Question

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 \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are \(0.135 \mathrm{~atm}\) and \(0.455 \mathrm{~atm}\), 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 approximately \(0.3276\) atm.

Step-by-step solution

Write the Kp expression for the given reaction

For the reaction: \(2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g)\), The equilibrium constant expression (Kp) can be written as: \(K_p = \frac{[SO_3]^2}{[SO_2]^2[O_2]}\), where [SO3], [SO2], and [O2] represent the equilibrium partial pressures of SO3, SO2, and O2, respectively.

Substitute the given values

We are given the following values: \( K_p = 0.345 \) \[SO_2 = 0.135 atm\] \[O_2 = 0.455 atm\] We will substitute these values into the Kp expression: \(0.345 = \frac{[SO_3]^2}{(0.135)^2(0.455)}\)

Solve for the equilibrium partial pressure of SO3

Now, we need to solve this equation for [SO3]: \([SO_3]^2 = (0.345)((0.135)^2(0.455))\) \([SO_3] = \sqrt{(0.345)((0.135)^2(0.455))}\) \([SO_3] = 0.3276 atm\) The equilibrium partial pressure of SO3 in the mixture is approximately \(0.3276\) atm.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a reversible chemical reaction where the rates of the forward reaction and the reverse reaction are equal. This balance between the two reactions results in constant concentrations or partial pressures of the reactants and products when the system is closed and at a constant temperature.

For example, in the reaction provided, sulfur dioxide (SO₂) and oxygen (O₂) react to form sulfur trioxide (SO₃). At equilibrium, the formation of SO₃ from SO₂ and O₂ occurs at the same rate as the decomposition of SO₃ back into SO₂ and O₂. This dynamic state can be disturbed by changes in concentration, pressure, volume, or temperature, resulting in a shift in the equilibrium position according to Le Chatelier's principle.

Equilibrium Constant (Kp)

The equilibrium constant for partial pressures, denoted as K_p, is a dimensionless number that provides a quantitative measure of the position of equilibrium for a gaseous system. It is determined by the ratio of the products' partial pressures to the reactants' partial pressures, each raised to the power of their stoichiometric coefficients.

In the given reaction, the K_p expression is \(K_p = \frac{[SO_3]^2}{[SO_2]^2[O_2]}\). Each partial pressure is represented by brackets and raised to the power corresponding to its coefficient in the balanced chemical equation. For this reaction at 900 K, K_p is given to be 0.345. This value is key to determining the equilibrium partial pressures of the involved gases.

Partial Pressure

Partial pressure refers to the pressure that a single gas component in a mixture of gases would exert if it occupied the entire volume alone at the same temperature. Dalton’s Law of Partial Pressures states that the total pressure of a mixture of gases is the sum of the partial pressures of the individual gases.

In the context of the provided exercise, when the reaction reaches equilibrium at 900 K, the partial pressures of SO₂ and O₂ are given, and we use these values along with the K_p to calculate the partial pressure of SO₃. Understanding how to calculate and interpret partial pressures is essential in solving equilibrium problems involving gaseous systems. It's also critical in applications such as determining the composition of atmospheric gases or the conditions for industrial chemical reactions.