Question

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 .\) L/min at \(1.50\) atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is \(21\) mole percent \(\mathrm{O}_{2}\) and \(79\) mole percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

Short answer

The flow rate of air necessary to deliver the required amount of oxygen for complete combustion of methane is approximately 8524 L/min. The composition of the exhaust gas in terms of mole fractions is approximately CO: 0.0017, CO₂: 0.0321, O₂: 0.1353, N₂: 0.7631, and H₂O: 0.0677.

Step-by-step solution

a. Flow rate of air required for complete combustion of methane

Step 1: Write the balanced chemical equation for complete combustion of methane \[CH_4 (g) + 2O_2 (g) -> CO_2 (g) + 2H_2O (g)\] Step 2: Determine the moles of oxygen required for complete combustion of methane Given that methane is flowing at 200 L/min at 1.50 atm and ambient temperature, we can use the ideal gas law to find the moles of methane flowing per minute. \(PV=nRT\) n = PV/RT Assume the ambient temperature is 298 K, and take the gas constant R = 0.0821 L.atm/mol.K. Methane moles (n\(CH_4\))= (200 L/min)(1.50 atm) / ((0.0821 L.atm/mol.K)(298 K)) ≈ 12.19 mol/min Using the stoichiometric ratio of the balanced equation above (1 mol\(CH_4\): 2 mol\(O_2\)): Moles of oxygen required (n\(O_2\)) = 12.19 mol/min * 2 = 24.38 mol/min Step 3: Calculate the flow rate of air necessary to deliver the required amount of oxygen Three times the amount of oxygen is necessary, therefore, n\(O_2\) required in air = 24.38 * 3 = 73.14 mol/min Since air is 21% oxygen, we can find the total moles of air needed per minute. Air moles (n\(air\)) = 73.14 / 0.21 ≈ 348.28 mol/min Now use the ideal gas law again to find the flow rate of air at 1.00 atm and 298 K: Flow rate of air (V\(air\))= n\(air\)RT/P ≈ (348.28 mol/min)(0.0821 L.atm/mol.K)(298 K)/(1.00 atm) ≈ 8524 L/min The flow rate of air necessary to deliver the required amount of oxygen is approximately 8524 L/min.

b. Composition of the exhaust gas in terms of mole fractions

Step 1: Calculate the moles of CO and CO₂ in the exhaust gas 95% of carbon in the exhaust gas is present as CO₂, remaining 5% as CO. Since 1 mol of CH₄ produces 1 mol of carbon: n\(CO_2\) = 0.95 * 12.19 = 11.58 mol/min n\(CO\) = 0.05 * 12.19 = 0.61 mol/min Step 2: Calculate the moles of unreacted O₂ and unreacted N₂ in the exhaust gas Moles of O₂ consumed in the formation of CO₂ and CO = 2 * 12.19 = 24.38 mol/min 3 * n\(O_2\) = 73.14 mol/min is present in the air, therefore, the unreacted moles of O₂ are: Unreacted O₂ = 73.14 - 24.38 = 48.76 mol/min Since air is 79% N₂, unreacted \(N_2\) = 0.79 * 348.28 = 275.14 mol/min Step 3: Calculate the moles of produced H₂O From the balanced equation, 1 mol of CH₄ produces 2 mol of H₂O: n\(H_2O\) = 2 * 12.19 = 24.38 mol/min Step 4: Calculate the mole fractions of the exhaust gas components Total moles in the exhaust gas = n\(CO_2\) + n\(CO\) + unreacted O₂ + unreacted N₂ + n\(H_2O\) Total moles = 11.58 + 0.61 + 48.76 + 275.14 + 24.38 = 360.47 mol/min Mole fractions of each component in the exhaust gas: X\(CO_2\) = n\(CO_2\)/total moles = 11.58 / 360.47 ≈ 0.0321 X\(CO\) = n\(CO\)/total moles = 0.61 / 360.47 ≈ 0.0017 X\(O_2\) = unreacted O₂/total moles = 48.76 / 360.47 ≈ 0.1353 X\(N_2\) = unreacted N₂/total moles = 275.14 / 360.47 ≈ 0.7631 X\(H_2O\) = n\(H_2O\)/total moles = 24.38 / 360.47 ≈ 0.0677 The composition of the exhaust gas in terms of mole fractions is approximately CO: 0.0017, CO₂: 0.0321, O₂: 0.1353, N₂: 0.7631 and H₂O: 0.0677

Key concepts

Combustion Reaction

A combustion reaction is a chemical process that commonly involves the burning of a fuel in the presence of oxygen. The primary goal of combustion is typically to release energy in the form of heat.
The typical products of complete combustion are water ( H_2O ) and carbon dioxide ( CO_2 ). Incomplete combustion can occur when there is insufficient oxygen, producing carbon monoxide ( CO ) and other compounds instead.

• Complete Combustion: Occurs when a fuel burns in the presence of plenty of oxygen, resulting in carbon dioxide and water.
• Incomplete Combustion: Happens when the oxygen supply is limited, resulting in carbon monoxide and water as well as potentially other by-products.

Understanding these types of reactions is important in various fields such as energy production and environmental science, where achieving complete combustion helps reduce harmful emissions and increase efficiency.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture.
It is calculated as the ratio of the number of moles of a specific component to the total number of moles of all components in the mixture.
Mole fraction is a dimensionless quantity, meaning it does not carry any units. It is essential in calculating the composition of gases, especially in chemical reactions like combustion.

• Mole fraction is crucial when dealing with gases, as it helps determine the concentrations of the various components in a gaseous mixture.
• In the combustion process of methane, calculating the mole fraction of each component in the exhaust gas helps understand the efficiency of the combustion.

Ideal Gas Law

The ideal gas law is a fundamental principle used to relate the pressure, volume, temperature, and amount of gas in moles.
The law is expressed by the equation PV = nRT , where:

• P is the pressure of the gas.
• V is the volume of the gas.
• n represents the moles of the gas.
• R is the ideal gas constant ( 0.0821 L.atm/mol.K).
• T is the temperature in Kelvin.

The ideal gas law assumes gases behave ideally, meaning they follow the law under most conditions of temperature and pressure, although real gases can sometimes deviate under extreme conditions.
The ideal gas law is instrumental in predicting the behavior of gases during reactions, including combustion. In our example, it was used to determine the flow rates and moles of reactants and products involved in methane combustion.

Stoichiometry

Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction.
It is pivotal in determining how much of a reactant is needed or how much of a product will be formed during a reaction.

• In combustion reactions, stoichiometry helps ensure the correct proportions of fuel and oxygen for complete combustion.
• This involves balancing chemical equations to reflect the conservation of mass. For methane combustion, the balanced equation CH_4 + 2O_2 → CO_2 + 2H_2O guides the stoichiometry calculations.
• Stoichiometry ensures that the moles of each reactant match the ratios set by the balanced equation, optimizing efficiency and minimizing waste.

Mastering stoichiometry is vital for chemists and engineers to predict reaction outcomes and adjust reaction conditions accordingly.