Question

A system at equilatium contains \(I_{2}(\xi)\) at a pressure of 0.21 atm and \(1(g)\) at a pressure of 0.23 atm. The system is then comptessed to half its volume. Find the pressure of each gas when the system returns to equilabrium.

Short answer

The pressure of \(I_{2}(\theta)\) will be 0.42 atm, and the pressure of \(I(g)\) will be 0.46 atm when the system returns to equilibrium.

Step-by-step solution

State the Combined Gas Law

Understand that the Combined Gas Law, which relates pressure, volume, and temperature of a gas, is given by the formula: \(P_1V_1 = P_2V_2\) for a given constant temperature and amount of gas. Since the temperature and the amount of the two gases do not change, we can use this law to relate the initial and final pressures and volumes.

Apply the Combined Gas Law for each gas

Use the formula to find the final pressures of each gas. If the volume is halved \(V_2 = \frac{V_1}{2}\), the pressure of each gas doubles assuming the temperature is constant. Calculate the final pressure for each gas using \(P_2 = \frac{P_1V_1}{V_2}\) where \(V_2 = \frac{1}{2}V_1\).

Calculate the final pressure for Iodine

Calculate the final pressure for \(I_2\) gas: \(P_{2,I_2} = \frac{P_{1,I_2}V_1}{V_2} = \frac{0.21 \text{ atm} \cdot V_1}{\frac{1}{2}V_1} = 0.21 \text{ atm} \cdot 2 = 0.42 \text{ atm}\).

Calculate the final pressure for Iodine derivatives

Calculate the final pressure for \(I\) gas: \(P_{2,I} = \frac{P_{1,I}V_1}{V_2} = \frac{0.23 \text{ atm} \cdot V_1}{\frac{1}{2}V_1} = 0.23 \text{ atm} \cdot 2 = 0.46 \text{ atm}\).

Key concepts

Understanding Gas Pressure

Gas pressure is a measure of the force that gas molecules exert on the walls of their container. It's a result of the many small collisions between the gas molecules and the container surfaces. The more frequent or forceful these collisions, the higher the pressure. In the context of our exercise, we started with an iodine molecule, represented as \( I_{2}(g) \) at a pressure of 0.21 atm, and iodine atoms \( I(g) \) at a pressure of 0.23 atm.

When the volume is decreased, the gas molecules have less space to move around in, which leads to an increase in the frequency of collisions between the gas molecules and the container. According to the Combined Gas Law, which we'll discuss in more detail, reducing the volume of a gas, while keeping its temperature constant, causes its pressure to increase. This is precisely what happens when the system is compressed to half its volume, leading to a predicted doubling of the pressure for each gas if the system returns to equilibrium.

Exercise Improvement Advice: Remember, it's critical to assume that temperature remains unchanged when applying the Combined Gas Law, and don't forget to consider changes in volume can have large impacts on pressure.

Gas Volume and Its Importance

Gas volume, which is the space that a gas occupies, plays a key role in determining the gas pressure as well. The Combined Gas Law, where \( P_1V_1 = P_2V_2 \) for constant temperature and amount of gas, shows the inverse relationship between pressure and volume.

If you start with a certain gas volume and compress the gas to half that volume (without changing the temperature or the amount of gas), this law tells us that the pressure will double, as observed in Step 2 of the solution. This insight isn't just important for solving textbook problems; it's fundamental to understanding how gases behave in real-world applications, from car engines to weather balloons.

Exercise Improvement Advice: To clearly understand this part of the exercise, visualize or conduct a simple physical experiment if possible, where reducing the volume available to a gas—like compressing a syringe—leads to an increase in pressure.

Chemical Equilibrium and its Dynamics

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of the reactants and products. It's a dynamic state, meaning that the reactions are still occurring, but the overall concentrations remain consistent over time.

In our exercise, the system containing \( I_{2}(g) \) and \( I(g) \) is at equilibrium, with their initial pressures given. When the volume is halved, it disrupts the equilibrium by increasing the pressures. However, the system will eventually adjust to reach a new equilibrium at the new volume, which is reflected in the higher pressures calculated in Steps 3 and 4.

Exercise Improvement Advice: To grasp this concept better, think of equilibrium as a balanced scale. When something changes (like the volume), the scale temporarily tips one way (increased pressure), but it will eventually find a new balance (a new equilibrium pressure).