Question

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of 200. L/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 \(79 \mathrm{~mole}\) 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

a. The flow rate of air necessary to deliver the required amount of oxygen for complete combustion is 347.1 mol/min. b. The exhaust gas composition in terms of mole fraction is: - CO: 0.00195 - CO2: 0.0369 - O2: 0.00471 - N2: 0.879 - H2O: 0.0779

Step-by-step solution

1. Calculate moles of methane per minute

Given the flow rate of methane at 200 L/min and the pressure and temperature being 1.5 atm and ambient, respectively, we can use the ideal gas law to find the moles of methane flowing per minute: \[PV=nRT\] \[n=\frac{PV}{RT}\] Let's use the value of R = 0.0821 L atm/mol K. Assuming the ambient temperature to be 298 K, we can now calculate the moles of methane per minute: \[n_{\mathrm{CH}_{4}}=\frac{(1.5\,\mathrm{atm})(200\,\mathrm{L/min})}{(0.0821\,\mathrm{L\,atm/mol\,K})(298\,\mathrm{K})} = 12.15\,\mathrm{mol/min}\]

2. Calculate the moles of oxygen required

To ensure complete combustion of methane, the balanced chemical equation is: \[\mathrm{CH}_{4}+\mathrm{2O}_{2}\rightarrow \mathrm{CO}_{2}+\mathrm{2H}_{2}\mathrm{O}\] According to the equation, one mole of methane requires two moles of oxygen. Since three times the required amount is reacted, the moles of oxygen required per minute are: \[n_{\mathrm{O}_{2}}=12.15\,\mathrm{mol/min}\times 2\times 3=72.90\,\mathrm{mol/min}\]

3. Calculate the flow rate of air

Air is 21 mole percent \(\mathrm{O}_{2}\), so the amount of air required to deliver the moles of oxygen is: \[\mathrm{Air\,flow\,rate}=\frac{72.90\,\mathrm{mol/min}}{0.21}=347.1\,\mathrm{mol/min}\]

4. Calculate the composition of the exhaust gas

Since \(95.0\%\) of carbon is present in the form of \(\mathrm{CO}_{2}\), we can calculate the moles of \[\mathrm{CO}_{2}\] and \[\mathrm{CO}\] produced using the conversion as follows: \[n_{\mathrm{CO}_{2}}=12.15\,\mathrm{mol/min}\times 0.95 = 11.54\,\mathrm{mol/min}\] \[n_{\mathrm{CO}}=12.15\,\mathrm{mol/min}\times 0.05 = 0.608\,\mathrm{mol/min}\] Calculate the moles of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) left unreacted: \[n_{\mathrm{O}_{2,unreacted}}=347.1\,\mathrm{mol/min}\times 0.21 - 72.90\,\mathrm{mol/min} = 1.47\,\mathrm{mol/min}\] \[n_{\mathrm{N}_{2,unreacted}}=347.1\,\mathrm{mol/min}\times 0.79 = 274.2\,\mathrm{mol/min}\] Calculate the moles of \(\mathrm{H}_{2} \mathrm{O}\) produced: \[n_{\mathrm{H}_{2} \mathrm{O}}=12.15\,\mathrm{mol/min}\times 2 = 24.3\,\mathrm{mol/min}\] Calculate the mole fraction of each component: \[x_{\mathrm{CO}}=\frac{0.608\,\mathrm{mol/min}}{312.1\,\mathrm{mol/min}}=0.00195\] \[x_{\mathrm{CO}_{2}}=\frac{11.54\,\mathrm{mol/min}}{312.1\,\mathrm{mol/min}}=0.0369\] \[x_{\mathrm{O}_{2}}=\frac{1.47\,\mathrm{mol/min}}{312.1\,\mathrm{mol/min}}=0.00471\] \[x_{\mathrm{N}_{2}}=\frac{274.2\,\mathrm{mol/min}}{312.1\,\mathrm{mol/min}}=0.879\] \[x_{\mathrm{H}_{2} \mathrm{O}}=\frac{24.3\,\mathrm{mol/min}}{312.1\,\mathrm{mol/min}}=0.0779\] So, the exhaust gas composition is: - \(\mathrm{CO}: 0.00195\) - \(\mathrm{CO}_{2}: 0.0369\) - \(\mathrm{O}_{2}: 0.00471\) - \(\mathrm{N}_{2}: 0.879\) - \(\mathrm{H}_{2} \mathrm{O}: 0.0779\)

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. In the combustion of methane (\textbf{CH}\(_4\)), stoichiometry allows us to determine the exact amount of oxygen needed to react completely with a given volume of methane gas.

The stoichiometric equation for the combustion of methane is: \[\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\] In this equation, we see that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water vapor. However, in practice, a surplus of oxygen is often used to ensure complete combustion. This is why in our example, three times the stoichiometric amount of oxygen is used. Utilizing stoichiometry is crucial for calculating flow rates and understanding the composition of the reaction products.

Ideal Gas Law

The ideal gas law is an equation of state that relates the pressure, volume, temperature, and number of moles of a gas. It is usually written as: \[PV = nRT\]Where

• \(P\) is the pressure in atmospheres,
• \(V\) is the volume in liters,
• \(n\) is the number of moles of the gas,
• \(R\) is the ideal gas constant, and
• \(T\) is the temperature in Kelvin.

In the context of our exercise, the ideal gas law allows us to calculate the number of moles of methane flowing per minute given the volume flow rate, pressure, and temperature. This is crucial to ensure that adequate amounts of reactants are being used for the combustion process. Understanding the ideal gas law is fundamental to solving many problems in chemistry, such as determining the conditions of gases involved in reactions.

Partial Pressure

Partial pressure is a measure of the pressure that a single component of a gas mixture would exert if it occupied the entire volume of the mixture at the same temperature. It is directly proportional to the mole fraction of the gas in the mixture. When dealing with combustion, knowing the partial pressures of reactants like oxygen is important for ensuring the right conditions for the reaction to proceed efficiently.

In our example, air is added to the combustion chamber at a constant pressure, and so the partial pressure of oxygen in the air is a key factor in determining the flow rate of air required to supply enough oxygen for the combustion of methane.

Mole Fraction

Mole fraction, denoted as \(x\), is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the number of moles of a particular substance to the total number of moles of all substances present. The mole fraction is a unitless number between 0 and 1.

In the latter part of our exercise, we utilized mole fractions to determine the composition of the exhaust gas after the combustion process. Understanding mole fractions allows us to calculate the proportion of each gas present in the mixture, which is fundamental when examining the products formed in chemical reactions and in processes like the environmental assessment of emissions.