Problem 47 - Chemistry The Central Science In Si Units

Question

At \(1285^{\circ} \mathrm{C},\) the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is \(K_{c}=1.04 \times 10^{-3} .\) A 1.00-L vessel containing an equilibrium mixture of the gases has \(1.50 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\operatorname{Br}(g)\) in the vessel?

Step-by-Step Solution

Answer

The mass of \( \mathrm{Br}(g) \) in the vessel is 0.25 g.

Step 1: Convert Mass to Moles for \( \mathrm{Br}_2 \)

First, calculate the number of moles of \( \mathrm{Br}_{2}(g) \). Use the molecular weight of \( \mathrm{Br}_{2} \), which is approximately 159.8 g/mol, to convert grams to moles: \[ \text{moles of } \mathrm{Br}_2 = \frac{1.50\, \text{g}}{159.8\, \text{g/mol}} = 0.00939\, \text{mol} \]

Step 2: Set up the Equilibrium Expression for \( K_c \)

Write the equilibrium expression using the given equilibrium constant, \( K_c \) for the reaction: \[ K_c = \frac{[\mathrm{Br}]^2}{[\mathrm{Br}_2]} \] where \([\mathrm{Br}_2]\) is the concentration of \( \mathrm{Br}_2(g) \) and \([\mathrm{Br}]\) is the concentration of \( \mathrm{Br}(g) \).

Step 3: Calculate the Concentrations at Equilibrium

Since the volume of the container is 1.00 L, the concentration of \( \mathrm{Br}_{2}(g) \) can be directly calculated from moles:\[ [\mathrm{Br}_2] = 0.00939 \text{ mol/L} \]Use this in the equilibrium expression:\[ K_c = \frac{[\mathrm{Br}]^2}{0.00939} \]

Step 4: Solve the Equilibrium Equation for \([\mathrm{Br}]\)

Substitute \( K_c = 1.04 \times 10^{-3} \) and solve for \([\mathrm{Br}]\) using:\[ [\mathrm{Br}]^2 = 1.04 \times 10^{-3} \times 0.00939 \]\[ [\mathrm{Br}]^2 = 9.76 \times 10^{-6} \]\[ [\mathrm{Br}] = \sqrt{9.76 \times 10^{-6}} = 0.00312 \text{ mol/L} \]

Step 5: Convert \([\mathrm{Br}]\) to Mass

Find the moles of \( \mathrm{Br}(g) \) in the vessel since the solution concentration is in moles per liter:\[ \text{moles of } \mathrm{Br} = 0.00312 \text{ mol/L} \times 1.00 \text{ L} = 0.00312 \text{ mol} \]Convert moles to grams using the atomic weight of \( \mathrm{Br} \), approximately 79.9 g/mol:\[ \text{mass of } \mathrm{Br} = 0.00312 \text{ mol} \times 79.9 \text{ g/mol} = 0.25 \text{ g} \]

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_c \), is a crucial concept in chemical equilibrium. It helps us understand how far a reaction will proceed before reaching its equilibrium state. In this instance, the equilibrium constant for the dissociation of bromine gas \( \mathrm{Br}_{2}(g) \) into bromine atoms \( \mathrm{Br}(g) \) is given by \( K_c = 1.04 \times 10^{-3} \).
This value tells us about the ratio of product and reactant concentrations when the system has reached equilibrium.
For the given reaction, the expression for the equilibrium constant is formulated as:

• \( K_c = \frac{[\mathrm{Br}]^2}{[\mathrm{Br}_2]} \)

At higher \( K_c \) values, products are favored, whereas lower \( K_c \) values indicate that reactants are more prevalent at equilibrium. The equilibrium constant is crucial for solving problems involving equilibrium concentrations, like the mass of \( \mathrm{Br}(g) \) in this case.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. A balanced chemical equation is key for stoichiometry and provides the mole ratio needed for calculations.
For the reaction \( \mathrm{Br}_{2}(g) \rightarrow 2 \mathrm{Br}(g) \), stoichiometry informs us that one mole of \( \mathrm{Br}_2 \) produces two moles of \( \mathrm{Br} \).
This mole ratio helps us determine the concentrations of materials at equilibrium:

• Each mole of \( \mathrm{Br}_2 \) dissociates to form two moles of \( \mathrm{Br} \).

Using stoichiometry, we know that even small amounts of \( \mathrm{Br}_{2} \) can increase the concentration of \( \mathrm{Br}(g) \) due to the 1:2 ratio. Therefore, stoichiometry is not only about balancing equations but also about leveraging these ratios to solve equilibrium problems.

Molar Concentration

Molar concentration, also known as molarity, describes how many moles of a substance are present per liter of solution. It is crucial for understanding how concentrated a substance is in a reaction mixture.
In our exercise, the molar concentration of \( \mathrm{Br}_2(g) \) is calculated using the formula:

• \( [\mathrm{Br}_2] = \frac{\text{moles of } \mathrm{Br}_2}{\text{Volume of Solution}} = 0.00939 \text{ mol/L} \)

Determining the molar concentration is essential before setting up the equilibrium expression. For \( \mathrm{Br}(g) \), it is calculated using the equilibrium constant, \( K_c \), and then converted to grams to find the mass. The use of molar concentration simplifies these calculations and is a vital skill in solving chemical equilibrium problems.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In reversible reactions, both the forward and backward reactions occur, moving towards a state of balance or equilibrium.
For the equation \( \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g) \), the double arrow indicates it is reversible. Here, \( \mathrm{Br}_2(g) \) dissociates into \( \mathrm{Br}(g) \), and \( \mathrm{Br}(g) \) can also recombine to form \( \mathrm{Br}_2(g) \).
In such systems, equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction. Being able to set up and solve the expressions for these reactions, as detailed with the equilibrium constant, allows scientists and students to predict and calculate the extent and concentrations of a reaction at equilibrium. Understanding these dynamics is fundamental for exploring the balance in chemical reactions.