Question

Imagine that the reaction \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) occurs in a container that has a piston that moves to maintain a constant pressure 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) As a result of the reaction, the volume of the container decreases because the number of moles of gas decreases from 3 to 2 and the piston moves inward to maintain constant pressure. The ratio of the product volume to the reactant volume is \(\frac{2}{3}\). (b) If the piston is not allowed to move, the pressure inside the container decreases as a result of the reaction because there are fewer moles of gas produced and the volume remains constant. The ratio of the product pressure to the reactant pressure is \(\frac{2}{3}\).

Step-by-step solution

Analyze the reaction and compare the number of moles on both sides.

The given reaction is: 2 CO(g) + O₂(g) → 2 CO₂(g) On the left side (reactants), we have 2 moles of CO gas and 1 mole of O₂ gas: total of 3 moles of gas. On the right side (products), we have 2 moles of CO₂ gas. We can see that the number of moles decreases from 3 to 2 as the reaction proceeds.

Apply Ideal Gas Law to analyze changes under constant pressure.

The Ideal Gas Law is: PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. (a) Under constant pressure, we can compare the volume change by using n and T as a constant multiplier. Since n = 3 for reactants and n = 2 for products, the ratio of the volume of products to the volume of reactants would be: \[ \frac{V_{products}}{V_{reactants}}=\frac{n_{products}}{n_{reactants}}=\frac{2}{3} \] This means that the volume of the container decreases as a result of the reaction, because there are fewer moles of gas produced, and the piston moves inward to maintain a constant pressure.

Apply Ideal Gas Law to analyze changes in pressure if the piston is not allowed to move.

(b) If the piston is not allowed to move, the volume remains constant. We can compare the pressure changes using n and T as a constant multiplier. Since n = 3 for reactants and n = 2 for products, the ratio of the pressure of products to the pressure of reactants would be: \[ \frac{P_{products}}{P_{reactants}}=\frac{n_{products}}{n_{reactants}}=\frac{2}{3} \] This means that the pressure inside the container decreases as a result of the reaction when the piston is stuck since there are fewer moles of gas produced and the volume remains constant.

Key concepts

Ideal Gas Law

Understanding the Ideal Gas Law is crucial when analyzing how the volume and pressure of a gas change during chemical reactions. The Ideal Gas Law, represented by the equation \( PV = nRT \), establishes a relationship between pressure (P), volume (V), number of moles of gas (n), the ideal gas constant (R), and temperature (T).

At a constant temperature, the Ideal Gas Law simplifies how we predict changes in volume and pressure based on the amount of gas, or moles, present. When the number of moles of gas decreases during a reaction, as seen in the provided exercise, the volume will also decrease if the pressure is kept constant. This is because the pressure, temperature, and gas constant act as fixed values, so the change in the number of moles directly affects the volume. Similarly, if the volume is not allowed to change (because the piston cannot move), a decrease in moles will result in a decrease in pressure. The law thereby acts as a gateway to predict how gases will behave under different conditions in a closed system.

By applying the ratio of the moles of products to reactants from the given chemical equation, you can also discern the corresponding changes in volume or pressure, provided that other variables remain constant. This is an essential concept for understanding gas behavior in various scientific and industrial applications.

Chemical Stoichiometry

Chemical stoichiometry involves the quantification of products and reactants in a chemical reaction. It relies on the principle of conservation of mass and the mole concept to establish ratios, allowing chemists to predict the amounts of substances consumed and produced in a reaction.

In the example of the carbon monoxide and oxygen reaction to form carbon dioxide, stoichiometry is applied to determine the change in the number of moles of gas before and after the reaction. Starting with 2 moles of CO and 1 mole of O₂, the reactants tally up to 3 moles. The reaction yields 2 moles of CO₂. Here, the stoichiometry tells us we have a reduction in moles.

This stoichiometric relationship provides essential insights into the expected changes in volume and pressure discussed previously. An important note for students to grasp is the direct connection between the stoichiometry of a balanced equation and the changes in gas volume and pressure. The ability to do these calculations accurately has far-reaching implications, from laboratory experiments to industrial-scale chemical manufacturing.

Gas Volume and Pressure Relationships

Gas volume and pressure relationships are fundamentally tied to how gases react to changes in their physical conditions. Using the concepts from the Ideal Gas Law, we know that at constant temperature, the volume of a gas is inversely proportional to its pressure (Boyle's Law).

In a system like the one in our exercise where a piston moves to keep the pressure constant, a decrease in the volume of gas (due to a reduction in moles after a chemical reaction) results in the piston moving inward. If the piston is locked and the volume can't change, the pressure must decrease as the number of gas moles goes down. It’s a push-and-pull relationship where the volume increases, the pressure decreases, and vice versa, all while keeping other factors like temperature the same.

Understanding these relationships is key to a variety of scientific disciplines, such as chemistry, physics, and engineering. It also has practical implications in everyday life, such as understanding how a car engine works or how to properly inflate a balloon to avoid popping. Recognising how volume and pressure are connected can help explain and predict the behavior of gases under different conditions.