Question

Consider the following reaction: $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ Imagine that this reaction occurs in a container that has a piston that moves to allow a constant pressure to be maintained when the reaction occurs at constant temperature. (a) What happens to the volume of the container as a result of the reaction? Explain. (b) If the piston is not allowed to move, what happens to the pressure as a result of the reaction? [Sections \(10.3\) and 10.5]

Short answer

(a) The volume of the container remains constant during the reaction when the pressure and temperature are maintained constant. This is because the total number of moles of gas remains the same before and after the reaction, according to the ideal gas law. (b) If the piston is not allowed to move and the volume is held constant, the pressure will also remain constant. This is due to the number of moles for both reactants and products remaining constant during the reaction.

Step-by-step solution

(Step 1: Analyzing the reaction and stoichiometry)

First, we analyze the given balanced chemical equation: \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) For every 2 moles of CO and 1 mole of O2, 2 moles of CO2 are produced.

(Step 2: Determine the initial and final moles of gas in the reaction)

We start by calculating the number of moles of each species before and after the reaction. Before the reaction: Let's assume there are n(CO) moles of CO and n(O2) moles of O2. After the reaction: Since 2 moles of CO react with 1 mole of O2, the moles of CO and O2 that react will be 2x and x, respectively. To find x, we consider that we need twice as many moles of CO as O2 for the reaction. Therefore, x = min(n(CO)/2, n(O2)). After the reaction, there will be n(CO)-2x moles of CO left, n(O2)-x moles of O2 left, and 2x moles of CO2 formed.

(Step 3: Applying the ideal gas law and analyzing the volume change for constant pressure)

The ideal gas law is given by the equation: PV = nRT where P represents pressure, V represents volume, n represents the number of moles, R is the gas constant, and T is the temperature. For the first scenario, where the temperature and pressure remain constant: Before the reaction: total moles of gas = n(CO) + n(O2) After the reaction: total moles of gas = (n(CO)-2x) + (n(O2)-x) + 2x = n(CO) + n(O2) As the total number of moles of gas remains the same and pressure and temperature are constant, the volume of the container must also remain constant according to the ideal gas law.

(Step 4: Analyzing the pressure change for a constant volume with a fixed piston)

Now consider the scenario where the piston is not allowed to move, and the volume is constant. Before the reaction, the pressure of the gas mixture: \(P_{1} = \frac{(n(CO) + n(O2)) RT}{V}\) After the reaction, the pressure of the gas mixture: \(P_{2} = \frac{(n(CO) - 2x + n(O2) - x + 2x ) RT}{V} = \frac{(n(CO) + n(O2)) RT}{V}\) Since the volume is constant in this scenario, the pressure will remain the same because the number of moles for both reactants and products is constant. #Summary# (a) When maintaining constant pressure and temperature, the volume of the container remains constant throughout the reaction. (b) When the piston is not allowed to move and the volume is constant, the pressure remains constant as well.

Key concepts

Ideal Gas Law

The ideal gas law is fundamental to understanding the behavior of gases in chemical reactions. It is expressed by the equation:
\[ PV = nRT \]
where:

• (P) stands for pressure of the gas,
• (V) is the volume,
• (n) is the number of moles,
• (R) is the universal gas constant, and
• (T) is the temperature in Kelvin.

This law explains how pressure, volume, temperature, and the amount of gas are interrelated. For stoichiometric calculations in chemical reactions involving gases, it allows us to predict how changes in these conditions will affect gas properties. If one variable changes, the others may adjust to maintain the equality—if the temperature and the number of moles of gas are held constant, then the product of pressure and volume must remain constant. This is key in understanding how changes during a reaction, like in the textbook exercise involving carbon monoxide and oxygen, affect the system.

Volume and Pressure Relationship

The relationship between volume and pressure of a gas is an essential aspect of the ideal gas law. For a given amount of gas at constant temperature, the pressure and volume are inversely proportional to each other—a concept known as Boyle's Law.
\[ P_1V_1 = P_2V_2 \]
This means if the volume decreases, the pressure increases, provided the temperature and the number of moles of gas remain unchanged. In the context of the provided exercise, when a reaction occurs in a container with a movable piston maintaining constant pressure, the volume does not change since the number of moles of reactants and products is the same. On the other hand, if the piston is not allowed to move, the volume is fixed, and since the number of moles remains constant in our scenario, the pressure stays the same.

Stoichiometric Calculations

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It's the calculation of the quantities of substances involved in a reaction based on balanced chemical equations. The stoichiometric coefficients—in the case of our exercise, the numbers showing the ratios of CO to O₂ to CO₂—determine the proportions needed to complete the reaction.
In a stoichiometric calculation, we assess the amount of reactants and predict the amount of products formed. Taking the exercise as an example, the stoichiometry shows a 2:1:2 relationship, ensuring that the number of moles of gas before and after the reaction remains constant—if we start with a combined total of 3 moles (2 moles CO and 1 mole O₂), we end up with 2 moles CO₂. This insight is crucial when applying the ideal gas law, as changes in the amount of gas directly affect volume and pressure under constant temperature conditions.