Question

At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=\) \(0.487 M\). The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(4.63 \times 10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless, and \(\mathrm{NO}_{2}(g)\) has a brown color.]

Short answer

The equilibrium concentrations of the gases are calculated, and the color becomes darker due to increased \( \mathrm{NO}_{2} \) concentration.

Step-by-step solution

Write the equilibrium equation

The equilibrium reaction is given as \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \). This means that \( \mathrm{N}_{2} \mathrm{O}_{4} \) dissociates into \( \mathrm{NO}_{2} \) molecules.

Write the expression for the equilibrium constant

The equilibrium constant expression for the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \) is \( K_{c} = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2} \mathrm{O}_{4}]} = 4.63 \times 10^{-3} \).

Calculate initial molar concentrations after volume change

Since the volume is halved, the initial concentrations will be doubled: \([\mathrm{NO}_{2}]_i = 2 \times 0.0475 \ M = 0.095 \ M\) and \([\mathrm{N}_{2} \mathrm{O}_{4}]_i = 2 \times 0.487 \ M = 0.974 \ M\).

Set up the ICE table

Set up an ICE (Initial, Change, Equilibrium) table to keep track of the changes in concentrations:- Initial concentrations: - \([\mathrm{N}_{2} \mathrm{O}_{4}] = 0.974 \ M \) - \([\mathrm{NO}_{2}] = 0.095 \ M \)- Change in concentration: let \( x \) be the change in \([\mathrm{N}_{2} \mathrm{O}_{4}] \), then \([-2x] \) change for \([\mathrm{NO}_{2}]\) because of the stoichiometric coefficient.- Equilibrium concentrations: - \([\mathrm{N}_{2} \mathrm{O}_{4}] = 0.974 - x \ M\) - \([\mathrm{NO}_{2}] = 0.095 + 2x \ M\)

Substitute into the equilibrium constant expression

Substitute the equilibrium concentrations into the \( K_c \) expression:\[ 4.63 \times 10^{-3} = \frac{(0.095 + 2x)^2}{0.974 - x} \]Solve this equation for \( x \).

Solve the quadratic equation

Rearrange the equation to solve for \( x \):\[4.63 \times 10^{-3} (0.974 - x) = (0.095 + 2x)^2\]Expand and simplify to form a quadratic equation, and solve for \( x \) using the quadratic formula or another suitable method.

Calculate new equilibrium concentrations

Once \( x \) is found, substitute it back into the expressions to get the concentrations at equilibrium:- \([\mathrm{N}_{2} \mathrm{O}_{4}]_{eq} = 0.974 - x \)- \([\mathrm{NO}_{2}]_{eq} = 0.095 + 2x \).

Analyze the color change

Determine the color change based on the concentrations of \( \mathrm{NO}_{2} \) and \( \mathrm{N}_{2} \mathrm{O}_{4} \). Since \( \mathrm{NO}_{2} \) is brown and its concentration increases, the color will become darker.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle helps us understand how a system at equilibrium responds to changes. When there's a change in concentration, pressure, or temperature, the system shifts to counteract the change and re-establish equilibrium. In this exercise, by halving the volume, the pressure increases. The principle suggests the equilibrium will shift towards the side with fewer gas molecules to reduce pressure. This reaction involves the equilibrium between \(\mathrm{N}_2\mathrm{O}_4\) (1 molecule) and \(\mathrm{NO}_2\) (2 molecules). By increasing pressure, the system will shift towards \(\mathrm{N}_2\mathrm{O}_4\), the side with fewer molecules, to minimize this effect.
Le Chatelier's Principle helps predict the direction of the shift. Since the forward reaction produces more moles of gas, and pressure is increased by decreasing volume, the equilibrium shifts to favor the reaction that decreases gas moles. The concentration of \(\mathrm{NO}_2\) increases making the gas mixture appear darker.

equilibrium constant (Kc)

The equilibrium constant \(K_c\) provides a measure of the ratio between the concentration of products and reactants in a reaction at equilibrium. It is unique to every reaction under a specific temperature and doesn't change unless the temperature changes. For the dissociation reaction \(\mathrm{N}_2\mathrm{O}_4(g) \rightleftharpoons 2\mathrm{NO}_2(g)\), \(K_c\) helps quantify where the equilibrium lies.
In our reaction, \(K_c = 4.63 \times 10^{-3}\). A small \(K_c\) value indicates that at equilibrium, a significant amount of \(\mathrm{N}_2\mathrm{O}_4\) remains, meaning the reaction favors the formation of \(\mathrm{N}_2\mathrm{O}_4\) over \(\mathrm{NO}_2\). After the volume change, even though concentrations change, equilibrium can still be re-calculated using \(K_c\) to find new concentrations.

ICE table

An ICE table is a helpful tool in chemical equilibrium calculations. ICE stands for Initial, Change, and Equilibrium. This table helps organize and calculate the concentrations of reactants and products at equilibrium. It provides clear guidance on how concentrations shift during reactions. Here's how it was used in this example:

• Initial: Represent the starting concentrations of reactants and products.
• Change: Represent the changes occurring as the reaction approaches equilibrium. Denoted by \(x\). For example, as \(\mathrm{N}_2\mathrm{O}_4\) dissociates, \(\mathrm{NO}_2\) concentration increases.
• Equilibrium: The new concentrations when equilibrium is re-established. It's calculated by adding initial concentrations and changes observed.

Using these steps makes it easy to write equilibrium expressions and solve for unknown concentrations.

reaction quotient

The reaction quotient \(Q_c\) is used to determine the current state of the reaction and predict the direction in which it will shift to reach equilibrium. While \(K_c\) tells us about the system at equilibrium, \(Q_c\) can be calculated at any point in time. Here's how it works:

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

In the discussed exercise, calculating \(Q_c\) after doubling the concentrations helps predict the shift direction. Since changes in pressure affect concentration, this comes handy to check system adjustments needed to re-establish equilibrium.