Problem 86 - Chemistry and Chemical Reactivity

Question

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 supplicd to the burner (at a pressure of \(742 \mathrm{mm} \mathrm{Hg}\) and a temperature of \(26^{\circ} \mathrm{C}\) ) \(?\)

Step-by-Step Solution

Answer

Oxygen must be supplied at a rate of approximately 10.3 L/min.

Step 1: Write the balanced chemical equation

The combustion of methane (CH extsubscript{4}) with oxygen (O extsubscript{2}) produces carbon dioxide (CO extsubscript{2}) and water (H extsubscript{2}O). The balanced chemical equation is: \[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \] This equation shows that 1 mole of CH extsubscript{4} reacts with 2 moles of O extsubscript{2}.

Step 2: Calculate the moles of methane gas

Use the ideal gas law to calculate the moles of methane. Start by converting the given pressure from mm Hg to atm: \[ P_1 = \frac{773 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 1.0171 \, \text{atm} \] Now, convert the temperature from Celsius to Kelvin: \[ T_1 = 28^{\circ}C + 273.15 = 301.15 \, K \] Use the ideal gas law, \( PV = nRT \), to find the moles (n), where \( R = 0.0821 \, \text{L atm/mol K} \): \[ n_1 = \frac{P_1V}{RT_1} = \frac{1.0171 \, \text{atm} \times 5.0 \, \text{L/min}}{0.0821 \, \text{L atm/mol K} \times 301.15 \, K} = 0.2054 \, \text{mol/min} \]

Step 3: Determine the required moles of oxygen

Use the stoichiometry from the balanced chemical equation. For every mole of methane, 2 moles of oxygen are needed: \[ n_{\text{O}_2} = 2 \times n_1 = 2 \times 0.2054 \, \text{mol/min} = 0.4108 \, \text{mol/min} \]

Step 4: Calculate the volume of oxygen needed

Now, use the ideal gas law to find the volume required at the specified conditions for oxygen. First, convert the pressure of oxygen to atm: \[ P_2 = \frac{742 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 0.9763 \, \text{atm} \] Convert the temperature of oxygen to Kelvin: \[ T_2 = 26^{\circ}C + 273.15 = 299.15 \, K \] Use \( PV = nRT \) to solve for the volume (V): \[ V_2 = \frac{n_rt_2}{p_2} = \frac{0.4108 \, \text{mol/min} \times 0.0821 \, \text{L atm/mol K} \times 299.15 \, K}{0.9763 \, \text{atm}} \approx 10.3 \, \text{L/min} \]

Step 5: Conclusion

The rate at which oxygen must be supplied to the burner is approximately 10.3 L/min.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of ideal gases. It is expressed as \( PV = nRT \), where:

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

This law allows us to calculate any one of these variables if the other three are known. When dealing with gases in chemical reactions, like in our methane combustion exercise, knowing how to manipulate this equation is crucial.
For instance, to find how many moles of methane gas are used, we can rearrange the equation to find \( n \) (moles): \( n = \frac{PV}{RT} \). This calculation tells us how much material we have to work with in a reaction.

Combustion Reaction

A combustion reaction is a type of chemical reaction where a substance combines with oxygen to release energy. In our exercise, methane \( (\text{CH}_4) \) is burned in a laboratory Bunsen burner in the presence of oxygen to produce carbon dioxide \((\text{CO}_2)\) and water vapor \((\text{H}_2\text{O})\). This is represented by the balanced chemical equation:\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]This equation shows the stoichiometric proportions of the reactants and products. For every one mole of methane, two moles of oxygen are required for complete combustion. This type of reaction is exothermic, meaning it releases energy in the form of heat and sometimes light, which is why it's commonly used in burners and engines.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants that go into a reaction and the products that result from it. A balanced chemical equation, like the one used in our example, ensures that the same number of each type of atom is present on both sides of the equation.
In the equation \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \), the coefficients (the numbers in front of molecules) are important. They tell us how many moles of each substance are involved. Balancing the equation ensures that mass is conserved in the reaction.
Understanding how to work with chemical equations is vital for predicting the outcome of a reaction, calculating reactant and product quantities, and ensuring safety and efficiency in chemical processes.

Gas Laws

Gas Laws are a set of rules that describe how gases behave under various conditions of pressure, volume, and temperature. Beyond the Ideal Gas Law, there are others such as Boyle's Law, Charles's Law, and Avogadro's Law, each connecting two of these variables while keeping the third constant.
For example, Boyle's Law states that the pressure of a gas is inversely proportional to its volume when temperature is constant. Charles's Law states that the volume of a gas is directly proportional to its temperature when pressure is constant.
Understanding these laws allows us to predict and calculate changes in gas behavior, which is particularly useful for processes like combustion. They complement the Ideal Gas Law and often provide the basis for more complex calculations like the ones done in the provided exercise.