Question

For the equilibrium $$2 \operatorname{IBr}(g) \Longrightarrow \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C} .\) If 0.025 atm of IBr is placed in a 2.0 -L container, what is the partial pressure of all substances after equilibrium is reached?

Short answer

At equilibrium with an initial pressure of 0.025 atm for IBr, the partial pressures of IBr, I2, and Br2 are approximately 0.0164 atm, 0.00431 atm, and 0.00431 atm, respectively.

Step-by-step solution

Write down the balanced reaction and set up the ICE table.

The balanced reaction is given in the exercise: $$2 \operatorname{IBr}(g) \Longrightarrow \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ We set up an ICE table to keep track of the initial pressures, changes in pressures, and final equilibrium pressures. | | IBr | ⇌ | I2 | + | Br2 | |--------|-----|----|-----|---|------| |Initial | 0.025 atm | | 0 atm | | 0 atm | |Change | -2x | | +x | | +x | |Equilibrium | 0.025-2x | | x | | x | We know that the total pressure of IBr at the beginning is 0.025 atm. Since the reaction is at equilibrium, we need to find x.

Write the equation for Kp and substitute the equilibrium pressures.

The equation for Kp is given as: $$K_{p}=\frac{[\mathrm{I}_{2}] [\mathrm{Br}_{2}]}{[\operatorname{IBr}]^2}$$ Since we are dealing with partial pressures, we can just substitute the equilibrium pressures (ICE table) into the equation: $$8.5 \times 10^{-3}=\frac{(x)(x)}{(0.025-2x)^2}$$

Solve the equation for x.

To find the value of x, we need to solve this equation. It is a quadratic equation, which can be simplified as follows: $$8.5 \times 10^{-3} = \frac{x^2}{(0.025-2x)^2}$$ Rearrange the equation: $$x^2 = 8.5 \times 10^{-3}(0.025 - 2x)^2$$ Expand and solve for x: $$x^2 = 8.5 \times 10^{-3}(4x^2 - 0.1x + 0.000625)$$ $$x^2=34x^4 - 8.5x^3 + 5.31 \times 10^{-6}x^2$$ $$x^2 - 34x^4 + 8.5x^3 - 5.31 \times 10^{-6}x^2 = 0$$ Solving this equation for x, we find two possible values, one positive and one negative. Since x represents changes in pressure, the negative value doesn't make sense in this context so we only consider the positive value: $$x \approx 0.00431 $$

Determine the partial pressures of all substances at equilibrium.

With the value of x, we can now substitute it back into the ICE table and find the partial pressures of all substances at equilibrium: | | IBr | ⇌ | I2 | + | Br2 | |--------|-----|----|-----|---|------| |Initial | 0.025 atm | | 0 atm | | 0 atm | |Change | -2x | | +x | | +x | |Equilibrium | 0.025-2x | | x | | x | The equilibrium partial pressures are: IBr: 0.025 - 2(0.00431) = 0.0164 atm I2: 0.00431 atm Br2: 0.00431 atm So at equilibrium, the partial pressures of IBr, I2, and Br2 are approximately 0.0164 atm, 0.00431 atm, and 0.00431 atm, respectively.

Key concepts

Partial Pressures

In a gaseous chemical reaction, the term "partial pressure" refers to the pressure that each gas exerts in a mixture of gases. When gases share a container, their individual pressures add up to the total pressure. In the context of chemical equilibrium, understanding partial pressures is crucial because it allows us to determine how much of each reactant and product is present at equilibrium.

Consider the reaction:

• 2 IBr(g) ⇌ I2(g) + Br2(g).

If we begin with a specific amount of IBr in a container initially, it will have a certain partial pressure based on its moles and the volume of the container. As the reaction proceeds, IBr breaks down into I2 and Br2, and the partial pressure of each of these gases will change accordingly. The sum of these pressures will be equal to the total pressure.

Determining the partial pressures of gases at equilibrium is essential for calculating other key parameters, such as the equilibrium constant using the pressures expressed in atmospheres. Knowing these partial pressures helps in predicting how changes in conditions can impact the reaction dynamics.

ICE Table

An "ICE Table" is an essential tool for solving equilibrium problems step by step. ICE stands for Initial, Change, and Equilibrium. Using an ICE table helps to visualize how the concentrations (or pressures in terms of gases) of your chemicals change from start to equilibrium.

Let's set it up for our reaction:

• 2 IBr(g) ⇌ I2(g) + Br2(g).

1. **Initial**: Start by noting the initial partial pressures of each species. For instance, initially, IBr is the only one present. 2. **Change**: Define changes in terms of a variable, commonly 'x', based on the stoichiometry of the reaction. For the decomposition of IBr, if 'x' amount of I2 and Br2 form, 2x amount of IBr must decompose because of the 2 in front of IBr in the reaction. 3. **Equilibrium**: Add the change to your initial pressures to deduce equilibrium pressures.

In essence, an ICE table effectively organizes the data, facilitates calculations, and enhances understanding of equilibrium dynamics by providing a clear picture of the changes occurring through the reaction process.

Equilibrium Constant (Kp)

The Equilibrium Constant, denoted as Kp when using partial pressures, is a numerical value that relates the concentrations of reactants and products at equilibrium for a specific temperature. This constant is specific to a particular reaction and dictates how far a reaction will proceed towards products or remain with reactants at equilibrium.

For the reaction

• 2 IBr(g) ⇌ I2(g) + Br2(g),

Kp can be defined by the formula: \[ K_p = \frac{P_{I_2} \cdot P_{Br_2}}{P_{IBr}^2} \]where \(P_{I_2}\), \(P_{Br_2}\), and \(P_{IBr}\) are the partial pressures of I2, Br2, and IBr, respectively.

When calculating Kp, it provides insight into the extent of a reaction. A small Kp value (like in our problem, which is \(8.5 \times 10^{-3}\)) suggests that at equilibrium, the reaction favors the reactants, meaning only a small portion turns into the products. This concept is essential because it connects kinetic principles with thermodynamic stability, impacting how chemical systems are manipulated in practical scenarios.