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. Which of the following statements describes how the volume of the container changes due to the reaction: (a) the volume increases by \(50 \%,(\mathbf{b})\) the volume increases by \(33 \%,\) (c) the volume remains constant, (d) the volume decreases by \(33 \%,(\) e) the volume decreases by 50\(\% .[\) Sections 10.3 and 10.4\(]\)

Short answer

The correct answer is option (d) the volume decreases by 33%. This is because the initial total moles of gas are 3 moles (2 CO and 1 O₂), and after the reaction, there are 2 moles of CO₂ gas. The new volume is \(\frac{2}{3}\) of the original volume, resulting in a 33.3% decrease in volume.

Step-by-step solution

Count the moles of gas before and after the reaction

In the reaction, we have 2 moles of CO and 1 mole of O₂ reacting to form 2 moles of CO₂. Before the reaction: Total moles of gas = 2 (CO) + 1 (O₂) = 3 moles After the reaction: Total moles of gas = 2 (CO₂) = 2 moles

Analyze the change in moles

From the reaction, we can observe that the total moles of gas have decreased from 3 moles to 2 moles. This will result in a change in the volume of the container since the number of moles of gas and volume are directly proportional, given a constant temperature and pressure.

Calculate the change in volume

Since the relationship between the number of moles and volume is direct(ideal gas law), we can find the change in volume by looking at the ratio of moles after to before the reaction: Change in volume: \(\frac{2\text{ moles}}{3\text{ moles}} = \frac{2}{3} = 0.6667\) Here, we can see that the new volume is 0.6667 of the original volume

Find the percentage change in volume

To find the percentage decrease in volume, we can compare the new volume to the original volume as follows: Percentage decrease in volume = \(\left(1 - \frac{2}{3}\right) × 100\% = 33.3\% \)

Compare the result to the given options

Comparing the result to the given options, we can see that our calculated percentage decrease in volume of 33.3% matches option d) the volume decreases by 33%. Thus, the correct answer is option (d) the volume decreases by 33%.

Key concepts

Ideal Gas Law

Understanding the behavior of gases in chemical reactions often requires familiarity with the Ideal Gas Law, expressed as the equation \( PV = nRT \). This fundamental relation connects the pressure (P), volume (V), and temperature (T) of a given amount of ideal gas to the number of moles (n) of the gas present, with R being the ideal gas constant.

The Ideal Gas Law predicts how a gas will change under different conditions. For instance, if the temperature and pressure of a certain quantity of gas remain consistent, any change in the number of moles directly affects the volume. This means if the amount of gas increases while pressure and temperature stay the same, the volume expands. Conversely, when the amount of gas decreases, the volume contracts.

In the context of our exercise, the Ideal Gas Law helps us understand why the volume of the container drops when the number of gas moles decreases after the chemical reaction. Since the pressure and temperature are constant, a decrease in moles results in a proportional decrease in volume, encapsulating the intuitive relationship that serves as the cornerstone for interpreting gas behavior in chemical reactions.

Moles and Gas Volume Relationship

The relationship between moles and gas volume is integral to gas stoichiometry and chemical reactions involving gases. According to Avogadro's law, equal volumes of all gases at the same temperature and pressure contain an equal number of molecules—or moles. This implies a direct, proportional relationship between the volume of a gas and the amount of substance in moles, given that temperature and pressure are held constant.

To visualize this, imagine a balloon being inflated; as more air (molecules) is added, the balloon (volume) increases accordingly. This same principle applies to gas in a reaction, where if the number of gas moles decreases, as seen in our example with CO and \( O_2 \) gases changing to \( CO_2 \) gas, the volume within the container would also decrease following the proportions indicated by Avogadro's law.

The exercise solution directly applies this relationship by comparing the moles before and after the reaction to determine the change in volume. The reduction in moles from 3 to 2 corresponds to a decrease in volume, underlining the straightforward linear relationship between the two properties for an ideal gas.

Gas Stoichiometry

Gas stoichiometry involves using the quantitative relationships between reactants and products in a chemical reaction to calculate the volumes, moles, or masses of gases involved. It essentially bridges the gap between chemical equations and measurable quantities of gases.

Most stoichiometric calculations for gases can be facilitated by the Ideal Gas Law and Avogadro's principle. These principles permit the conversion between moles of gases and their respective volumes, offering a practical route to solving problems involving gas-producing or gas-consuming reactions.

In our exercise, gas stoichiometry is essential to determining the volume change in the container. By counting the moles of reactants and products, we can predict the volume before and after the reaction to find the percentage change in volume. The step-by-step solution walks us through this by translating the stoichiometric coefficients in the balanced equation into a change in the volume of the gases. This highlights the practical utility of stoichiometry in connecting chemical equations with real-world physical changes in a system, such as the volume change in a gas reaction.