Question

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of 200\. \(\mathrm{L} / \mathrm{min}\) at \(1.50 \mathrm{~atm}\) and ambient temperature. Air is added to the chamber at \(1.00 \mathrm{~atm}\) and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(\mathrm{~g})\) 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 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 short answer to the question is: a. To ensure complete combustion of methane, the required flow rate of air is 348.6 mol/min. b. Under the given conditions, the exhaust gas has a composition of mole fractions of CO, CO2, O2, N2, and H2O, approximately 0.0017, 0.0322, 0.1353, 0.7632, and 0.0676, respectively.

Step-by-step solution

Write the balanced chemical equation for complete combustion of methane

For complete combustion of methane to produce carbon dioxide and water vapor, the balanced chemical equation is: \[ \mathrm{CH}_4 + 2\ \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\ \mathrm{H}_2\mathrm{O} \]

Calculate the molar flow rate of methane

Using the ideal gas law, we can find the molar flow rate of methane. The ideal gas law is given by \(PV=nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the moles, \(R\) is the gas constant, and \(T\) is the temperature. Here, \(P = 1.50\ \mathrm{atm}\), \(V = 200\ \mathrm{L/min}\), and \(T\) is ambient temperature. To find the molar flow rate of methane, we need to rearrange the ideal gas law as follows: \[n_{\mathrm{CH_4}} = \frac{PV}{RT}\] Assuming the ambient temperature to be 298 K (25°C) and using the appropriate gas constant (\(0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}\)), we get: \[n_{\mathrm{CH_4}} = \frac{1.50\ \mathrm{atm} \times 200\ \mathrm{L/min}}{(0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}})(298\ \mathrm{K})}\] \[n_{\mathrm{CH_4}} = 12.2\ \mathrm{mol/min}\]

Calculate the molar flow rate of oxygen required

Using the stoichiometry of the balanced chemical equation, 2 moles of oxygen are required for every mol of methane. So, the molar flow rate of oxygen required (\(n_{\mathrm{O_2, required}}\)) is: \[n_{\mathrm{O_2, required}} = 2 \times n_{\mathrm{CH_4}}\] \[n_{\mathrm{O_2, required}} = 2 \times 12.2\ \mathrm{mol/min}\] \[n_{\mathrm{O_2, required}} = 24.4\ \mathrm{mol/min}\]

Calculate the flow rate of pure oxygen required

It is given that three times as much oxygen as necessary is reacted. Therefore, the molar flow rate of oxygen supplied (\(n_{\mathrm{O_2, supplied}}\)) is: \[n_{\mathrm{O_2, supplied}} = 3 \times n_{\mathrm{O_2, required}}\] \[n_{\mathrm{O_2, supplied}} = 3 \times 24.4\ \mathrm{mol/min}\] \[n_{\mathrm{O_2, supplied}} = 73.2\ \mathrm{mol/min}\]

Calculate the flow rate of air required

It is given that the air is 21 mole percent oxygen and 79 mole percent nitrogen. So, the flow rate of air (\(V_{air}\)) necessary to deliver the required amount of oxygen can be calculated by rearranging the flow rate of oxygen supplied as follows: \[V_{air} = \frac{n_{\mathrm{O_2, supplied}}}{0.21}\] \[V_{air} = \frac{73.2\ \mathrm{mol/min}}{0.21}\] \[V_{air} = 348.6\ \mathrm{mol/min}\] b. Calculating the composition of the exhaust gas under incomplete combustion

Determine the amount of CO and CO2 produced

It is given that 95% of the carbon in the exhaust gas is present in CO2 and the remainder is present as carbon in CO. Therefore, for every mol of methane reacted, 0.95 mol CO2 and 0.05 mol CO are produced.

Determine the molar flow rates of CO2, CO, O2, N2, and H2O in the exhaust gas

We will first determine the molar flow rates of CO2, CO, O2, N2, and H2O in the exhaust gas using the stoichiometry of the reactions. \[ n_{\mathrm{CO_2}} = 0.95 \times n_{\mathrm{CH_4}} \] \[ n_{\mathrm{CO_2}} = 0.95 \times 12.2\ \mathrm{mol/min} \] \[ n_{\mathrm{CO_2}} = 11.6\ \mathrm{mol/min} \] Similarly, for CO: \[ n_{\mathrm{CO}} = 0.05 \times n_{\mathrm{CH_4}} \] \[ n_{\mathrm{CO}} = 0.05 \times 12.2\ \mathrm{mol/min} \] \[ n_{\mathrm{CO}} = 0.61\ \mathrm{mol/min} \] For O2, we know that the molar flow rate of oxygen supplied is 73.2 mol/min and 24.4 mol/min is consumed in the reaction, so the remaining oxygen in the exhaust gas is: \[ n_{\mathrm{O_2}} = 73.2\ \mathrm{mol/min} - 24.4\ \mathrm{mol/min} \] \[ n_{\mathrm{O_2}} = 48.8\ \mathrm{mol/min} \] Now calculate the molar flow rate of nitrogen (N2) in the exhaust gas: \[n_{\mathrm{N_2}} = 0.79 \times V_{air}\] \[n_{\mathrm{N_2}} = 0.79 \times 348.6\ \mathrm{mol/min}\] \[n_{\mathrm{N_2}} = 275.4\ \mathrm{mol/min}\] For H2O, we know that 2 mol of H2O is produced for every mol of methane, so: \[ n_{\mathrm{H_2O}} = 2 \times n_{\mathrm{CH_4}} \] \[ n_{\mathrm{H_2O}} = 2 \times 12.2\ \mathrm{mol/min} \] \[ n_{\mathrm{H_2O}} = 24.4\ \mathrm{mol/min} \]

Calculate the mole fraction of each component in the exhaust gas

Find the total moles in the exhaust gas: \[n_{total} = n_{\mathrm{CO_2}} + n_{\mathrm{CO}} + n_{\mathrm{O_2}} + n_{\mathrm{N_2}} + n_{\mathrm{H_2O}}\] \[n_{total} = 11.6\ \mathrm{mol/min} + 0.61\ \mathrm{mol/min} + 48.8\ \mathrm{mol/min} + 275.4\ \mathrm{mol/min} + 24.4\ \mathrm{mol/min}\] \[n_{total} = 360.8\ \mathrm{mol/min}\] Finally, calculate the mole fractions of each component in the exhaust gas: \[y_{\mathrm{CO_2}} = \frac{n_{\mathrm{CO_2}}}{n_{total}}\] \[y_{\mathrm{CO_2}} = \frac{11.6\ \mathrm{mol/min}}{360.8\ \mathrm{mol/min}}\] \[y_{\mathrm{CO_2}} = 0.0322\] Similarly, \[y_{\mathrm{CO}} = \frac{n_{\mathrm{CO}}}{n_{total}}\] \[y_{\mathrm{CO}} = \frac{0.61\ \mathrm{mol/min}}{360.8\ \mathrm{mol/min}}\] \[y_{\mathrm{CO}} = 0.0017\] \[y_{\mathrm{O_2}} = \frac{n_{\mathrm{O_2}}}{n_{total}}\] \[y_{\mathrm{O_2}} = \frac{48.8\ \mathrm{mol/min}}{360.8\ \mathrm{mol/min}}\] \[y_{\mathrm{O_2}} = 0.1353\] \[y_{\mathrm{N_2}} = \frac{n_{\mathrm{N_2}}}{n_{total}}\] \[y_{\mathrm{N_2}} = \frac{275.4\ \mathrm{mol/min}}{360.8\ \mathrm{mol/min}}\] \[y_{\mathrm{N_2}} = 0.7632\] \[y_{\mathrm{H_2O}} = \frac{n_{\mathrm{H_2O}}}{n_{total}}\] \[y_{\mathrm{H_2O}} = \frac{24.4\ \mathrm{mol/min}}{360.8\ \mathrm{mol/min}}\] \[y_{\mathrm{H_2O}} = 0.0676\] Thus, the composition of the exhaust gas in terms of mole fraction of CO, CO2, O2, N2, and H2O is approximately 0.0017, 0.0322, 0.1353, 0.7632, and 0.0676, respectively.

Key concepts

Methane Combustion

Methane, a simple hydrocarbon with the chemical formula \( \mathrm{CH}_4 \), is a primary component of natural gas. When methane combusts in the presence of oxygen, it typically produces carbon dioxide \( \mathrm{CO}_2 \) and water \( \mathrm{H}_2\mathrm{O} \). This reaction releases energy, making methane a valuable fuel.

• The balanced chemical equation for the complete combustion of methane is:

\[\mathrm{CH}_4 + 2\ \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\ \mathrm{H}_2\mathrm{O} \]

• Complete combustion occurs when there is enough oxygen to convert all the methane into \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \). In real-world scenarios, ensuring ample oxygen is crucial for efficient energy production.

When not enough oxygen is present, methane can undergo incomplete combustion. This leads to the production of carbon monoxide \( \mathrm{CO} \) along with \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \).
While \( \mathrm{CO}_2 \) is a greenhouse gas contributing to climate change, carbon monoxide is a toxic gas that poses serious health risks.

Chemical Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In the methane combustion scenario, stoichiometry helps determine the precise amounts of oxygen needed for complete combustion.

• From the balanced equation, 2 moles of \( \mathrm{O}_2 \) are required for each mole of \( \mathrm{CH}_4 \).
• This means that for every mole of methane burned, two moles of oxygen are consumed.

To ensure complete combustion, it is often necessary to use excess oxygen. In our example, the requirement is to provide three times the needed oxygen:

• This excess ensures that all the fuel is burned, maximizing energy output and reducing emissions of unwanted byproducts like \( \mathrm{CO} \).

Gas Laws

Gas laws are essential for understanding how gases behave under different conditions of temperature, volume, and pressure. The Ideal Gas Law is a pivotal concept applied during the exercise to calculate the molar flow rate of methane. The law is given by:\[ PV = nRT \]Where:

• \( P \): Pressure of the gas (atm)
• \( V \): Volume of the gas (L)
• \( n \): Moles of gas
• \( R \): Universal gas constant \( (0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}) \)
• \( T \): Temperature in Kelvin

Using the conditions specified in the problem, it was possible to find that the molar flow of \( \mathrm{CH}_4 \) was 12.2 moles/minute. This involves rearranging the Ideal Gas Law to solve for \( n \).
The gas laws allow us to translate real-world measurements like pressure and volume into moles, making them invaluable tools in chemical calculations.

Incomplete Combustion

Incomplete combustion occurs when a fuel burns in insufficient oxygen. This results in the production of \( \mathrm{CO} \) in addition to \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \), with potential leftover \( \mathrm{O}_2 \).

• In the given problem, 95% of the carbon ends up as \( \mathrm{CO}_2 \), and the remaining 5% forms \( \mathrm{CO} \).
• This shows that although three times the theoretical amount of oxygen was supplied, not all the carbon was fully oxidized.

Incomplete combustion is undesirable not only because it is less energy-efficient but also because \( \mathrm{CO} \) is harmful if inhaled. Monitoring the exhaust and adjusting combustion processes can help minimize these issues and improve safety and efficiency.