Question

A sample of \(S_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{K}\) at an initial pressure of 1.00 atm, where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)$$. At equilibrium, the partial pressure of \(S_{8}\) is 0.25 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(1325 \mathrm{K}\)

Short answer

The equilibrium constant K_p for the decomposition reaction \(S_8(g) \rightleftharpoons 4 S_2(g)\) at 1325 K is 324.

Step-by-step solution

Identify the Balanced Reaction

Given the decomposition reaction: \[ S_8(g) \rightleftharpoons 4 S_2(g) \]

Set the Initial Conditions

The initial conditions are: - Pressure of S_8(g) = 1.00 atm - Since the container is empty at first, we assume the initial pressure of S_2(g) = 0 atm.

Write the Expressions for Pressure Changes

As the reaction proceeds, the partial pressure of S_8 will decrease, while that of S_2 will increase. Let x be the change in the pressure of S_8 during the reaction: - Change in pressure of S_8 = -x - Change in pressure of S_2 = +4x (due to the stoichiometry of the reaction)

Determine the Equilibrium Pressures

At equilibrium, the partial pressure of S_8 is given as 0.25 atm. Therefore, we can write the change in pressure for S_8 as: 1.00 atm - x = 0.25 atm So, x = 0.75 atm Now, we can find the equilibrium pressure of S_2: Equilibrium pressure of S_2 = 4x = 4(0.75 atm) = 3 atm

Calculate the Equilibrium Constant, K_p

The expression for K_p for the given reaction is: \[ K_p =\frac{[S_2]^4}{[S_8]} \] We are given that the partial pressures at equilibrium are: - 0.25 atm for S_8 - 3 atm for S_2 Therefore, the equilibrium constant at 1325 K can be calculated as: \[ K_p = \frac{(3)^4}{0.25} =\frac{81}{0.25} = 324 \] Thus, the equilibrium constant K_p for this reaction at 1325 K is 324.

Key concepts

Partial Pressure Equilibrium

Understanding the concept of partial pressure equilibrium is essential for comprehending how gases behave in a chemical reaction. Partial pressure, in simple terms, refers to the pressure contributed by a single gas in a mixture of gases. At equilibrium in a chemical reaction involving gases, the partial pressures of the reactants and products reach stable values that no longer change with time.

In the context of the given exercise, a mixture of sulfur gas species exists within a confined space, and we monitor the partial pressures until they achieve equilibrium. This state is defined by the condition where the rate at which the sulfur molecules react to form the diatomic gas is equal to the rate at which the diatomic gas decomposes back into its original form. The ability to calculate these pressures and relate them using the equilibrium constant is crucial to predicting the behavior of gases under different conditions.

Equilibrium Constant Calculation

The equilibrium constant, denoted as K_p for reactions involving gases, provides a numerical value to express the concentrations (or partial pressures) of reactants and products at equilibrium. When calculating K_p, we use the partial pressures of the gaseous components and arrange them as indicated by the balanced chemical equation. For the reaction given in the exercise, the equilibrium constant K_p is expressed as the partial pressures of sulfur gas to the fourth power divided by the partial pressure of octasulfur gas.

K_p illustrates the position of the equilibrium - a higher value indicates a reaction favoring the products, while a lower value suggests a reaction favoring the reactants. Hence, in the exercise provided, calculating K_p allows us to quantitatively understand the extent to which S₈ decomposes into S₂ at the given temperature.

Le Chatelier's Principle

Le Chatelier's principle is an invaluable tool for predicting how a system at equilibrium will respond to changes in conditions such as pressure, temperature, or concentration. According to this principle, if external conditions are changed, the equilibrium will shift in a direction that counteracts the change. For instance, increasing the pressure of a gaseous reaction will favor the shift towards the side with fewer gas molecules.

Applying this concept to the provided exercise, if the pressure within the rigid container were to increase, the equilibrium would shift toward the formation of S₈, which has fewer moles of gas compared to the products. This principle goes beyond solving textbook exercises—it's a concept that helps chemists control outcomes of reactions in industry and research.

Reaction Stoichiometry

Reaction stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. It is directly based on the balanced chemical equation and permits the precise calculation of product and reactant quantities. In stoichiometry, coefficients in the balanced equation translate into molar ratios, which indicate how much of one substance reacts or forms in relation to another.

Within our exercise, the stoichiometry indicates that one molecule of S₈ produces four molecules of S₂. Therefore, for every decrease in pressure due to the consumption of S₈ (g), there is an associated fourfold increase in the pressure of S₂ (g). Mastering stoichiometry allows students to bridge the gap between theoretical equations and real-world chemical applications.