Problem 16 - Chemistry and Chemical Reactivity

Question

The equilibrium constant, \(K_{c}\), for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g}) $$ at \(25^{\circ} \mathrm{C}\) is \(170 .\) Suppose \(15.6 \mathrm{g}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is placed in a \(5.000-\mathrm{L}\) flask at \(25^{\circ} \mathrm{C} .\) Calculate the following: (a) the amount of \(\mathrm{NO}_{2}\) (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) 0.315 mol of NO2; (b) 92.92% dissociation.

Step 1: Convert mass of N2O4 to moles

First, we need to find the number of moles of \( \mathrm{N}_2\mathrm{O}_4 \) using the given mass. The molar mass of \( \mathrm{N}_2\mathrm{O}_4 \) is \( 92.02 \text{ g/mol} \). Convert the mass of \( \mathrm{N}_2\mathrm{O}_4 \) to moles using the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Thus, \( \frac{15.6 \text{ g}}{92.02 \text{ g/mol}} = 0.1695 \text{ mol} \).

Step 2: Set up the initial and change of concentrations

The initial concentration (\( \mathrm{N}_2\mathrm{O}_4 \)) is the initial moles divided by the volume of the flask, \( 0.1695 \text{ mol} / 5.000 \text{ L} = 0.0339 \text{ M} \). As the reaction reaches equilibrium, \( \mathrm{N}_2\mathrm{O}_4 \) dissociates to form \( \mathrm{NO}_2 \). The change in concentration can be represented as \( -x \) for \( \mathrm{N}_2\mathrm{O}_4 \) and \( +2x \) for \( \mathrm{NO}_2 \).

Step 3: Write and solve the equilibrium concentration expression

At equilibrium, the concentration of \( \mathrm{N}_2\mathrm{O}_4 \) is \( 0.0339 - x \) and of \( \mathrm{NO}_2 \), it is \( 2x \). The equilibrium constant \( K_c = 170 = \frac{(2x)^2}{0.0339 - x} \). Simplify this equation: \( 170(0.0339 - x) = 4x^2 \).

Step 4: Solve the quadratic equation

This simplifies to \( 170 \times 0.0339 - 170x = 4x^2 \). Rearrange to form \( 4x^2 + 170x - 5.763 = 0 \). Solve using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 170 \), and \( c = -5.763 \). On solving, we find \( x \approx 0.0315 \).

Step 5: Calculate moles of NO2 at equilibrium

The concentration of \( \mathrm{NO}_2 \) is \( 2x \), so \( 2 \times 0.0315 = 0.063 \text{ M} \). Convert this to moles: \( 0.063 \text{ M} \times 5.000 \text{ L} = 0.315 \text{ mol} \). So, the moles of \( \mathrm{NO}_2 \) present at equilibrium is \( 0.315 \text{ mol} \).

Step 6: Calculate percentage dissociation of N2O4

Convert \( x = 0.0315 \) M back to moles: \( 0.0315 \text{ M} \times 5.000 \text{ L} = 0.1575 \text{ mol} \). Percentage dissociation is \( \frac{0.1575}{0.1695} \times 100 \approx 92.92\% \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state of a chemical reaction where the concentrations of reactants and products remain constant over time. This occurs when the forward and reverse reactions occur at the same rate, so no net change is observed. In our case, the reaction between dinitrogen tetroxide (\(\mathrm{N}\_2\mathrm{O}\_4\)) and nitrogen dioxide (\(\mathrm{NO}\_2\)) reaches equilibrium when the rate at which \(\mathrm{N}\_2\mathrm{O}\_4\) dissociates into \(\mathrm{NO}\_2\) matches the rate at which \(\mathrm{NO}\_2\) recombines to form \(\mathrm{N}\_2\mathrm{O}\_4\).The equilibrium constant, \(K\_c\), is a crucial value that quantifies the equilibrium position of a chemical reaction at a given temperature. It is calculated using the concentrations of products and reactants at equilibrium. For our reaction, the equilibrium constant is given as 170, indicating that at equilibrium, the concentration of \(\mathrm{NO}\_2\) is significantly higher compared to \(\mathrm{N}\_2\mathrm{O}\_4\). Understanding chemical equilibrium helps predict the extent of a reaction and determine the concentrations of various species involved.

Dissociation Percentage

The dissociation percentage is the proportion of a reactant that has broken down into its components at equilibrium. It's a crucial metric to assess the extent to which a reaction has proceeded. In our example, the percentage of \(\mathrm{N}\_2\mathrm{O}\_4\) that dissociates into \(\mathrm{NO}\_2\) is determined using the concentration changes at equilibrium. This is calculated by finding the decrease in \(\mathrm{N}\_2\mathrm{O}\_4\) concentration as it converts to \(\mathrm{NO}\_2\), and comparing that to the initial concentration of \(\mathrm{N}\_2\mathrm{O}\_4\). Here’s how you calculate it:

• Determine the initial moles of \(\mathrm{N}\_2\mathrm{O}\_4\) and its concentration at equilibrium.
• Find the moles of \(\mathrm{N}\_2\mathrm{O}\_4\) that have dissociated by subtracting the equilibrium concentration from the initial concentration.
• Convert this value to moles and calculate the percentage by dividing the dissociated moles by the initial moles and multiplying by 100%.

In our example, approximately 92.92% of \(\mathrm{N}\_2\mathrm{O}\_4\) dissociates, indicating a high degree of reaction in reaching equilibrium.

Quadratic Equation Solving

Solving quadratic equations is a mathematical technique utilized to determine the unknowns, often used in chemical equilibrium problems. In reactions such as \(\mathrm{N}\_2\mathrm{O}\_4\) dissociation, the change in concentration due to the reaction proceeds through a quadratic expression derived from the equilibrium condition.The equilibrium constant expression for this reaction equates to a form where the relation between concentrations results in a quadratic equation: \[170(0.0339 - x) = 4x^2\]To solve it:

• Rearrange the terms to form the standard quadratic equation \(ax^2 + bx + c = 0\).
• In our case, \(a = 4\), \(b = 170\), and \(c = -5.763\).
• Apply the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the value of \(x\), representing the change in concentration.

Solving these equations accurately is essential to determine the concentrations of involved species at equilibrium, confirming how the equilibrium constant governs the reaction.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of reactants and products present in a reaction mixture at equilibrium. These concentrations are determined by starting with initial concentrations and applying the concept of equilibrium constant. For the reaction in question, calculating equilibrium concentrations involves:

• Identifying the initial concentration of reactants.
• Setting a change variable \(x\) representing the amount dissociated.
• Using \(K_c\) to set up an expression that connects these concentrations at equilibrium.
• Solving the derived quadratic equation to find \(x\).

For \(\mathrm{N}\_2\mathrm{O}\_4\), the equilibrium concentration is \(0.0339 - x\). For \(\mathrm{NO}\_2\), it is \(2x\) due to its formation from one \(\mathrm{N}\_2\mathrm{O}\_4\) molecule producing two \(\mathrm{NO}\_2\) molecules. Understanding equilibrium concentrations helps predict quantities of each reactant and product, thus assisting in controlling and optimizing chemical reactions.