Problem 88 - Chemistry and Chemical Reactivity

Question

A Methane is burned in a laboratory Bunsen burner to give \(\mathrm{CO}_{2}\) and water vapor. Methane gas is supplied to the burner at the rate of \(5.0 \mathrm{L} / \mathrm{min}\) (at a temperature of \(28^{\circ} \mathrm{C}\) and a pressure of \(773 \mathrm{mm} \mathrm{Hg}\) ). At what rate must oxygen be supplied to the burner (at a pressure of \(\left.742 \mathrm{mm} \mathrm{Hg} \text { and a temperature of } 26^{\circ} \mathrm{C}\right) ?\)

Step-by-Step Solution

Answer

Supply oxygen at approximately 10.27 L/min.

Step 1: Write the Chemical Equation

The combustion of methane (CH₄) is as follows: \[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) \] This means that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water vapor.

Step 2: Calculate the Moles of Methane

Use the ideal gas law to find the moles of methane being supplied per minute. \[ PV = nRT \]Rearrange to solve for \(n\) (moles):\[ n = \frac{PV}{RT} \]Convert the units of pressure from mmHg to atm:\[ P = \frac{773 \text{ mmHg}}{760 \text{ mmHg/atm}} = 1.0171 \text{ atm} \]Using \(R = 0.0821 \text{ L atm/mol K} \) and \(T = 28^\circ \text{C} + 273 = 301 \text{ K} \), calculate \(n\):\[ n = \frac{1.0171 \times 5.0}{0.0821 \times 301} \approx 0.204 \text{ moles/min} \]

Step 3: Determine the Required Moles of Oxygen

From the chemical equation, 1 mole of methane requires 2 moles of oxygen. Therefore, for the 0.204 moles of methane supplied per minute:\[ n_{O_2} = 2 \times 0.204 = 0.408 \text{ moles of } \text{O}_2 \text{ per minute} \]

Step 4: Calculate the Volume of Oxygen Required

Use the ideal gas law again to find the volume of oxygen per minute at the given conditions. \[ PV = nRT \]Rearrange to solve for \(V\) (volume):\[ V = \frac{nRT}{P} \]Convert the pressure of oxygen from mmHg to atm:\[ P = \frac{742 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.9763 \text{ atm} \]Using \(R = 0.0821 \text{ L atm/mol K} \) and \(T = 26^\circ \text{C} + 273 = 299 \text{ K} \), calculate \(V\):\[ V = \frac{0.408 \times 0.0821 \times 299}{0.9763} \approx 10.27 \text{ L/min} \]

Step 5: Conclusion: Rate of Oxygen Supply

Oxygen must be supplied to the burner at a rate of approximately \(10.27 \text{ L/min}\) to completely combust the methane gas supplied at this rate and conditions.

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry used to relate the pressure, volume, temperature, and number of moles of a gas. It is expressed as: \[ PV = nRT \] where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume the gas occupies.
• \( n \) is the number of moles of gas.
• \( R \) is the ideal gas constant \( (0.0821 \, \text{L atm/mol K}) \).
• \( T \) is the temperature in Kelvin.

To apply this law, it’s essential to ensure that all units are compatible, typically using atmosphere (atm) for pressure, liters (L) for volume, and Kelvin (K) for temperature. This equation helps calculate any one of these variables if the others are known, making it a vital tool in understanding gas behavior under various conditions.

Mole Calculation

Mole calculation is crucial in understanding chemical reactions, especially when dealing with gases. In this context, the ideal gas law helps us calculate the number of moles of a gas given its pressure, volume, and temperature. This is useful because moles act as a bridge between the micro (atoms/molecules) and macro (grams/volume) world.Using the equation \( n = \frac{PV}{RT} \), we can substitute known values of pressure \( P \), volume \( V \), temperature \( T \), and the gas constant \( R \) to find \( n \), the number of moles. This calculation allows us to understand how much of each reactant we have in chemical reactions, such as the combustion of methane to determine how much oxygen is needed.

Chemical Equation Balancing

Chemical equation balancing is vital to ensuring that the number of atoms on the reactant side matches those on the product side of a reaction. In the combustion reaction of methane, balancing the equation \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \) demonstrates stoichiometry, indicating that one molecule of methane reacts with two molecules of oxygen to form carbon dioxide and water.Balancing equations allows us to determine the exact ratios of reactants needed and products formed. In our methane combustion example, it's this balanced equation that informs us that 2 moles of oxygen are needed for every mole of methane combusted. This understanding is key for mole calculations and assessing the required conditions for a complete reaction.

Volumetric Flow Rate Calculation

Volumetric flow rate refers to the volume of gas that moves through a given area per unit of time. It combines the concepts of volume and time to quantify how much gas is being used or produced in reactions, which is especially useful in industrial or laboratory settings.To find the volumetric flow rate needed for an element like oxygen in our methane combustion example, we utilize the ideal gas law rearranged to solve for volume: \[ V = \frac{nRT}{P} \] Given the moles of oxygen needed per minute and the specific conditions such as pressure and temperature, this formula provides the required volumetric flow rate. This calculation helps ensure that the right amount of oxygen is supplied to maintain an efficient and complete combustion reaction.