Question

\(\mathrm{At} 900 \mathrm{K},\) the following reaction has \(K_{p}=0.345 :\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ In an equilibrium mixture the partial pressures of \(S \mathrm{O}_{2}\) and \(\mathrm{O}_{2}\) are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

Short answer

The equilibrium partial pressure of SO₃ in the mixture is 0.0494 atm.

Step-by-step solution

Write the equilibrium expression for the reaction

Based on the balanced chemical equation, we can write the equilibrium constant expression for Kp as: \[K_p = \frac{(\mathrm{P}_{\mathrm{SO}_3})^2}{(\mathrm{P}_{\mathrm{SO}_2})^2 \cdot (\mathrm{P}_{\mathrm{O}_2})}\] Where P_SO₃, P_SO₂, and P_O₂ are the partial pressures of SO₃, SO₂, and O₂ respectively.

Substitute the given values into the equilibrium constant expression

We are given Kp = 0.345, P_SO₂ = 0.135 atm, and P_O₂ = 0.455 atm. We can now substitute these values into the equilibrium constant expression: \[0.345 = \frac{(\mathrm{P}_{\mathrm{SO}_3})^2}{(0.135)^2 \cdot (0.455)}\]

Solve for the equilibrium partial pressure of SO₃

We now have an equation with one unknown, which is P_SO₃. To solve for this, perform the following steps: 1. Multiply both sides of the equation by \((0.135)^2\) and \(0.455\) to isolate P_SO₃²: \[(0.345) \cdot (0.135)^2 \cdot (0.455) = (\mathrm{P}_{\mathrm{SO}_3})^2\] 2. Calculate the numerical value: \[(0.345) \cdot (0.135)^2 \cdot (0.455) = 0.0024416\] 3. Take the square root of both sides to find the value of P_SO₃: \[\sqrt{0.0024416} = \mathrm{P}_{\mathrm{SO}_3}\] 4. Calculate the value of P_SO₃: \[\mathrm{P}_{\mathrm{SO}_3} = 0.0494 \, \mathrm{atm}\]

State the answer

The equilibrium partial pressure of SO₃ in the mixture is 0.0494 atm.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentrations of reactants and products. Equilibrium can be reached from either direction of the reaction and does not mean that the reactants and products are present in equal amounts, but rather that their ratios are constant. In the exercise provided, the equilibrium state is for the reaction involving sulfur dioxide (SO2), oxygen (O2), and sulfur trioxide (SO3). In this state, the system has reached a balance where the production of SO3 from SO2 and O2 equals the breakdown of SO3 back into SO2 and O2.

Understanding equilibrium allows us to predict how a reaction mixture will respond to changes in conditions, such as changes in pressure, temperature, or the concentration of reactants and products. This is an important concept in chemistry, particularly in industrial processes where the maximum yield of a product needs to be achieved efficiently. The reaction in the exercise, for example, represents a simplified model of the industrial process used to produce sulfuric acid, one of the most important chemicals in chemical manufacturing.

Equilibrium Constant Expression

The equilibrium constant expression (Kc for concentration or Kp for pressure) is a formula that expresses the ratio of the concentrations (or pressures) of products to reactants at chemical equilibrium, each raised to the power of their respective stoichiometric coefficients in the balanced equation. For gaseous systems, as is the case in the given exercise, the equilibrium constant in terms of partial pressures (Kp) is used.

Understanding Kp

The value of Kp provides insight into the extent of a reaction; a large Kp indicates a greater concentration of products, whereas a small Kp indicates a higher concentration of reactants. It's important to remember that Kp is only affected by temperature; changes in pressure or the amounts of reactants and products do not affect its value.

In the provided step-by-step solution, we applied the equilibrium constant expression by using the given Kp and the known partial pressures of SO2 and O2 to solve for the partial pressure of the product SO3. This calculation is fundamental in determining the conditions needed to achieve a desired yield of products in chemical reactions.

Partial Pressures in Chemistry

Partial pressure is a concept in chemistry that refers to the pressure exerted by a single gas in a mixture of gases. It is directly proportional to its concentration in the mixture and can be determined using Dalton's Law, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas.

Application in Equilibrium

When dealing with chemical equilibria for reactions involving gases, the partial pressures are used in the equilibrium constant expression, just as concentrations are used for reactions in solution.

For the exercise we have been discussing, the equilibrium constant expression is calculated using the partial pressures of the gases SO2, O2, and SO3. Remember that each gas contributes to the total pressure in proportion to its molar fraction in the gas mixture. In industrial processes, the control of partial pressures is vital for optimizing yields and understanding the practical applications of chemical reactions under different pressure conditions. By solving for the equilibrium partial pressure of SO3, as was done in the step-by-step exercise, chemists can determine how effective a reaction is under given conditions and adjust variables accordingly to optimize production or reduce waste.