Question

For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), K_{p}=3.0 \times 10^{4}\) at \(700 \mathrm{~K}\). In a \(2.00-\mathrm{L}\) vessel, the equilibrium mixture contains \(1.17 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(0.105 \mathrm{~g}\) of \(\mathrm{O}_{2}\). How many grams of \(\mathrm{SO}_{2}\) are in the vessel? .

Short answer

At equilibrium, there are approximately \(0.452\hspace{2pt}g\) of SO₂ in the 2.00-L vessel.

Step-by-step solution

Convert masses to moles

Given the mass of SO₃ and O₂, we will first convert them to moles using their molar masses. The molar masses of SO₃ and O₂ are approximately 80 g/mol and 32 g/mol, respectively. Moles of SO₃ = mass / molar mass = \(1.17\hspace{2pt}g\) / \(80\hspace{2pt}g/mol\) ≈ \(0.0146\hspace{2pt}mol\) Moles of O₂ = mass / molar mass = \(0.105\hspace{2pt}g\) / \(32\hspace{2pt}g/mol\) ≈ \(0.00328\hspace{2pt}mol\) Now, let's convert these moles into molar concentrations, considering the volume of the vessel (2.00 L): [SO₃] = moles / volume = \(0.0146\hspace{2pt}mol\) / \(2.00\hspace{2pt}L\) = \(0.00730\hspace{2pt}M\) [O₂] = moles / volume = \(0.00328\hspace{2pt}mol\) / \(2.00\hspace{2pt}L\) = \(0.00164\hspace{2pt}M\)

Define a variable for moles of SO₂

Let's denote the moles of SO₂ at equilibrium as x. We can relate the change in moles of reactants and products by using the stoichiometry of the balanced chemical equation. For each mole of O₂ consumed, 2 moles of SO₂ are consumed and 2 moles of SO₃ are produced. Therefore, we can set up the following relationships at equilibrium: Moles of SO₂ = \(2x\) Moles of O₂ = \(0.00328\hspace{2pt}mol\) Moles of SO₃ = \(0.0146\hspace{2pt}mol + x\) Now, let's find the molar concentrations for these species: [SO₂] = (2x) / αL [O₂] = \(0.00164\hspace{2pt}M\) [SO₃] = \((0.0146\hspace{2pt}mol + x)\) / \(2.00\hspace{2pt}L\)

Apply the Kₚ expression

Now, we will use the Kₚ expression and the equilibrium concentrations to find the value of x. The Kₚ expression for the given reaction is: Kₚ = [\(SO_3\)^2] / ([\(SO_2\)^2] [O₂]) We know that Kₚ = \(3.0 \times 10^4\), and we have expressions for [O₂], [SO₂], and [SO₃], so we can plug those in to solve for x: \(3.0 \times 10^4\) = [(\(0.0146\hspace{2pt}mol + x\))^2] / ([(2x)^2] [\(0.00164\hspace{2pt}M\)])

Solve for x

Now we need to solve the equation to find the value of x: x ≈ 0.00707 mol

Convert moles of SO₂ back to grams

As we have found the moles of SO₂ at equilibrium, we will now convert this number back to grams using the molar mass of SO₂ (approximately 64 g/mol). Mass of SO₂ = moles x molar mass = \(0.00707\hspace{2pt}mol\) x \(64\hspace{2pt}g/mol\) = \(0.452\hspace{2pt}g\) So at equilibrium, there are approximately \(0.452\hspace{2pt}g\) of SO₂ in the vessel.

Key concepts

Equilibrium Constant (Kp)

The equilibrium constant for a reaction involving gases, denoted as Kp, is a critical figure in chemistry that indicates the extent to which reactants are converted to products. It is linked with the partial pressures of the gases involved in the reaction at a constant temperature. When a reaction has reached equilibrium, the ratio of the products' concentrations to the reactants' concentrations, each raised to the power of their stoichiometric coefficients, remains constant.

In the practice problem, the reaction between sulfur dioxide and oxygen to form sulfur trioxide has a Kp value of 3.0 x 10⁴ at 700 K. This high value suggests that, at equilibrium, the concentration of products is much greater than that of the reactants. Understanding Kp is essential for predicting the position of equilibrium and determining the quantities of substances at equilibrium.

Molar Mass Conversion

Converting from grams to moles and vice versa is a fundamental skill in chemistry used for stoichiometry calculations, known as molar mass conversion. The molar mass, which is the mass of one mole of a substance, serves as the conversion factor. It is determined by the sum of the atomic masses of all the atoms in the molecular formula of the substance and is expressed in grams per mole (g/mol).

For instance, the molar masses of SO₃ and O₂ given in the problem are 80 g/mol and 32 g/mol, respectively. To find the number of moles, you divide the given mass of the substance by its molar mass. This step is crucial for further stoichiometry calculations.

Stoichiometry

Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in chemical reactions. It is based on the conservation of mass and the concept of the mole. By using the coefficients from a balanced chemical equation, stoichiometry allows chemists to predict the amounts of substances consumed and formed.

In our textbook problem, the stoichiometry of the balanced equation tells us that two moles of SO₂ react with one mole of O₂ to produce two moles of SO₃. These ratios are essential for understanding how to set up equilibrium calculations and balance all substances involved both before and after the reaction reaches equilibrium.

Equilibrium Calculations

Equilibrium calculations encompass the numerical methods used to find the concentrations of reactants and products at equilibrium in a chemical reaction. By using the equilibrium constant (Kp or Kc) expression, we can calculate unknown concentrations or partial pressures. These calculations often involve setting up an equation based on the change in concentrations or moles of reactants and products as they shift toward equilibrium, as shown by the reaction's stoichiometry.

In the problem at hand, we formulated a relationship based on the stoichiometry given to find the moles of SO₂ at equilibrium. After setting up the Kp expression, we inserted the known quantities and solved for the unknown, in this case 'x', which represented the moles of SO₂. After determining 'x', we converted the moles back to grams, giving us the desired mass of SO₂ present in the vessel.