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 \(0.200-\mathrm{L}\) vessel containing an equilibrium mixture of the gases has \(0.245 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel?

Short answer

The mass of \(\mathrm{Br}(g)\) in the vessel can be found by following these steps: 1. Convert the mass of \(\mathrm{Br}_{2}\) to moles: \(\frac{0.245\mathrm{g}}{159.8\mathrm{g/mol}}\) 2. Set up an ICE table and represent the change in moles with x. 3. Write the equilibrium constant expression and solve for x: \(1.04 \times 10^{-3} = \frac{(2x)^{2}}{\frac{0.245}{159.8}-x}\) 4. Find the equilibrium concentration of both species and calculate the mass of \(\mathrm{Br}(g)\): Mass of \(\mathrm{Br}(g)\) = moles of \(\mathrm{Br}\) \(\times\) molar mass of \(\mathrm{Br}\) = \(2x \times 79.9\) g

Step-by-step solution

Convert \(\mathrm{Br}_{2}\) mass to moles

To find the number of moles for \(\mathrm{Br}_{2}\), divide the given mass by the molar mass of \(\mathrm{Br}_{2}\). The molar mass of \(\mathrm{Br}_{2}\) is approximately 159.8 g/mol (by adding the molar mass of two bromine atoms). moles of \(\mathrm{Br}_{2}\) = \(\frac{0.245\mathrm{g}}{159.8\mathrm{g/mol}}\)

Set up an ICE table

Use the moles of \(\mathrm{Br}_{2}\) we just calculated as the initial amount and set up an ICE table. When the decomposition reaction reaches equilibrium, there will be an increase in the number of moles of the \(\mathrm{Br}\) and a decrease in the number of moles of \(\mathrm{Br}_{2}\). Let the increase in moles for \(\mathrm{Br}\) and decrease for \(\mathrm{Br}_{2}\) be x. \( \begin{array}{c|c|c|c} & \mathrm{Br}_{2}(g) & \rightleftharpoons & 2\mathrm{Br}(g) \\ \hline I & \frac{0.245}{159.8} & & 0 \\ C & -x & & +2x \\ E & \frac{0.245}{159.8}-x & & 2x \\ \end{array} \)

Write the equilibrium constant expression and solve for x

Write the equilibrium constant expression in terms of x and the equilibrium concentrations of \(\mathrm{Br}_{2}\) and \(\mathrm{Br}\). \(K_{c}=\frac{[\mathrm{Br}]^{2}}{[\mathrm{Br}_{2}]}\) To write the expression using x, substitute equilibrium values from the ICE table: \(1.04 \times 10^{-3} = \frac{(2x)^{2}}{\frac{0.245}{159.8}-x}\) Now solve for x. It is a quadratic equation.

Find the equilibrium concentration of both species and calculate the mass of \(\mathrm{Br}(g)\)

Once we have the value of x, we can find the equilibrium concentrations of both \(\mathrm{Br}_{2}\) and \(\mathrm{Br}\) using the ICE table. Make sure to convert moles into concentrations by dividing the equilibrium moles by the volume of the vessel, which is 0.200 L. Equilibrium concentration of \(\mathrm{Br}_{2}\): \([\mathrm{Br}_{2}] = \frac{0.245}{159.8}-x\) Equilibrium concentration of \(\mathrm{Br}\): \([\mathrm{Br}] = 2x\) Mass of \(\mathrm{Br}(g)\): moles of \(\mathrm{Br}\) \(\times\) molar mass moles of \(\mathrm{Br}\) = \(2x\) Mass of \(\mathrm{Br}(g)\) = moles of \(\mathrm{Br}\) \(\times\) molar mass of \(\mathrm{Br}\) = \(2x \times 79.9\) g

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_c \), is a critical concept in chemical equilibrium. It provides a measure of the concentrations of reactants and products at equilibrium for a reversible reaction at a specific temperature. In this exercise, we examine the reaction \( \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g) \). The equilibrium constant expression for this reaction is derived from the balanced chemical equation:

• \( K_{c} = \frac{[\mathrm{Br}]^{2}}{[\mathrm{Br}_{2}]} \)

Here, \([\mathrm{Br}]\) and \([\mathrm{Br}_{2}]\) represent the molar concentrations of bromine atoms and bromine molecules, respectively.
At \( 1285^{\circ} \mathrm{C} \), the equilibrium constant \( K_c \) is given as \( 1.04 \times 10^{-3} \). This small value indicates that the reaction favors the reactants over products, meaning there are more \( \mathrm{Br}_{2} \) molecules than \( \mathrm{Br} \) atoms at equilibrium. Understanding how to use this constant in calculations is fundamental to predicting the behavior of chemical systems in equilibrium.

ICE Table

An ICE table is a structured format that organizes the Initial concentrations, the Change that occurs as the reaction proceeds, and the Equilibrium concentrations of reactants and products. It is a crucial tool for solving equilibrium problems in chemistry.
For the reaction \( \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g) \), we set up the ICE table using the initial moles of \( \mathrm{Br}_{2} \), calculated from its mass. This step allows us to keep track of any change in moles as the reaction moves towards equilibrium. The amount \( x \) is introduced to signify the change:

• Decrease in moles of \( \mathrm{Br}_{2}: -x \)
• Increase in moles of \( \mathrm{Br}: +2x \)

At equilibrium:

• \( [\mathrm{Br}_{2}] = \frac{0.245}{159.8} - x \)
• \( [\mathrm{Br}] = 2x \)

Utilizing the ICE table is essential for systematically solving the equilibrium problem and efficiently applying the equilibrium expression.

Mass to Moles Conversion

Converting mass to moles is an important operation in chemistry needed to solve problems involving chemical reactions. It relates the mass of a substance to the number of particles, allowing us to use stoichiometric calculations effectively.
In this exercise, we begin by determining the number of moles of \( \mathrm{Br}_{2} \) given its mass of 0.245 g. The molar mass of \( \mathrm{Br}_{2} \) is approximately 159.8 g/mol:

• Moles of \( \mathrm{Br}_{2} = \frac{0.245\mathrm{~g}}{159.8\mathrm{~g/mol}} \)

This conversion sets the stage for further calculations, such as setting up the ICE table, to determine the equilibrium concentrations in the vessel. Having accurate mole values is critical for moving to the subsequent steps in equilibrium calculations.

Quadratic Equations in Chemistry

Quadratic equations often arise in chemistry when dealing with equilibrium problems, especially when the change in moles \( x \) must be solved based on the equilibrium constant expression. To find the value of \( x \) in this case, we need to rearrange and manipulate the equilibrium expression into a quadratic form:

• \( 1.04 \times 10^{-3} = \frac{(2x)^{2}}{\frac{0.245}{159.8}-x} \)

After expanding and simplifying, this leads to a quadratic equation in the standard form\( ax^2 + bx + c = 0 \), which can be solved using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This will provide the value of \( x \), which can then be used to find the equilibrium concentrations of the species involved. Understanding how to solve quadratic equations is vital for finding solutions to many equilibrium problems in chemistry.