Question

For the equilibrium $$ 2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) $$ \(K_{p}=8.5 \times 10^{-3}\) 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, the partial pressures of all substances in the given reaction are: IBr, \(0.00942\) atm; I2, \(0.00829\) atm; and Br2, \(0.00829\) atm.

Step-by-step solution

Write the expression for Kp

To start, we need to write the expression for Kp for the given equilibrium: $$ K_p = \frac{[\mathrm{I_2}][\mathrm{Br}_2]}{[\mathrm{IBr}]^2} $$

Find the initial concentrations

Given a 2.0 L container and an initial IBr pressure of 0.025 atm, we can find the initial concentrations of all species. At the initial state, the concentrations of I2 and Br2 are 0, since the IBr has not dissociated yet. For IBr, we have: $$ [\mathrm{IBr}] \, \text{initial} = \frac{0.025 \, \text{atm}}{(2.0 \, \text{L})(0.0821 \, \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}})(423 \, \text{K})} \approx 7.45 \times 10^{-4} \, \text{M} $$

Set up an ICE table

Create an ICE table to track the changes in the concentration of the species: ``` IBr → I2 + Br2 Initial: 7.45E-4 → 0 0 Change: -2x → x x Equil: 7.45E-4-2x → x x ```

Write the Kp expression in terms of equilibrium concentrations

Using the ICE table, we can express the equilibrium concentrations in terms of x (the change in concentration). We can substitute these expressions back into our Kp equation: $$ 8.5 \times 10^{-3} = \frac{x^2}{(7.45 \times 10^{-4}-2x)^2} $$

Solve for x

Now, we need to solve for x, which represents the change in concentration at equilibrium. This is a quadratic equation, but since Kp is small, we can make an assumption that x is small compared to the initial concentration of IBr, so 2x << 7.45×10^(-4). Therefore, we can simplify the equation and solve for x: $$ 8.5 \times 10^{-3} \approx \frac{x^2}{(7.45 \times 10^{-4})^2} $$ $$ x \approx 6.07 \times 10^{-4}\,\text{M} $$

Convert to partial pressures and answer

Finally, we will update the ICE table with the value of x and convert the equilibrium concentrations back into partial pressures. This will give us the partial pressures at equilibrium: $$ P_{\mathrm{IBr}} = (7.45\times10^{-4}-2(6.07\times10^{-4}))\times(2\,\text{L})\times 0.0821\,\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\times 423\,\text{K} \approx 0.00942\,\text{atm} $$ $$ P_{\mathrm{I_2}} = P_{\mathrm{Br_2}} = 6.07\times10^{-4}\,\text{M}\times(2\,\text{L})\times 0.0821\,\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\times 423\,\text{K} \approx 0.00829\,\text{atm} $$ At equilibrium, the partial pressures of all substances are: IBr, 0.00942 atm; I2, 0.00829 atm; and Br2, 0.00829 atm.

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_p \) when dealing with gases, plays a pivotal role in understanding chemical reactions at equilibrium, especially in calculating the concentrations or partial pressures of reactants and products. It expresses the ratio of the product of concentrations of products to the product of concentrations of reactants, each raised to the power of their stoichiometric coefficients at equilibrium.
For the given reaction:

• \[ 2 \text{IBr}(g) \rightleftharpoons \text{I}_2(g) + \text{Br}_2(g) \]

The equilibrium constant formula is represented as:

• \[ K_p = \frac{[\text{I}_2][\text{Br}_2]}{[\text{IBr}]^2} \]

In this particular reaction, \( K_p \) is given as \( 8.5 \times 10^{-3} \) at \( 150^{\circ} \text{C} \). This implies that at this temperature, the system will naturally self-adjust until the ratio adheres to this value, maintaining equilibrium. A small \( K_p \) value indicates that the reactants are favored over the products at equilibrium under the given conditions.

Partial Pressure

Partial pressure is a critical concept used in gaseous equilibrium to express the pressure a gas exerts if it alone occupies the given volume. In the context of chemical equilibrium involving gases, it is essential to know each component's partial pressure to understand their reaction dynamics.
When we begin with a specific initial pressure, as in the case of IBr set at 0.025 atm, it is crucial to calculate how these pressures will change as the system reaches equilibrium. The partial pressure at equilibrium of each gas can be deduced from the change in the concentrations of reactants and products.

• After determining \( x \), the change in concentration, you can calculate the partial pressures using:
  • \[ P_{\text{IBr}} = (7.45 \times 10^{-4} - 2x) \times R \times T \]
  • \[ P_{\text{I}_2} = P_{\text{Br}_2} = x \times R \times T \]

Here, \( R \) denotes the universal gas constant, and \( T \) is the temperature in Kelvin. This calculation tells us how much each gas contributes to the total pressure at equilibrium.

ICE Table

An ICE table is a systematic way to follow the changes in concentration and partial pressures of reactants and products as a chemical system reaches equilibrium. ICE stands for Initial, Change, and Equilibrium:

• **Initial**: Start with the known initial concentrations or pressures of reactants and products. For instance, before any reaction, the initial pressure of IBr is 0.025 atm, while I\(_2\) and Br\(_2\) are zero.
• **Change**: Represent the changes occurring as the system moves towards equilibrium. Typically, this is expressed in terms of \( x \), where each component changes by its stoichiometric relationship.
• **Equilibrium**: Determine the equilibrium concentrations by applying the changes to the initial values. This stage utilizes the calculated \( x \) to deduce the end concentrations or pressures.
  • The equilibrium concentrations are key as they are substituted back into the expression for \( K_p \) to ensure it equals the known value, verifying the correctness of calculations.

By following the ICE table method, you can systematically solve for unknowns and analyze equilibrium systems with clarity and precision.