Question

(a) 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? (b)For the reaction \(\mathrm{H}_{2}(g)+\mathrm{l}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at \(700 \mathrm{~K}\). In a 2.00-L flask containing an equilibrium mixture of the three gases, there are \(0.056 \mathrm{~g} \mathrm{H}_{2}\) and \(4.36 \mathrm{~g} \mathrm{I}_{2}\). What is the mass of HI in the flask?

Short answer

For part (a), the mass of \(\mathrm{Br}(g)\) in the vessel is 0.045 g. For part (b), the mass of \(\mathrm{HI}(g)\) in the flask is 658 g.

Step-by-step solution

Write the reaction and find the molar masses of the species involved.

The reaction is given as: \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g) \) Molar masses of \(\mathrm{Br}_{2}\) and \(\mathrm{Br}\) are 159.808 g/mol and 79.904 g/mol.

Calculate the initial concentration of Br2.

We are given a 0.200 L vessel containing 0.245 g of \(\mathrm{Br}_{2}\). The concentration of \(\mathrm{Br}_{2}\) can be calculated using its mass and the volume of the vessel: \([Br_2] = \dfrac{\text{moles of } Br_2}{\text{volume of the vessel}}\) \([Br_2] = \dfrac{0.245\,\text{g}}{159.808\,\text{g/mol} \times 0.200\,\text{L}} = 7.65 \times 10^{-3} \,\text{M}\)

Use the equilibrium constant to find the concentration of Br.

The equilibrium constant for the reaction at the given temperature is \(K_c = 1.04 \times 10^{-3}\). For the reaction, the relation is: \(K_c = \dfrac{[Br]^2}{[Br_2]}\) We can now solve for the concentration of Br: \([Br]^2 = K_c \times [Br_2]\) \([Br]^2 = 1.04 \times 10^{-3} \times 7.65 \times 10^{-3}\) \([Br]=\sqrt{7.96 \times 10^{-6}}= 2.82 \times 10^{-3}\,\text{M}\)

Calculate the mass of Br.

With the concentration of Br, we can now calculate its mass: \(\text{Mass of Br}=\text{moles of Br} \times \text{molar mass of Br}\) \(\text{Mass of Br}=[Br] \times \text{volume of the vessel} \times \text{molar mass of Br}\) \(\text{Mass of Br} = 2.82 \times 10^{-3}\,\text{M} \times 0.200\,\text{L} \times 79.904\,\text{g/mol} = 0.045\,\text{g}\) So, the mass of Br in the vessel is 0.045 g. For part (b):

Write the reaction and find the molar masses of the species involved.

The reaction is given as: \(\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) Molar masses of \(\mathrm{H}_{2}\), \(\mathrm{I}_{2}\), and \(\mathrm{HI}\) are 2.016 g/mol, 253.808 g/mol, and 127.912 g/mol.

Calculate the initial concentrations of H2 and I2.

We are given a 2.00 L flask containing 0.056 g H2 and 4.36 g I2. Calculate their concentrations: \([H_2] = \dfrac{0.056\,\text{g}}{2.016\,\text{g/mol} \times 2.00\,\text{L}} = 1.39 \times 10^{-2} \,\text{M}\) \([I_2] = \dfrac{4.36\,\text{g}}{253.808\,\text{g/mol} \times 2.00\,\text{L}} = 8.58 \times 10^{-3} \,\text{M}\)

Use the equilibrium constant to find the concentration of HI.

The equilibrium constant for the reaction at the given temperature is \(K_c = 55.3\). For the reaction, the relation is: \(K_c = \dfrac{[HI]^2}{[H_2][I_2]}\) We can now solve for the concentration of HI: \([HI]^2 = K_c \times [H_2] \times [I_2]\) \([HI]^2 = 55.3 \times 1.39 \times 10^{-2} \times 8.58 \times 10^{-3}\) \([HI]=\sqrt{6.60 \times 10^{-3}}= 2.57\,\text{M}\)

Calculate the mass of HI.

With the concentration of HI, we can now calculate its mass: \(\text{Mass of HI} = \text{moles of HI} \times \text{molar mass of HI}\) \(\text{Mass of HI} = [HI] \times \text{volume of the flask} \times \text{molar mass of HI}\) \(\text{Mass of HI} = 2.57\,\text{M} \times 2.00\,\text{L} \times 127.912\,\text{g/mol} = 658\,\text{g}\) So, the mass of HI in the flask is 658 g.

Key concepts

Molar Concentration

Molar concentration, also known as molarity, is a measure of the concentration of a solute in a solution. It is expressed in terms of the amount of substance (in moles) per unit volume of the solution (in liters). Molarity is a convenient way of expressing chemical concentrations because it directly relates to the stoichiometry of chemical reactions.

To calculate molarity, use the formula:

• \([\text{Molarity}] = \frac{\text{moles of solute}}{\text{liters of solution}}\)

Let's consider an example given in the exercise. We are given 0.245 g of \(\text{Br}_2\), and the volume of the vessel is 0.200 L. To find the molarity:

• First, calculate the number of moles by dividing the mass of \(\text{Br}_2\) by its molar mass.
• Then, divide the moles by the volume of the solution to find the molarity: \(\frac{0.245\ g}{159.808\ g/mol} = 0.00153\ moles\)
• Molecular concentration of \(\text{Br}_2\) is then \(\frac{0.00153\ moles}{0.200\ L} = 0.00765\ M\)

Molar concentration is crucial for determining the proportion of reactants and products in a chemical reaction at equilibrium.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate. At this point, the concentrations of reactants and products remain constant over time. The equilibrium constant, \(K_c\), describes the relationship between the concentrations of the reactants and products at equilibrium.

For a general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant is given by:

• \(K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)

The value of \(K_c\) depends on the specific reaction and the temperature at which it occurs. It indicates the extent of the reaction:

• If \(K_c\) is much greater than 1, products are favored at equilibrium.
• If \(K_c\) is much smaller than 1, reactants are favored at equilibrium.

In the exercise, the equilibrium constant \(K_c\) helps us find the concentration of \(Br\) by setting up the equation based on stoichiometric relationships:

• For \(Br_2(g) \rightleftharpoons 2Br(g)\), \(K_c = \frac{[Br]^2}{[Br_2]}\)

This relationship is used to solve for the concentration of \(Br\), showcasing how equilibrium concepts are critical in chemical calculations.

Molar Mass Calculations

Molar mass is the mass of one mole of a given substance, measured in grams per mole (g/mol). It is a fundamental concept necessary for converting mass into moles and vice versa in chemical calculations.

Calculating molar mass involves summing the atomic masses of all atoms present in a molecule based on the periodic table.

• For diatomic bromine \(Br_2\), with an atomic mass of approximately 79.904 g/mol per bromine atom, the molar mass is \(2 \times 79.904\ g/mol = 159.808\ g/mol\).
• For hydrogen iodide \(HI\), with atomic masses of approximately 1.008 g/mol for hydrogen and 126.904 g/mol for iodine, the molar mass is \(1.008\ g/mol + 126.904\ g/mol = 127.912\ g/mol\).

Molar mass is a central component of stoichiometry and allows us to translate between the mass of a substance and the amount or volume it occupies. In the exercise, molar mass is essential for:

• Calculating the molar concentration from mass and volume of \(Br_2\) and \(I_2\).
• Determining the mass of \(Br\) and \(HI\) from their concentrations and the volume of the container.

Understanding molar mass calculations enhances the ability to perform accurate and meaningful equilibrium constant calculations in chemical reactions.