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 decreases as the reaction occurs, due to the reduction in the number of moles of gas from 3 (2 moles CO and 1 mole O_2) to 2 (2 moles CO_2). As the piston moves to maintain a constant pressure, the volume will decrease to accommodate the change in moles of gas. (b) If the piston is not allowed to move, the volume remains constant during the reaction. However, since the number of gas moles decreases, the pressure exerted by the gaseous mixture will also decrease at constant temperature and volume as the reaction takes place.

Step-by-step solution

Analyze the balanced chemical reaction

The given reaction is already balanced, and it is as follows: \[ 2\, \mathrm{CO}(g) + \mathrm{O}_2(g) \longrightarrow 2\, \mathrm{CO}_2(g) \] We can see that there are 3 moles of reactant gas (2 moles CO and 1 mole O_2) and 2 moles of product gas (2 moles CO_2) in the balanced chemical equation.

Determine volume changes due to the reaction (Part a)

As a result of the reaction, the total number of moles of gases decreases from 3 moles to 2 moles. At constant temperature and pressure, by applying the Ideal Gas Law (PV = nRT) we deduce that the volume of gas in the container will decrease as the reaction takes place since the number of moles decreases. In this situation, the piston moves to maintain constant pressure. So, when the reaction takes place, the piston will compress, and the volume of the container will decrease.

Determine Pressure changes if the volume is constant (Part b)

If the piston is not allowed to move, the volume of the container remains constant during the reaction. Since the number of moles of gas decreases during the reaction, and P∝n (from the Ideal Gas Law), the pressure exerted by the gaseous mixture will decrease at constant temperature and volume as the reaction proceeds. In this case, since the piston does not move, the decrease in gaseous moles will result in a decrease in overall pressure.

Key concepts

Balanced Chemical Equation

In chemistry, a balanced chemical equation is pivotal to understanding reactions. It ensures that the number of atoms for each element is equal on both the reactant and product sides. Here, the equation used is:\[ 2\, \mathrm{CO}(g) + \mathrm{O}_2(g) \longrightarrow 2\, \mathrm{CO}_2(g) \]Breaking this down, there are 2 moles of carbon monoxide (CO) and 1 mole of oxygen gas (O\(_2\)) that combine to form 2 moles of carbon dioxide (CO\(_2\)). This equation underscores the conservation of mass, as no atoms are lost, just rearranged into new compounds. It's crucial in assessing how different factors like pressure, volume, and temperature affect the products of a reaction.

Moles of Gas

Moles are a fundamental unit in chemistry that represent a quantity of particles, such as atoms or molecules. In gases, moles are directly related to gas volume, especially when the conditions of pressure and temperature are held constant. For the reaction \[ 2\, \mathrm{CO}(g) + \mathrm{O}_2(g) \rightarrow 2\, \mathrm{CO}_2(g) \]we start with 3 moles of gas in the reactants (2 moles of CO and 1 mole of O\(_2\)). After the reaction, we end up with 2 moles of product gas (2 moles of CO\(_2\)). This decrease in moles from 3 to 2 implies that the gas will occupy less space, demonstrating a direct link between the moles of gas and volume under controlled conditions.

Constant Pressure and Temperature

When reactions occur under constant pressure and temperature, the relationship among pressure, volume, and moles is essential. The Ideal Gas Law, given by \[ PV = nRT \]helps explain these changes. In this reaction, where the piston allows constant pressure to be maintained, changes in the number of moles of gas directly lead to changes in volume. Here, as the number of moles decreases from 3 to 2, the volume of gas will compress to maintain the constant pressure. This indicates that maintaining constant pressure and temperature is key to predicting how each component of the gas changes during a reaction.

Volume and Pressure Changes

The interplay between volume and pressure is governed by the number of gas moles present. In this scenario, when the piston moves, volume decreases because fewer moles of gas mean less space is needed. The reaction \[ 2\, \mathrm{CO}(g) + \mathrm{O}_2(g) \rightarrow 2\, \mathrm{CO}_2(g) \]shows this effect as it proceeds in conditions of constant pressure, compressing the volume as a result of mole reduction.However, if the volume remains constant, as with an immovable piston, the pressure will decrease since the number of gas molecules decreases and no additional space is provided to accommodate this change. In both cases, the relationship is clear: when conditions like volume or pressure are fixed, the other property adjusts in response to the change in mole number.