Problem 29 - Chemistry and Chemical Reactivity

Question

These questions are not designated as to type or location in the chapter. They may combine several concepts. Suppose 0.086 mol of \(\mathrm{Br}_{2}\) is placed in a \(1.26-\mathrm{L}\). flask and heated to \(1756 \mathrm{K},\) a temperature at which the halogen dissociates to atoms $$ \mathrm{Br}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{Br}(\mathrm{g}) $$ If \(\mathrm{Br}_{2}\) is \(3.7 \%\) dissociated at this temperature, calculate \(K\)

Step-by-Step Solution

Answer

The equilibrium constant \( K \) is approximately 0.000389.

Step 1: Determine initial conditions

Initially, we have 0.086 mol of \( \mathrm{Br}_2 \) in a 1.26 L flask. Calculate the initial concentration of \( \mathrm{Br}_2 \) using \( \text{Concentration} = \frac{\text{moles}}{\text{volume}} \). Thus, the initial concentration is \( \frac{0.086 \text{ mol}}{1.26 \text{ L}} = 0.0683 \text{ mol/L}. \)

Step 2: Calculate change due to dissociation

Since \( \mathrm{Br}_2 \) is 3.7% dissociated, 0.037 of the initial \( \mathrm{Br}_2 \) concentration converts to \( \mathrm{Br} \). Calculate the change: \( 0.0683 \times 0.037 = 0.00253 \text{ mol/L} \). Thus, \( [\mathrm{Br}_2] \) decreases by this amount and \( [\mathrm{Br}] \) increases by twice this amount.

Step 3: Calculate equilibrium concentrations

At equilibrium, the concentration of \( \mathrm{Br}_2 \) is \( 0.0683 - 0.00253 = 0.06577 \text{ mol/L} \). The concentration of \( \mathrm{Br} \), produced is twice the change, so \( 2 \times 0.00253 = 0.00506 \text{ mol/L} \).

Step 4: Write the expression for the equilibrium constant

The equilibrium expression for the reaction \( \mathrm{Br}_2(g) \rightleftarrows 2 \mathrm{Br}(g) \) is \( K = \frac{[ \mathrm{Br} ]^2}{[ \mathrm{Br}_2 ]} \).

Step 5: Substitute equilibrium concentrations into expression

Substitute the equilibrium concentrations found in Step 3 into the expression: \( K = \frac{(0.00506)^2}{0.06577} \).

Step 6: Calculate the value of K

Calculate \( K \) by evaluating the expression: \( K = \frac{(0.00506)^2}{0.06577} \approx 0.000389 \).

Key Concepts

Dissociation Reaction

Understanding dissociation reactions is vital when discussing chemical equilibrium. A dissociation reaction occurs when a compound breaks into smaller parts, typically molecules or ions. In our example, bromine gas (\(\mathrm{Br}_2(g)\) breaks into two bromine atoms (\(2 \mathrm{Br}(g)\)).Here are a few key takeaways about dissociation in chemical reactions:

• Not all reactants will dissociate to the same extent at a given temperature or pressure. Each system has unique equilibrium conditions.
• In the provided exercise, 3.7% dissociation means that only this fraction of the original \(\mathrm{Br}_2\) molecules break up into separate atoms.
• After dissociation occurs, the reaction can proceed in both directions as it reaches equilibrium.

Understanding these fundamentals helps in analyzing the extent of a reaction and predicting concentrations of components at equilibrium.

Equilibrium Constant (K)

The equilibrium constant, known as \(K\), is a central theme in the study of chemical reactions reaching equilibrium. It provides a measure of the position of equilibrium for a given reaction.For the dissociation of \(\mathrm{Br}_2\) into \(2\mathrm{Br}\), the equilibrium constant \(K\) is expressed by the formula:\[K = \frac{[ \mathrm{Br} ]^2}{[ \mathrm{Br}_2 ]}\]This formula considers the concentrations of the products and reactants:

• The \([ \mathrm{Br} ]\) represents the equilibrium concentration of bromine atoms.
• The \([ \mathrm{Br}_2 ]\) is the equilibrium concentration of bromine molecules.

To calculate \(K\):1. Find the equilibrium concentrations of all species involved (as we did in the exercise).2. Insert these values into the equilibrium expression.With \(K\) calculated, chemists gain insight into whether reactants or products are favored at equilibrium. A larger \(K\) suggests the formation of more products at equilibrium, whereas a smaller\(K\) indicates a greater amount of reactants remaining.

Concentration Calculations

Concentration calculations are essential for interpreting results and predicting the outcome of a chemical reaction.The initial concentration of a substance is calculated using the formula:\[\text{Concentration} = \frac{\text{moles}}{\text{volume}}\]In the provided task, the initial concentration of \(\mathrm{Br}_2\) was calculated using the flask's volume and the moles of \(\mathrm{Br}_2\).After calculating the initial concentration, modifications occur due to dissociation. When a reaction reaches equilibrium, these changes need to be considered:

• The concentration of the dissociated species must be calculated by determining the percentage dissociation.
• The equilibrium concentrations are used in subsequent calculations, such as finding the equilibrium constant.

Each change in concentration is crucial in establishing the equilibrium position and calculating \(K\). Having these calculations mastered will further one's understanding of the dynamic balance in chemical reactions.