Problem 16 - Chemistry and Chemical Reactivity

Question

The equilibrium constant for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g}) $$ at \(25^{\circ} \mathrm{C}\) is \(5.88 \times 10^{-3} .\) Suppose \(15.6 \mathrm{g}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is placed in a \(5.00-\) L. flask at \(25^{\circ}\) C. Calculate the following: (a) the amount of \(\mathrm{NO}_{2}(\mathrm{mol})\) present at equilibrium (b) the percentage of the original \(\mathrm{N}_{2} \mathrm{O}_{4}\) that is dissociated

Step-by-Step Solution

Answer

(a) 0.0626 mol NO_2; (b) 3.69% dissociated.

Step 1: Determine Initial Concentrations

First, calculate the initial concentration of \( \mathrm{N}_2\mathrm{O}_4 \) before any reaction occurs. Use the formula \( C = \frac{n}{V} \), where \( n \) is the number of moles and \( V \) is the volume. The molar mass of \( \mathrm{N}_2\mathrm{O}_4 \) is 92.02 g/mol. Calculate the moles of \( \mathrm{N}_2\mathrm{O}_4 \) as \( n = \frac{15.6\, \text{g}}{92.02\, \text{g/mol}} = 0.1696\, \text{mol} \). Therefore, the initial concentration \( [\mathrm{N}_2\mathrm{O}_4]_0 = \frac{0.1696 \text{ mol}}{5.00 \text{ L}} = 0.03392\, \text{M}\).

Step 2: Define Equilibrium Expressions

Let \( x \) be the change in concentration of \( \mathrm{N}_2\mathrm{O}_4 \) that dissociates into \( 2 \mathrm{NO}_2 \) at equilibrium. Then at equilibrium, the concentrations are \([\mathrm{N}_2\mathrm{O}_4] = 0.03392 - x \) and \([\mathrm{NO}_2] = 2x \).

Step 3: Apply the Equilibrium Constant Expression

The equilibrium constant \( K_c \) is given by \( K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} = 5.88 \times 10^{-3} \). Substitute the equilibrium concentrations into the expression to get \( \frac{(2x)^2}{0.03392 - x} = 5.88 \times 10^{-3} \).

Step 4: Solve for \( x \)

Solve the equation \((2x)^2 = 5.88 \times 10^{-3} (0.03392 - x)\) for \( x \). This simplifies to \(4x^2 = 0.0001997 - 5.88 \times 10^{-3}x\). Rewrite as \( 4x^2 + 5.88 \times 10^{-3}x - 0.0001997 = 0 \). Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 4, b = 0.00588, c = -0.0001997 \). Calculating gives \( x \approx 0.00626 \).

Step 5: Find Moles of \( \mathrm{NO}_2 \) at Equilibrium

Since \( [\mathrm{NO}_2] = 2x \), at equilibrium \( [\mathrm{NO}_2] = 2(0.00626) = 0.01252 \text{ M} \). Therefore, the moles of \( \mathrm{NO}_2 \) in the 5.00 L flask are \( 0.01252 \text{ M} \times 5.00 \text{ L} = 0.0626 \text{ mol} \).

Step 6: Calculate Percentage of \( \mathrm{N}_2\mathrm{O}_4 \) Dissociation

Initially, there were 0.1696 moles of \( \mathrm{N}_2\mathrm{O}_4 \). At equilibrium, the change \( x \) is 0.00626 mol, representing the amount that decomposed. Thus, the percentage dissociation is \( \frac{0.00626 \text{ mol}}{0.1696 \text{ mol}} \times 100 \% \approx 3.69\% \).

Key Concepts

Equilibrium Constant

The concept of the equilibrium constant, denoted as \( K_c \), is pivotal in understanding chemical equilibrium. It represents a ratio expressing the relative concentrations of products and reactants at equilibrium for a reversible chemical reaction. This constant provides insights into the composition of the reaction mixture when the reaction has reached equilibrium. For instance, in the reaction \( \mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_2(\mathrm{g}) \), the equilibrium constant \( K_c \) is given by the formula:

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

Here, square brackets denote the concentration of the respective chemical species in moles per liter (Molarity). At equilibrium, the concentration of \( \mathrm{NO}_2 \) and \( \mathrm{N}_2\mathrm{O}_4 \) stabilize in such a way that their ratio will always equal the given \( K_c \), illustrating the unique point of equilibrium for that reaction under the same conditions of temperature and pressure. In this scenario, \( K_c \) is \( 5.88 \times 10^{-3} \), indicating the extent to which \( \mathrm{N}_2\mathrm{O}_4 \) dissociates into \( \mathrm{NO}_2 \), favoring reactants in this case since \( K_c \) is much less than 1.

Reaction Quotient

The reaction quotient, denoted as \( Q_c \), is a measure used to determine the direction in which a reaction will proceed to reach equilibrium. It uses the same formula as the equilibrium constant but involves the concentrations at any point before achieving equilibrium. To calculate \( Q_c \) for the reaction \( \mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_2(\mathrm{g}) \), one can apply the formula:

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

By comparing \( Q_c \) to \( K_c \), we can predict the reaction's tendency. If \( Q_c < K_c \), the reaction shifts towards the products to reach equilibrium. Conversely, if \( Q_c > K_c \), it suggests a shift toward the reactants. When \( Q_c = K_c \), equilibrium is established. At the start of this exercise, only \( \mathrm{N}_2\mathrm{O}_4 \) was present, meaning \( Q_c = 0 \), causing the reaction to move forward to produce \( \mathrm{NO}_2 \) until equilibrium is reached.

Mole Calculation

Mole calculations are fundamental to understanding chemical reactions and involve converting masses of substances to moles, a basic unit in chemistry representing the amount of a substance. For the reaction \( \mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_2(\mathrm{g}) \), the initial task involves determining the moles of \( \mathrm{N}_2\mathrm{O}_4 \). Given a mass of \( 15.6 \) g and its molar mass of \( 92.02 \) g/mol, the calculation is:

• \( n = \frac{\text{mass}}{\text{molar mass}} = \frac{15.6 \ \text{g}}{92.02 \ \text{g/mol}} = 0.1696 \ \text{mol} \)

This conversion forms the basis for subsequent calculations in chemical equilibria, as it allows us to determine initial concentrations and examine how quantities shift as equilibrium is approached. In this exercise, the initial number of moles of \( \mathrm{N}_2\mathrm{O}_4 \) helped establish the starting point for analyzing how much of it dissociates.

Dissociation Percentage

Dissociation percentage refers to how much of a substance is decomposed into its constituent parts relative to the initial amount. In the context of the reaction \( \mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_2(\mathrm{g}) \), it quantifies how much \( \mathrm{N}_2\mathrm{O}_4 \) breaks down into \( \mathrm{NO}_2 \) at equilibrium. Given initial moles of \( 0.1696 \) and the change in moles \( 0.00626 \), the percentage is calculated as:

• \( \text{Dissociation percentage} = \left( \frac{0.00626 \ \text{mol}}{0.1696 \ \text{mol}} \right) \times 100\% \approx 3.69\% \)

This value indicates that a small fraction of \( \mathrm{N}_2\mathrm{O}_4 \) has dissociated under the given equilibrium conditions, aligning with the relatively low \( K_c \) value that suggests limited conversion to products. Understanding dissociation percentage is crucial for predicting equilibrium composition and adjusting conditions to favor desired outcomes in chemical processes.