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 \(0.025 \mathrm{~atm}\) of \(\mathrm{IBr}\) is placed in a \(2.0\) - \(\mathrm{L}\) container, what is the partial pressure of this substance after equilibrium is reached?

Short answer

The partial pressure of IBr at equilibrium is 0.01268 atm.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction is given as: \( 2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \)

Write the expression for Kp

The expression for Kp, based on the balanced chemical equation, is: \( K_p = \dfrac{[I_2] [Br_2]}{[IBr]^2} \)

Set up an ICE table (Initial, Change, Equilibrium)

An ICE table helps to organize the initial pressures, changes in pressures, and equilibrium pressures. Let x be the change in pressure for IBr(g) at equilibrium: | | IBr | I2 | Br2 | |---------|------|----|------| | Initial | 0.025| 0 | 0 | | Change | -2x | x | x | | Equilibrium| 0.025-2x|x |x |

Substitute the pressures in Kp expression

Replace the pressures in Kp expression with the equilibrium pressures represented in the ICE table: \( K_p = \dfrac{x \cdot x}{(0.025 - 2x)^2} \) We are given that the value of Kp is \(8.5 \times 10^{-3}\). Substitute this value in the expression: \( 8.5 \times 10^{-3} = \dfrac{x^2}{(0.025 - 2x)^2} \)

Solve for x

To find the value of x, we can first rearrange the equation and then find the square root: \( x^2 = 8.5 \times 10^{-3} \times (0.025 - 2x)^2 \) Take the square root on both sides: \(|x|= \sqrt{8.5 \times 10^{-3} \times (0.025 - 2x)^2} \) Now, solve this equation for x. You may use a calculator or suitable numeric method to find the root. The value of x = 0.00616

Find the partial pressure of IBr at equilibrium

Using the equilibrium expression in the ICE table for IBr: Partial pressure of IBr at equilibrium = \(0.025 - 2x\) Substitute the value of x found above: Partial pressure of IBr at equilibrium = \(0.025 - 2(0.00616)\) Partial pressure of IBr at equilibrium = 0.01268 atm

Key concepts

Equilibrium Constant

When you encounter a chemical reaction that reaches equilibrium, the ratio of the concentration of products to the concentration of reactants is a significant figure known as the equilibrium constant, represented by the symbol K. In particular, for reactions involving gases, the equilibrium constant relating to the partial pressures of the gases is denoted as Kp. This key concept is critical in understanding how reactions behave when they are in balance, and the value of Kp indicates the extent of the reaction; a large Kp suggests products are favored, whereas a small Kp implies reactants are favored.

The equilibrium constant for a reaction is determined by the specific reaction conditions, such as temperature. In the given exercise, Kp is provided as 8.5 x 10^-3, which implies that at 150°C, the reaction heavily favors the reactants IBr.

ICE Table

Studying equilibrium also calls for a structured approach to track the concentration changes over the course of the reaction. This is where the ICE table (Initial, Change, Equilibrium) comes into play. It is a simple yet powerful tool for organizing and calculating the shifts in concentrations or partial pressures of reactants and products as they move towards equilibrium.

To use an ICE table, we start by entering the initial concentrations or pressures. Then, we define the change that occurs as the system reaches equilibrium, often expressed in terms of a variable 'x'. Finally, we derive the equilibrium concentrations or pressures by combining the initial measurements with the change. In the provided problem, we applied the ICE table to effectively modulate the information and solve for the change in pressure, 'x', which leads us to the equilibrium state of the reactants and products.

Partial Pressure

In gas-phase reactions, it’s essential to consider the partial pressure of each gas, which is the pressure that gas would exert if it alone occupied the whole volume. Partial pressure plays a crucial role in calculating equilibrium constants for gaseous reactions (Kp), and understanding the individual pressure exerted by each gas helps to predict the direction of the reaction and the quantities of each gas at equilibrium.

In our exercise, we were asked to determine the partial pressure of IBr after the system has reached equilibrium. Using the balanced chemical equation, the ICE table, and the known value of Kp, we calculated the change in the partial pressures and further estimated the partial pressure of IBr, using the logic and values derived from our systematic approach. This demonstrates the interconnectedness of the chemical equilibrium concepts and their practical application in predicting the behavior of chemical systems.