Question

A 1.604 -g sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas and 6.400 g oxygen gas are sealed into a 2.50 -L vessel at \(411^{\circ} \mathrm{C}\) and are allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium, the pressure of oxygen is 0.326 atm, and the pressure of water vapor is 4.45 atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.

Short answer

The pressures of carbon monoxide and carbon dioxide at equilibrium are 3.77 atm and 0.0224 atm, respectively.

Step-by-step solution

1. Identify reactions and write balanced equations

The balanced equations for each reaction are given below: Reaction 1: \(\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\) Reaction 2: \(\mathrm{CH}_{4} + \frac{3}{2}\mathrm{O}_{2} \rightarrow \mathrm{CO} + 2\mathrm{H}_{2}\mathrm{O}\)

2. Calculate initial moles of each substance

We will use the given masses of methane and oxygen to find the initial moles of each substance, using their respective molar masses. For methane (\(\mathrm{CH}_{4}\)): Molar mass = 12.01 g/mol (C) + 4×1.01 g/mol (H) = 16.05 g/mol Initial moles = \(\frac{1.604\,\text{g}}{16.05\,\text{g/mol}} = 0.100\,\text{mol}\) For oxygen (\(\mathrm{O}_{2}\)): Molar mass = 2×16.00 g/mol = 32.00 g/mol Initial moles = \(\frac{6.400\,\text{g}}{32.00\,\text{g/mol}} = 0.200\,\text{mol}\)

3. Convert equilibrium pressure to moles

We will convert the equilibrium pressures of oxygen and water vapor to moles using the ideal gas law equation: \(PV = nRT\), where P is the pressure, V is the volume, n is the moles, R is the ideal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin. First, let's convert the given temperature to Kelvin: T = \(411^{\circ}\text{C} + 273.15 = 684.15\,\text{K}\) Now, we can find the moles of oxygen and water vapor at equilibrium: For oxygen (\(\mathrm{O}_{2}\)): Equilibrium moles = \(\frac{P \times V}{R \times T} = \frac{0.326\,\text{atm} \times 2.50\,\text{L}}{0.0821\,\text{L atm/mol K} \times 684.15\,\text{K}} = 0.0153\,\text{mol}\) For water vapor (\(\mathrm{H}_{2}\mathrm{O}\)): Equilibrium moles = \(\frac{P \times V}{R \times T} = \frac{4.45\,\text{atm} \times 2.50\,\text{L}}{0.0821\,\text{L atm/mol K} \times 684.15\,\text{K}} = 0.2020\,\text{mol}\)

4. Determine the remaining moles of methane

Since we know the initial moles of methane and moles of water vapor at equilibrium, we can determine the remaining moles of methane. Let's subtract the moles of water vapor formed in response to the moles of methane consumed: Remaining moles of \(\mathrm{CH}_{4}\) = \(0.100\,\text{mol} - \frac{1}{2} \times 0.2020\,\text{mol} = 0.0010\,\text{mol}\)

5. Find the moles of carbon dioxide and carbon monoxide

We can use the stoichiometry of each of the balanced equations to find the moles of carbon dioxide and carbon monoxide: Moles of \(\mathrm{CO}_{2}\) = Remaining moles of \(\mathrm{CH}_{4}\) = 0.0010 mol Moles of \(\mathrm{CO}\) = \(\frac{3}{2}\) × (Initial moles of \(\mathrm{O}_{2}\) - Equilibrium moles of \(\mathrm{O}_{2}\)) = \(\frac{3}{2} \times (0.200\,\text{mol} - 0.0153\,\text{mol}) = 0.138\,\text{mol}\)

6. Convert moles of carbon dioxide and carbon monoxide to pressure

Now, we will use the ideal gas law to convert the moles of carbon dioxide and carbon monoxide to their respective equilibrium pressures: For carbon dioxide (\(\mathrm{CO}_{2}\)): Equilibrium pressure = \(\frac{n \times R \times T}{V} = \frac{0.0010\,\text{mol} \times 0.0821\,\text{L atm/mol K} \times 684.15\,\text{K}}{2.50\,\text{L}} = 0.0224\,\text{atm}\) For carbon monoxide (\(\mathrm{CO}\)): Equilibrium pressure = \(\frac{n \times R \times T}{V} = \frac{0.138\,\text{mol} \times 0.0821\,\text{L atm/mol K} \times 684.15\,\text{K}}{2.50\,\text{L}} = 3.77\,\text{atm}\) The pressures of carbon monoxide and carbon dioxide at equilibrium are 3.77 atm and 0.0224 atm, respectively.

Key concepts

Chemical Reactions

Chemical reactions are the processes where reactants are transformed into products through the breaking and forming of chemical bonds. Every reaction is governed by certain laws, and one fundamental law is the conservation of mass. This implies that in a closed system, the mass of reactants must equal the mass of the products.

For instance, in the given example, the reaction of methane and oxygen produces either carbon dioxide and water, or carbon monoxide and water, depending on the reaction pathway. It's crucial to understand that these reactions are balanced, meaning the number of atoms for each element is the same on both sides of the equation. This reflects the conservation of mass in chemical reactions and sets the stage for stoichiometric calculations, which we'll discuss later on.

Ideal Gas Law

The ideal gas law is a fundamental equation that relates the pressure (P), volume (V), amount (n) in moles, and temperature (T) of an ideal gas through the gas constant (R). Expressed as PV = nRT, it assumes that particles in an ideal gas do not attract or repel each other and occupy no volume in themselves.

In real-world situations like in our exercise, although gases are not perfect, the ideal gas law provides a close approximation that facilitates the determination of an unknown quantity if the others are known. The ability to manipulate this equation is indispensable for solving numerous problems in chemistry, particularly when working with gases at equilibrium, as seen in the calculation of moles of gases at equilibrium pressure.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves calculations that use molar ratios from balanced chemical equations to predict the amounts of products and reactants involved.

In the context of our example, stoichiometry allows us to deduce how many moles of each product are formed from a given amount of reactants, taking into account the balanced reaction equations. This is essential for determining the changes in the system as it moves towards equilibrium. By applying stoichiometric principles, we can also calculate which reactants are in excess, which are completely consumed, and how much of each product is generated.

Equilibrium Pressure

Equilibrium pressure is a term commonly used in the context of chemical reactions involving gases, where the system has reached a state where the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time.

In a sealed container, as seen in the problem, equilibrium pressures are important because they relate to the amounts of each gas present at equilibrium. By using the ideal gas law, we can determine the moles of gases from these pressures and subsequently find the pressures of other gases in the mixture using stoichiometry. Understanding how to calculate equilibrium pressures is vital for students as it connects the conceptual understanding of dynamic equilibrium in chemical systems with the quantitative aspect of gas behavior.