Problem 79 - Chemistry The Central Science In Si Units

Question

For the equilibrium $$ 2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\operatorname{Br}_{2}(g) $$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If \(5.07 \mathrm{kPa}\) of IBr is placed in a 10.0-L container, what is the partial pressure of all substances after equilibrium is reached?

Step-by-Step Solution

Answer

The equilibrium pressures are approximately 0.372 kPa for \(\mathrm{I}_2\) and \(\operatorname{Br}_2\), and 4.326 kPa for \(\operatorname{IBr}\).

Step 1: Write down the equilibrium expression

The equilibrium constant expression for a gas-phase reaction involving partial pressures is given by: \[ K_p = \frac{P_{I_2} \cdot P_{Br_2}}{P_{IBr}^2} \] The problem states that \(K_p = 8.5 \times 10^{-3}\) at \(150^\circ C\). The task is to find the partial pressures of \( \operatorname{IBr} \), \(\mathrm{I}_2\), and \(\operatorname{Br}_2\) at equilibrium.

Step 2: Establish initial conditions

Initially, \(P_{IBr} = 5.07\) kPa, and \(P_{I_2} = P_{Br_2} = 0\) kPa because there are no \(\mathrm{I}_2\) or \(\operatorname{Br}_2\) initially present in the container.

Step 3: Define the change in pressures

Suppose \(x\) kPa of \(\operatorname{IBr}\) dissociates to form \(\mathrm{I}_2\) and \(\operatorname{Br}_2\). From stoichiometry, - The decrease in \(P_{IBr}\) is \(2x\) due to the 2:1 stoichiometry.- Both \(P_{I_2}\) and \(P_{Br_2}\) increase by \(x\) due to the 1:1 stoichiometry.

Step 4: Write expressions for equilibrium pressures

At equilibrium, the partial pressures are:\[ P_{IBr} = 5.07 - 2x \]\[ P_{I_2} = x \]\[ P_{Br_2} = x \]

Step 5: Solve for x using the equilibrium expression

Substitute the equilibrium pressures into the equilibrium expression:\[ K_p = \frac{x \cdot x}{(5.07 - 2x)^2} = 8.5 \times 10^{-3} \]\[ x^2 = 8.5 \times 10^{-3} \cdot (5.07 - 2x)^2 \] This equation needs to be solved for \(x\) using algebraic methods.

Step 6: Simplify and solve the quadratic equation

Expanding and simplifying,\[ x^2 = 8.5 \times 10^{-3} \cdot (25.7049 - 20.28x + 4x^2) \]\[ x^2 = 0.21849265 - 0.17238x + 0.034x^2 \]Rearrange as: \[ 0.966x^2 - 0.17238x + 0.21849265 = 0 \]Use the quadratic formula to solve for \(x\):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 0.966\), \(b = -0.17238\), \(c = 0.21849265\).

Step 7: Calculate the values of x

Substitute values into the quadratic formula:\[ x = \frac{-(-0.17238) \pm \sqrt{(-0.17238)^2 - 4 \cdot 0.966 \cdot 0.21849265}}{2 \cdot 0.966} \]\[ x = \frac{0.17238 \pm \sqrt{0.029704244 - 0.84591334}}{1.932} \]Solve under the square root to find feasible \(x\) considering positive values only.

Step 8: Calculate final equilibrium pressures

Using the solved value of \(x\), calculate:\[ P_{I_2} = x \]\[ P_{Br_2} = x \]\[ P_{IBr} = 5.07 - 2x \]Complete calculations for exact numerical pressures of each gas.

Key Concepts

Understanding the Equilibrium Constant

In chemical reactions involving gases, like the reaction of iodine monobromide (\( \operatorname{IBr}\)) dissociating into \( \mathrm{I}_{2}(g)\) and \( \operatorname{Br}_{2}(g)\), an equilibrium state is reached where the concentrations of reactants and products no longer change over time. The equilibrium constant, \( K_{p}\), provides a numeric representation of the position of this equilibrium.

• It indicates whether products or reactants are more favoured at equilibrium.
• A large \( K_{p}\) (greater than 1) means more products, while a small one (less than 1) means more reactants.
• For the reaction \( 2 \, \operatorname{IBr} (g) \rightleftharpoons \mathrm{I}_{2}(g) + \operatorname{Br}_{2}(g)\), \( K_{p} = 8.5 \times 10^{-3}\)

This suggests that at 150°C, the system prefers to stay more as \( \operatorname{IBr}\) rather than converting completely into \( \mathrm{I}_{2}\) and \( \operatorname{Br}_{2}\). To calculate it, the relationship involves the partial pressures of these gases at equilibrium. Knowing how to manipulate \( K_p\) is essential when predicting or calculating the extent of a reaction.

Exploring Partial Pressure in Gas-Phase Reactions

Partial pressure is a crucial concept when dealing with gas-phase reactions. It describes the pressure exerted by an individual gas in a mixture of non-reacting gases. For each gas, it relates directly to its concentration. The total pressure of a gas mixture is the sum of the partial pressures of all gases present.
For the reaction of \( \operatorname{IBr}(g)\), \( \mathrm{I}_{2}(g)\), and \( \operatorname{Br}_{2}(g)\), \( P_{IBr}\) initially was given as 5.07 kPa.

• To reach equilibrium, \( \operatorname{IBr}\) dissociates, and this change affects its partial pressure.
• The partial pressures for \( \mathrm{I}_{2}\) and \( \operatorname{Br}_{2}\) start at zero and will increase as more \( \operatorname{IBr}\) decomposes.

These partial pressures are used in the equilibrium constant expression: \( K_p = \frac{P_{I_2} \cdot P_{Br_2}}{P_{IBr}^2}\). Understanding this helps to predict and calculate the equilibrium state, especially when starting conditions of gases change.

Mechanics of a Gas-Phase Reaction

A gas-phase reaction involves reactants and products in the gaseous state reacting within a confined space, like a container. The behavior and equilibrium of gases follow specific laws. Ideal gas law and Dalton’s Law are often used to describe and predict these reactions.
For the given reaction: \( 2 \operatorname{IBr} (g) \rightleftharpoons \mathrm{I}_{2}(g) + \operatorname{Br}_{2}(g)\), working with pressures complements our understanding instead of concentrations.

• The reaction starts with a certain amount of one gas and progresses towards an equilibrium where partial pressures reach specific values.
• In gas-phase reactions, the temperature and initial pressure play significant roles in determining the direction and extent of the reaction.

The fact that these are gases allows for a uniform mixture and swift equilibria, built upon the principles of kinetic molecular theory.

Role of Stoichiometry in Chemical Equilibrium

Stoichiometry plays a critical role in predicting how gas-phase reactions proceed to equilibrium. It refers to the ratios in which reactants get converted to products, which you can think of as a recipe for a chemical reaction. In the equilibrium exercise involving \( 2 \operatorname{IBr} (g) \rightleftharpoons \mathrm{I}_{2}(g) + \operatorname{Br}_{2}(g)\), understanding stoichiometry helps in mapping out pressure changes as the reaction progresses.

• The initial setup of 5.07 kPa of \( \operatorname{IBr}\) is altered as the reaction moves forward because of stoichiometric proportions.
• Stoichiometry allows us to relate these changes directly with the variables \( x\) we define for pressure variations.
• For example, for every mole of \( \operatorname{IBr}\) dissociating, half of that amount generates the same moles of \( \mathrm{I}_{2}\) and \( \operatorname{Br}_{2}\)

Utilizing stoichiometry helps in calculating exact pressures at equilibrium and ensures that you consider conservation of mass and energy consistently.