^{\circ} \mathrm{C} .\) Suppose \(\mathrm{O}_{2}\) gas is added to the flask until \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{O}_{2}\) are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of \(\mathrm{O}_{2}\) and what is the total pressure in the flask?

Step-by-Step Solution

Answer

(a) \(\mathrm{CO}_2, \mathrm{O}_2, \mathrm{C}_2\mathrm{H}_6, \mathrm{H}_2\mathrm{O}\). (b) \(\mathrm{O}_2: 1.194 \text{ atm}, \text{Total}: 1.531 \text{ atm}\).

Step 1: Identify the Molar Masses

Calculate the molar mass of each gas involved: \(\mathrm{C}_2\mathrm{H}_6\), \(\mathrm{O}_2\), \(\mathrm{CO}_2\), and \(\mathrm{H}_2\mathrm{O}\). Molar masses are approximately 30 g/mol for \(\mathrm{C}_2\mathrm{H}_6\), 32 g/mol for \(\mathrm{O}_2\), 44 g/mol for \(\mathrm{CO}_2\), and 18 g/mol for \(\mathrm{H}_2\mathrm{O}\).

Step 2: Determine RMS Speed Order

The root mean square (rms) speed of a gas is inversely proportional to the square root of its molar mass. Order the gases by their molar masses: \(\mathrm{CO}_2\), \(\mathrm{O}_2\), \(\mathrm{C}_2\mathrm{H}_6\), \(\mathrm{H}_2\mathrm{O}\). So, the order of increasing rms speed is \(\mathrm{CO}_2\), \(\mathrm{O}_2\), \(\mathrm{C}_2\mathrm{H}_6\), \(\mathrm{H}_2\mathrm{O}\).

Step 3: Calculate Moles of Ethane

Use the ideal gas law \(PV = nRT\) to find moles of ethane \(\mathrm{C}_2\mathrm{H}_6\). Plug in \(P = 256 \text{ mm Hg} = 0.337 \text{ atm}\), \(V = 3.26 \text{ L}\), \(T = 298 \text{ K}\), and \(R = 0.0821 \text{ L atm}\text{ mol}^{-1}\text{ K}^{-1}\). Find \(n\) as \(n = \frac{PV}{RT} = \frac{0.337 \times 3.26}{0.0821 \times 298} \approx 0.0452 \text{ mol}\).

Step 4: Calculate Required Moles of Oxygen

From the reaction equation, 2 moles of \(\mathrm{C}_2\mathrm{H}_6\) require 7 moles of \(\mathrm{O}_2\). Therefore, 0.0452 moles of \(\mathrm{C}_2\mathrm{H}_6\) require \(0.0452 \times \frac{7}{2} = 0.1582 \text{ mol}\) of \(\mathrm{O}_2\).

Step 5: Determine Partial Pressure of Oxygen

Using the ideal gas law again for \(\mathrm{O}_2\), find \(P = \frac{nRT}{V}\). Substitute \(n = 0.1582 \text{ mol}\), \(R = 0.0821 \text{ L atm}\text{ mol}^{-1}\text{ K}^{-1}\), \(T = 298 \text{ K}\), \(V = 3.26 \text{ L}\). Compute: \(P = \frac{0.1582 \times 0.0821 \times 298}{3.26} \approx 1.194 \text{ atm}\).

Step 6: Calculate Total Pressure

The total pressure in the flask is the sum of partial pressures of \(\mathrm{C}_2\mathrm{H}_6\) and \(\mathrm{O}_2\): \(0.337 + 1.194 = 1.531 \text{ atm}\).

Key Concepts

Combustion Reaction

Combustion reactions are chemical processes where a substance combines with oxygen, releasing energy in the form of light and heat. These reactions typically involve hydrocarbons, like ethane (\(\mathrm{C}_2\mathrm{H}_6\)), burning in the presence of oxygen (\(\mathrm{O}_2\)) to form carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). In the provided example, ethane undergoes combustion:

• Starting reactants: ethane and oxygen.
• Products formed: carbon dioxide and water.
• Energy released: in the form of heat or light, making it an exothermic process.

Understanding the stoichiometry of the combustion reaction is important. Here, the balanced equation \[2 \mathrm{C}_2 \mathrm{H}_6 + 7 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + 6 \mathrm{H}_2\mathrm{O}\] illustrates that two moles of ethane react with seven moles of oxygen to produce four moles of carbon dioxide and six moles of water.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation connecting pressure, volume, temperature, and amount of gas in moles through:\( PV = nRT \) where:

• \( P \): Pressure
• \( V \): Volume
• \( n \): Number of moles
• \( R \): Universal gas constant, 0.0821 L atm mol^-1 K^-1
• \( T \): Temperature in Kelvin

This formula was used to find the moles of ethane and oxygen. By rearranging it to solve for moles (\( n \): \( n = \frac{PV}{RT} \)), you can determine how much gas you have under certain conditions. Using the same equation helps predict pressures and volume changes when gases react, as seen when adding oxygen to the ethane at a calculated stoichiometric ratio for combustion.

Molar Mass

Molar mass is the mass of one mole of a given substance, expressed in grams per mole (\text{g/mol}). It is calculated by summing the atomic masses of all the atoms in a molecule. Here are the molar masses involved in the exercise:

• Ethane (\(\mathrm{C}_2\mathrm{H}_6\)): Approximately 30 g/mol
• Oxygen (\(\mathrm{O}_2\)): Approximately 32 g/mol
• Carbon dioxide (\(\mathrm{CO}_2\)): Approximately 44 g/mol
• Water (\(\mathrm{H}_2\mathrm{O}\)): Approximately 18 g/mol

Molar mass is crucial in determining the amount of each gas required for reactions and their respective contributions to the overall mass in any given chemical equation. Here, molar masses are used to understand root mean square (RMS) speed and stoichiometry in chemical reactions.

RMS Speed

Root Mean Square (RMS) Speed provides insight into the molecular speed distribution in a gas. It's a statistical measure of the speed of particles in a gaseous state, calculated using:\[\text{RMS speed} = \sqrt{\frac{3RT}{M}}\]where:

• \( R \): gas constant
• \( T \): temperature in Kelvin
• \( M \): molar mass of the gas in kg/mol

This relationship reveals that as the molar mass increases, the RMS speed decreases, assuming the temperature remains constant. Explain how the ordering of gases by their RMS speeds was determined by their molar masses, with the lightest molecules (water) moving fastest and the heaviest (carbon dioxide) moving slowest. Thus we understand the order: \(\mathrm{CO}_2\), \(\mathrm{O}_2\), \(\mathrm{C}_2\mathrm{H}_6\), \(\mathrm{H}_2\mathrm{O}\).