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 \(1.00-\mathrm{L}\) vessel containing an equilibrium mixture of the gases has \(1.50 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel?

Short answer

The mass of Br(g) in the vessel at equilibrium is approximately 0.72 g.

Step-by-step solution

Write the balanced chemical equation and Kc expression

The balanced chemical equation for the reaction is already given: \[ \mathrm{Br_2(g)} \rightleftharpoons 2\mathrm{Br(g)} \] The Kc expression for this reaction is: \[ K_c = \frac{[\mathrm{Br}]^2}{[\mathrm{Br_2}]} \] Where [Br] and [Br2] represent the molar concentrations of the bromine atoms and bromine gas at equilibrium, respectively.

Convert mass of Br2 to moles

We need to convert the given mass of Br2 (1.50 g) into moles using its molar mass. The molar mass of Br2 is: \[ \mathrm{M(Br_2)} = 2 \times \mathrm{M(Br)} = 2 \times 79.904 = 159.808\,\mathrm{g/mol} \] Now, calculate the moles of Br2: \[ \mathrm{moles\,of\,Br_2} = \frac{\mathrm{mass\,of\,Br_2}}{\mathrm{molar\,mass\,of\,Br_2}} = \frac{1.50\,\mathrm{g}}{159.808\,\mathrm{g/mol}} =9.39\,\mathrm{x\,10^{-3}\,mol} \]

Set up an ICE table

We will set up an ICE (Initial, Change, Equilibrium) table to find the molar concentrations of Br2 and Br at equilibrium. | | Br2 | -> | 2 Br | |-------|-----|------|-------| |Initial| x | | 0 | |Change |-y | | 2y | |Equilibrium| x-y | | 2y | Where x represents the initial moles of Br2 and y represents moles of Br2 that dissociate at equilibrium. We already calculated x as 9.39 x 10^-3 mol.

Calculate the molar concentrations and solve for y

Now, we will use the Kc expression and the ICE table to solve for y. \[ K_c = \frac{[\mathrm{Br}]^2}{[\mathrm{Br_2}]} = \frac{(2\,y)^2}{x - y} \] The vessel has a volume of 1.00 L, so the molar concentrations at equilibrium can be calculated like this: \[ 1.04 \times 10^{-3} = \frac{((2\,y)/1.00)^2}{(x - y)/1.00} \] Now, solve for y: \[ y = 0.0045 \,\text{mol} \]

Calculate the mass of Br at equilibrium

Since two moles of Br are produced for every mole of Br2 dissociated, the number of moles of Br at equilibrium is twice the value of y: \[ \mathrm{moles\,of\,Br} = 2 \times y = 2 \times 0.0045 \,\text{mol} = 0.0090\,\text{mol} \] Now, we can calculate the mass of Br using its molar mass: \[ \mathrm{mass\,of\,Br} = \mathrm{moles\,of\,Br} \times \mathrm{M(Br)} = 0.0090 \,\text{mol} \times 79.904 \,\text{g/mol} \approx 0.72\,\text{g} \] Thus, there is approximately 0.72 g of Br(g) in the vessel at equilibrium.

Key concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant (\( K_c \)) is a crucial parameter. It represents the ratio of the concentration of products to reactants at equilibrium for a reversible chemical reaction. This constant indicates how far a reaction proceeds before reaching equilibrium. Each reaction has a unique \( K_c \) value under specific conditions like temperature. A small \( K_c \), like \( 1.04 \times 10^{-3} \) in our exercise, suggests that at equilibrium, the concentration of reactants outweighs that of products. This means the reaction favors the reactants.For instance, in the example reaction \( \mathrm{Br_2(g)} \rightleftharpoons 2\mathrm{Br(g)} \), \( K_c \) is given by \( \frac{[\mathrm{Br}]^2}{[\mathrm{Br_2}]} \), demonstrating the relationship between the concentrations of Br and Brtwo. Understanding \( K_c \) helps predict the direction of the reaction under specific conditions.

ICE Table

An ICE table is a structured approach to solving equilibrium problems in chemistry that stands for Initial, Change, and Equilibrium. By setting up an ICE table, we can efficiently track changes in concentration or pressure across the different stages of a chemical reaction.Here's a brief guide on how to use an ICE table with our given example:

• **Initial:** Start by noting down the initial concentration or moles of reactants and products. For \( \mathrm{Br_2(g)} \), we have calculated this as \( 9.39 \times 10^{-3} \,\mathrm{mol} \).
• **Change:** Indicate the change in concentration as the reaction moves toward equilibrium. We assign \( -y \) to \( \mathrm{Br_2(g)} \) and \( +2y \) to \( \mathrm{Br(g)} \).
• **Equilibrium:** These values provide us with the concentrations at equilibrium, expressed as \( x-y \) and \( 2y \) respectively.

With this setup, we can solve for unknown variables like \( y \) using the equilibrium constant expression. This helps us find out how the substances' concentrations change over time until they reach equilibrium conditions.

Molar Concentration

Molar concentration, often referred to in chemistry as molarity, is a measure of the concentration of a solute in a solution. This is expressed in moles per liter (mol/L). It's vital for quantifying how much solute is present in a given volume of solution.Equilibrium problems often require converting masses to molar concentrations to solve for unknowns. Start by using the formula:\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]In our exercise, the mass of \( \mathrm{Br_2} \) was first converted to moles using its molar mass \( 159.808 \, \mathrm{g/mol} \). Given the reaction occurs in a 1.00 L container, this allows the direct calculation of molar concentration:\[\text{Molarity of \( \mathrm{Br_2} \)} = \frac{9.39 \times 10^{-3} \,\mathrm{mol}}{1.00\,\mathrm{L}} = 9.39 \times 10^{-3}\, \text{mol/L}\]This step is critical as it lays the foundation for applying the equilibrium constant and determining the changes in concentration during the reaction.

Balanced Chemical Equation

A balanced chemical equation is the representation of a chemical reaction with an equal number of atoms for each element on both sides of the equation. This is essential for conserving mass and charge in a reaction.For the reaction in our exercise, the balanced equation is:\[\mathrm{Br_2(g)} \rightleftharpoons 2\mathrm{Br(g)}\]This equation tells us that one molecule of bromine gas (\( \mathrm{Br_2} \)) dissociates into two bromine atoms (\( \mathrm{Br} \)). Balancing a chemical equation is crucial as it allows us to use stoichiometry to accurately determine the relationships between reactants and products. In this case, it shows that every dissociation of one \( \mathrm{Br_2} \) produces two \( \mathrm{Br} \) atoms. Hence, our calculations involving stoichiometry and the ICE table rely on having the correct balanced equation. Without balancing, it wouldn't be possible to accurately predict the outcome of a reaction or use equilibrium constants correctly.