Problem 58 - Chemistry The Molecular Science

Question

At \(503 \mathrm{~K}\) the equilibrium constant \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ has the value 40.0 . (a) Calculate the fraction of \(\mathrm{N}_{2} \mathrm{O}_{4}\) left undissociated when \(1.00 \mathrm{~mol}\) of this gas is heated to \(503 \mathrm{~K}\) in a \(10.0-\mathrm{L}\) sealed container. (b) If the volume is now reduced to \(2.0 \mathrm{~L},\) calculate the new fraction of \(\mathrm{N}_{2} \mathrm{O}_{4}\) that is undissociated. (c) Calculate all three equilibrium concentrations.

Step-by-Step Solution

Answer

(a) 0.75 undissociated, (b) 0.333 undissociated, (c) 0.075 M \( \mathrm{N}_2\mathrm{O}_4 \), 0.05 M \( \mathrm{NO}_2 \); then 0.125 M \( \mathrm{N}_2\mathrm{O}_4 \), 0.50 M \( \mathrm{NO}_2 \).

Step 1: Determine Initial Concentrations

The initial concentration of \( \mathrm{N}_2\mathrm{O}_4 \) before any dissociation is \( \frac{1.00 \text{ mol}}{10.0 \text{ L}} = 0.10 \text{ M} \). Initially, the concentration of \( \mathrm{NO}_2 \) is 0 M since no dissociation has occurred yet.

Step 2: Write the Expression for Equilibrium Concentrations

Let \( x \) be the amount of \( \mathrm{N}_2\mathrm{O}_4 \) that dissociates. Therefore, at equilibrium, the concentration of \( \mathrm{N}_2\mathrm{O}_4 \) is \( 0.10 - x \) and that of \( \mathrm{NO}_2 \) is \( 2x \).

Step 3: Write the Equilibrium Expression

The equilibrium constant \( K_c = 40 = \frac{(2x)^2}{0.10 - x} \). Solve this equation for \( x \) to find the concentrations at equilibrium.

Step 4: Solve the Quadratic Equation

Rewriting gives us a quadratic equation: \( 4x^2 = 40(0.10 - x) \). Simplify and rearrange: \( 4x^2 + 40x - 4 = 0 \). Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 4, b = 40, c = -4 \).

Step 5: Calculate Equilibrium Concentrations

Solve for \( x \): \( x = \frac{-40 \pm \sqrt{1600 + 16}}{8} = \frac{-40 \pm 40.2}{8} \). Calculate the positive root for \( x = 0.025 \text{ M} \). Then, \([\mathrm{N}_2\mathrm{O}_4]\) = \(0.10 - 0.025 = 0.075\text{ M} \), \([\mathrm{NO}_2]\) = \(2 \times 0.025 = 0.05 \text{ M}\).

Step 6: Calculate Fraction Undissociated

Fraction undissociated = \( \frac{0.075}{0.10} = 0.75 \). This is the fraction of \( \mathrm{N}_2\mathrm{O}_4 \) undissociated.

Step 7: Effect of Volume Change on Concentration

When the volume is reduced to \( 2.0 \text{ L} \), the concentration of \( \mathrm{N}_2\mathrm{O}_4 \) becomes \( \frac{0.075 \text{ mol}}{2.0 \text{ L}} = 0.375 \text{ M} \). Use the equilibrium expression \( 40 = \frac{(2x)^2}{0.375 - x} \) to find new \( x \).

Step 8: Solve for New Equilibrium Concentrations

Solve for \( x \) using \( 4x^2 = 40(0.375 - x) \). Simplify to \( 4x^2 + 40x - 15 = 0 \). Using the quadratic formula, solve for \( x = 0.25 \text{ M} \). Therefore, at new equilibrium, \([\mathrm{N}_2\mathrm{O}_4] = 0.375 - 0.25 = 0.125 \text{ M}\), \([\mathrm{NO}_2] = 2 \times 0.25 = 0.50 \text{ M}\).

Step 9: New Fraction Undissociated

With the reduced volume, the fraction undissociated is \( \frac{0.125}{0.375} \approx 0.333 \).

Step 10: Summary of Equilibrium Concentrations

For 10.0 L: \( [\mathrm{N}_2\mathrm{O}_4] = 0.075 \text{ M} \), \( [\mathrm{NO}_2] = 0.05 \text{ M} \). For 2.0 L: \( [\mathrm{N}_2\mathrm{O}_4] = 0.125 \text{ M} \), \( [\mathrm{NO}_2] = 0.50 \text{ M} \).

Key Concepts

equilibrium constant (Kc)

The equilibrium constant, denoted as \(K_c\), is a fundamental concept in chemical equilibrium. It represents the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients in the balanced chemical equation. For the dissociation of \(\mathrm{N}_2\mathrm{O}_4\) into \(\mathrm{NO}_2\), the equilibrium constant expression is given by:

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

Here, the concentration of \(\mathrm{NO}_2\) is squared because the balanced chemical equation has a coefficient of 2 for \(\mathrm{NO}_2\).
The value of \(K_c\) can provide insight into how far the reaction has proceeded toward products or remained with reactants. A large \(K_c\) value, like 40.0 in our example, indicates that the products are favored at equilibrium, meaning more \(\mathrm{NO}_2\) is formed compared to \(\mathrm{N}_2\mathrm{O}_4\) remaining undissociated.

Le Chatelier's Principle

Le Chatelier's Principle helps us understand how a system at equilibrium responds to external changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.
In our example, when the volume of the system is decreased from 10.0 L to 2.0 L, the concentration of the gases increases. According to Le Chatelier's Principle, the reaction will shift towards the side with fewer gas molecules to reduce the pressure.

• Since there are fewer moles of \(\mathrm{N}_2\mathrm{O}_4\) compared to 2 moles of \(\mathrm{NO}_2\), the equilibrium shifts to the left, increasing the concentration of \(\mathrm{N}_2\mathrm{O}_4\).
• This behavior is evident when calculating the new fraction undissociated, which reveals an increase in \(\mathrm{N}_2\mathrm{O}_4\) concentration.

Therefore, Le Chatelier's Principle provides a useful prediction tool for the direction of equilibrium shifts due to changes in concentration, pressure, or temperature.

quadratic equation solving

Quadratic equations frequently arise in the context of chemical equilibrium when calculating unknown concentrations. The equation is usually in the form of:

• \[ax^2 + bx + c = 0\]

Where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable representing the change in concentration of a substance at equilibrium.
For example, consider the dissociation of \(\mathrm{N}_2\mathrm{O}_4\), which leads to a quadratic equation when expressing the equilibrium constant:

• \[4x^2 + 40x - 4 = 0\]

This can be solved using the quadratic formula:

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Applying the quadratic formula appropriately will yield the concentration changes needed to determine equilibrium positions.
The simplicity of the quadratic formula makes it exceptionally powerful for solving these types of equilibrium problems, allowing for precise determination of equilibrium concentrations even in complex scenarios.