Question

You place 0.0300 mol of pure \(\mathrm{SO}_{3}\) in an \(8.00-\mathrm{L}\) flask at \(1150 \mathrm{K}\). At equilibrium, 0.0058 mol of \(\mathrm{O}_{2}\) has been formed. Calculate \(K_{c}\) for the reaction at \(1150 \mathrm{K}\) $$ 2 \mathrm{sO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{sO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$

Short answer

Kc at 1150 K is approximately 0.000147.

Step-by-step solution

Set up the initial conditions

Initially, we have 0.0300 mol of \(\mathrm{SO}_{3}\) in an 8.00 L flask, and the concentrations of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are zero. The initial concentration of \(\mathrm{SO}_3\) is \(\frac{0.0300 \text{ mol}}{8.00 \text{ L}} = 0.00375 \text{ M}\).

Determine change and equilibrium concentrations

The reaction \(2 \mathrm{SO}_3 \rightleftharpoons 2 \mathrm{SO}_2 + \mathrm{O}_2\) indicates that for every mole of \(\mathrm{O}_2\) formed, 2 moles of \(\mathrm{SO}_2\) are formed, and 2 moles of \(\mathrm{SO}_3\) are consumed. At equilibrium, \(0.0058\) mol of \(\mathrm{O}_2\) is formed in an 8.00 L flask, which means that \(\mathrm{O}_2\) concentration is \(\frac{0.0058}{8.00} = 0.000725 \text{ M}\).

Calculate equilibrium concentrations of SO3 and SO2

Since \(2 \times 0.0058 \text{ mol}\) of \(\mathrm{SO}_3\) is consumed, the concentration of \(\mathrm{SO}_3\) at equilibrium is \(0.00375 - 2 \times 0.000725 = 0.00230 \text{ M}\). The concentration of \(\mathrm{SO}_2\) at equilibrium is \(2 \times 0.000725 = 0.00145 \text{ M}\).

Write the expression for Kc

The equilibrium expression for the reaction is \(K_c = \frac{[\mathrm{SO}_2]^2[\mathrm{O}_2]}{[\mathrm{SO}_3]^2}\).

Calculate Kc with equilibrium concentrations

Substitute the equilibrium concentrations into the \(K_c\) expression: \(K_c = \frac{(0.00145)^2 (0.000725)}{(0.00230)^2}\). Calculating gives: \(K_c \approx 0.000147\).

Key concepts

Reaction Quotient

In the study of chemical equilibria, understanding the Reaction Quotient, denoted as \( Q_c \), is crucial. It helps in determining the direction a reaction will proceed to reach equilibrium. The Reaction Quotient is similar to the Equilibrium Constant, \( K_c \), but it is calculated using the current concentrations of the reactants and products.

When we are given a specific state of a reaction, such as the one where \( 0.0300 \) mol of \( \text{SO}_3 \) is placed in an \( 8.00 \) L flask, \( Q_c \) can be calculated to predict the progression towards equilibrium.

• If \( Q_c < K_c \), the reaction will proceed in the forward direction to form more products until equilibrium is reached.
• If \( Q_c > K_c \), the reaction will go in the reverse direction, forming more reactants.
• If \( Q_c = K_c \), the system is already at equilibrium.

Calculating \( Q_c \) involves using the same expression as \( K_c \), but with the initial concentrations instead of the equilibrium ones. This allows chemists to understand how far and in which direction a reaction must shift to reach equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle is a vital concept that describes how a chemical system at equilibrium responds to external changes.

When an external factor changes, such as concentration, temperature, or pressure, this principle helps predict the shift in the equilibrium position.

• **Concentration Changes:** If the concentration of a reactant or product is changed, the system will shift in a direction to counteract that change.
• **Temperature Changes:** If the temperature changes, the equilibrium will shift to absorb or release heat, depending on whether the reaction is exothermic or endothermic.
• **Pressure Changes:** Mainly affects gaseous equilibria. An increase in pressure will cause the system to shift towards the side with fewer gas molecules.

Let's apply Le Chatelier's Principle to the reaction of \( 2 \text{SO}_3 \rightleftharpoons 2 \text{SO}_2 + \text{O}_2 \). Assume there is an increase in the concentration of \( \text{SO}_3 \) at equilibrium. According to the principle, the equilibrium position will shift towards the right, forming more \( \text{SO}_2 \) and \( \text{O}_2 \), to minimize the disturbance and restore equilibrium.

Concentration Changes

Concentration changes play a significant role in determining the position of equilibrium in a chemical reaction. When dealing with reactions, it is important to understand how altering the concentration of reactants or products can affect the overall process.

In our given example, initial and equilibrium concentrations were used to find \( K_c \). Initially, \( \text{SO}_3 \) had a concentration of \( 0.00375 \text{ M} \) and no \( \text{SO}_2 \) or \( \text{O}_2 \) was present. At equilibrium, the concentration of \( \text{SO}_2 \) and \( \text{O}_2 \) were determined based on the amount of products formed.

Understanding how to calculate these concentrations is critical in predicting the behavior of reactions under different conditions:

• **Initial Concentration:** The starting point before any reaction has occurred, crucial for establishing a baseline.
• **Equilibrium Concentration:** Once the system has stabilized, these concentrations are used to determine \( K_c \).
• **Effect of Changes:** Adding more reactant generally shifts the equilibrium towards product formation, while increasing product concentration shifts the equilibrium towards the reactants.

These concentration dynamics help in predicting and controlling the outcomes of chemical reactions in laboratory and industrial settings.