Problem 40 - Chemistry The Molecular Science

Question

Assume you place \(0.010 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) in a sealed \(2.0-\mathrm{L}\) flask at \(50 .{ }^{\circ} \mathrm{C}\). After the system reaches equilibrium, \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.00090 \mathrm{M} .\) Calculate the value of \(K_{\mathrm{c}}\) for this reaction. $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$

Step-by-Step Solution

Answer

The value of \(K_c\) is approximately 0.075.

Step 1: Write the Balanced Equation

The given reaction is already balanced: \( \text{N}_{2} \text{O}_{4}(\text{g}) \rightleftharpoons 2\text{NO}_{2}(\text{g}) \).

Step 2: Determine Initial Concentration of N2O4

We start with \(0.010 \text{ mol} \) of \( \text{N}_{2} \text{O}_{4} \) in a \(2.0 \text{ L} \) flask. The initial concentration \([\text{N}_{2} \text{O}_{4}]_0\) is calculated as \( \frac{0.010 \text{ mol}}{2.0 \text{ L}} = 0.0050 \text{ M} \).

Step 3: Calculate Change in Concentration of N2O4

At equilibrium, \([\text{N}_{2} \text{O}_{4}] = 0.00090\text{ M}\). The change in concentration of \( \text{N}_{2} \text{O}_{4} \) is \(0.0050 - 0.00090 = 0.0041 \text{ M}\).

Step 4: Calculate Change in Concentration of NO2

Since \( \text{N}_{2} \text{O}_{4} \) dissociates into 2 \( \text{NO}_{2} \), the increase in concentration of \( \text{NO}_{2} \) is \( 2 \times 0.0041 = 0.0082 \text{ M}\). At equilibrium, \([\text{NO}_{2}] = 0.0082 \text{ M}\).

Step 5: Write the Equilibrium Expression

For the reaction \(\text{N}_{2} \text{O}_{4}(\text{g}) \rightleftharpoons 2\text{NO}_{2}(\text{g})\), \(K_{c} = \frac{[\text{NO}_{2}]^2}{[\text{N}_{2} \text{O}_{4}]}\).

Step 6: Plug in Equilibrium Concentrations

Substitute the equilibrium concentrations into the expression: \(K_{c} = \frac{(0.0082)^2}{0.00090}\) \(= \frac{0.00006724}{0.00090} \approx 0.075\).

Key Concepts

Equilibrium Constant

Chemical reactions often reach a state known as equilibrium, where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time. The equilibrium constant, denoted as \( K_c \), is a numerical value that describes the ratio of the concentrations of products to reactants at equilibrium. It is specific to a particular reaction at a given temperature. For the reaction \( \text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g}) \), the equilibrium expression is given by:

• \( K_{c} = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \)

This equation highlights how the equilibrium constant reflects the balance of concentrations; a large \( K_c \) indicates a higher concentration of products at equilibrium, while a small \( K_c \) suggests more reactants. Understanding \( K_c \) is crucial for predicting how a reaction behaves under different conditions.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept used to predict how a change in conditions can shift the position of equilibrium in a chemical reaction. According to this principle, if a system at equilibrium experiences a change in concentration, temperature, or pressure, the system adjusts to counteract the effect of the change and re-establish equilibrium.

• If the concentration of a reactant or product is changed, the equilibrium will shift to oppose that change.
• An increase in temperature usually favors the endothermic direction of a reaction.
• Changes in pressure mainly affect reactions involving gases with differing numbers of moles on each side.

For example, increasing the amount of \( \text{N}_2\text{O}_4 \) in the reaction \( \text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g}) \) will shift the equilibrium position to the right, favoring the production of more \( \text{NO}_2 \).Le Chatelier's Principle helps chemists control reactions to obtain the desired yield.

Equilibrium Concentrations

Equilibrium concentrations are the amounts of reactants and products present in a reaction mixture when the reaction has reached equilibrium. These concentrations are key to calculating the equilibrium constant \( K_c \).In the exercise, we start by calculating the initial concentration of \( \text{N}_2\text{O}_4 \):

• Initial concentration \([\text{N}_2\text{O}_4]_0 = \frac{0.010 \text{ mol}}{2.0 \text{ L}} = 0.0050 \text{ M} \)

As the reaction proceeds towards equilibrium, the concentration of \( \text{N}_2\text{O}_4 \) decreases, and that of \( \text{NO}_2 \) increases. At equilibrium, the concentrations are:

• \([\text{N}_2\text{O}_4] = 0.00090\text{ M}\)
• \([\text{NO}_2] = 0.0082\text{ M}\)

These values are then used in the equilibrium expression to calculate \( K_c \). Understanding how to determine equilibrium concentrations is crucial for analyzing the extent and direction of chemical reactions.

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a measure of the relative amounts of products and reactants present in a reaction at any point in time. It is similar to the equilibrium constant, but \( Q \) is not necessarily at equilibrium. The expression for \( Q \) is identical to that for \( K_c \). For the reaction \( \text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g}) \), it is written as:

• \( Q = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \)

Comparison between \( Q \) and \( K_c \) determines the direction in which a reaction will proceed:

• If \( Q = K_c \), the system is at equilibrium.
• If \( Q < K_c \), the reaction will shift to the right, forming more products.
• If \( Q > K_c \), the reaction will shift to the left, forming more reactants.

Understanding the reaction quotient is essential for predicting changes in concentration and managing chemical processes effectively.