Problem 81 - Chemistry: Principles and Reactions

Question

For the system $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ at \(1000 \mathrm{~K}, K=0.45 .\) Sulfur trioxide, originally at \(1.00 \mathrm{~atm}\) pressure, partially dissociates to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) at \(1000 \mathrm{~K}\). What is its partial pressure at equilibrium?

Step-by-Step Solution

Answer

Answer: The equilibrium partial pressure of sulfur trioxide (SO_3) is 0.676 atm.

Step 1: Write the equilibrium expression for the reaction

The equilibrium expression for the given reaction is written as: $$K = \frac{[SO_2][O_2]^{1/2}}{[SO_3]}$$ Here, the square brackets denote the partial pressure of the respective species at equilibrium.

Step 2: Set up the ICE table

Let's set up an ICE table using the initial pressure of SO_3 as 1 atm, and then find the changes in partial pressures for each species due to the reaction: | | SO_3 | SO_2 | O_2 | |---------|-------|--------|------| | Initial | 1.00 | 0 | 0 | | Change | -x | +x | +x/2 | | Equilib.|(1-x) | x | x/2 | Here, x is the change in the partial pressure of SO_3.

Step 3: Use the equilibrium expression to solve for x

Now, we can plug the equilibrium partial pressures from the ICE table into the K expression: $$0.45 = \frac{x(x/2)^{1/2}}{1-x}$$

Step 4: Solve for x

To solve for x, we can square both sides of the equation to get rid of the square root, and then rearrange terms to find x: $$0.2025 = \frac{x^2}{2(1-x)}$$ $$0.405(1-x) = x^2$$ $$0.405 - 0.405x = x^2$$ This is a quadratic equation, and solving for x gives us two possible values: x = 0.324 or x = 1.248 However, since x represents the change in the partial pressure of SO_3 and the initial pressure is only 1 atm, x cannot be more than 1. So, the valid value for x is 0.324.

Step 5: Find the equilibrium partial pressure of SO_3

Now that we have the value of x, we can find the equilibrium partial pressure of SO_3: Equilibrium pressure of SO_3 = Initial pressure - x = 1 - 0.324 = 0.676 atm So, the partial pressure of sulfur trioxide (SO_3) at equilibrium is 0.676 atm.

Key Concepts

Equilibrium Constant

The equilibrium constant, symbolized as K, is a numerical value that provides a measure of the extent of a chemical reaction at equilibrium. For gas-phase reactions, like the one involving sulfur trioxide and oxygen, the equilibrium constant is expressed in terms of partial pressures. This is designated as Kp.

When dealing with these reactions, it’s crucial to understand that a large Kp value indicates that the reaction favors the formation of products, while a small Kp value suggests that reactants are favored. In the exercise provided, K equals 0.45, which implies that the reaction does not heavily favor either the reactants or the products at 1000 K.

Equilibrium Expression

The equilibrium expression for a chemical reaction is derived by writing the concentration (or pressure) of products, raised to the power of their coefficients, divided by the concentration (or pressure) of reactants, similarly raised to the power of their coefficients. These expressions allow chemists to calculate the equilibrium constant for a reaction.

For the given reaction equilibrium, \(SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g)\),the equilibrium expression is \(K = \frac{[SO_2][O_2]^{1/2}}{[SO_3]}\),which reflects the stoichiometry of the reaction. Partial pressures are used in the expression because the substances involved are gases.

ICE Table

An ICE table, which stands for Initial, Change, and Equilibrium, is a systematic way to organize and calculate changes in concentration or pressure for a reaction reaching equilibrium. It is particularly useful to track the progress of the reaction and deduce unknown values.

To set up an ICE table, you list the initial pressures or concentrations of reactants and products (which may often be zero for products in a reaction starting from reactants only). Then you denote the changes that occur as the reaction moves towards equilibrium and finally compute the equilibrium concentrations or pressures. The exercise we're examining uses an ICE table effectively to approach the problem methodically.

Partial Pressure

The partial pressure of a gas is the pressure that it would exert if it occupied the whole volume of the mixture alone. In a mixture of gases, the total pressure is the sum of the partial pressures of all the individual gases. It’s an important concept in gas-phase equilibrium chemistry because it reflects the concentration of a gas within a mixture.

In our sample exercise, sulfur trioxide dissociates to sulfur dioxide and oxygen, and the individual partial pressures of these gases at equilibrium are crucial in calculating the equilibrium constant. The partial pressure is directly used in the equilibrium expression for gases, and the understanding of partial pressures is key to establishing the condition of the system at equilibrium.