Problem 28 - Chemistry: An Atoms-Focused Approach

Question

Analyses of an equilibrium mixture of gaseous \(\mathrm{N}_{2} \mathrm{O}_{4}\) and NO, gave the following results: \(\left[\mathrm{NO}_{2}\right]=4.2 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=2.9 \times 10^{-3} \mathrm{M} .\) What is the value of \(K_{e}\) for the following reaction at the temperature of the mixture? $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) $$

Step-by-Step Solution

Answer

Answer: The equilibrium constant (Ke) for the given reaction is 0.164.

Step 1: Write the balanced chemical equation

The given balanced chemical equation is: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) $$

Step 2: Write the expression for the equilibrium constant (Ke)

For the given reaction, the equilibrium constant expression can be written as: $$ K_{e} = \frac{[\mathrm{N}_{2}\mathrm{O}_{4}]_{eq}}{[\mathrm{NO}_{2}]_{eq}^2} $$ where \([\mathrm{N}_{2}\mathrm{O}_{4}]_{eq}\) is the equilibrium concentration of N2O4 and \([\mathrm{NO}_{2}]_{eq}\) is the equilibrium concentration of NO2.

Step 3: Find Ke using the given equilibrium concentrations

We are given that \([\mathrm{NO}_{2}]_{eq} = 4.2 \times 10^{-3} \mathrm{M}\) and \([\mathrm{N}_{2}\mathrm{O}_{4}]_{eq} = 2.9 \times 10^{-3} \mathrm{M}\). Now we can plug these values into the Ke expression: $$ K_{e} = \frac{2.9 \times 10^{-3}}{(4.2 \times 10^{-3})^2} $$

Step 4: Calculate the value of Ke

Now, we can calculate the value of Ke: $$ K_{e} = \frac{2.9 \times 10^{-3}}{(4.2 \times 10^{-3})^2} = 0.164 $$ The value of Ke for the given reaction at the temperature of the mixture is 0.164.

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state of a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This results in no net change in the concentrations of the reactants and products. However, it's crucial to understand that equilibrium does not mean the concentrations of reactants and products are equal, but rather, their rates of change stabilize, leading to constant concentrations in a closed system.
This balance is a dynamic one, meaning that atoms and molecules don't stop reacting; instead, they continue to react at a rate that maintains the same overall concentrations. This state can be reached under various conditions, including changes in pressure, temperature, or concentration.

• Forward Reaction: The process where reactants form products.
• Reverse Reaction: The conversion of products back to reactants.
• Equilibrium Constant (\(K_{e}\)): A numerical value derived from the concentration ratios at equilibrium, reflecting the extent to which a reaction occurs.

Reaction Quotient

The reaction quotient (\(Q\)) is a measure used to determine the direction in which a chemical reaction will proceed to reach equilibrium. It is calculated using the same expression as the equilibrium constant (\(K_{e}\)), but with initial concentrations rather than equilibrium concentrations.
The value of \(Q\) indicates whether a reaction is at equilibrium or if it needs to shift to reach it:

• If \(Q = K_{e}\), the system is at equilibrium.
• If \(Q < K_{e}\), the reaction will proceed in the forward direction to reach equilibrium.
• If \(Q > K_{e}\), the reaction will move in the reverse direction.

For the reaction given in the exercise (\(2 \mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}\)), calculating \(Q\) can help predict how the concentrations of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) will change before equilibrium is achieved.

Concentration Calculation

Concentration calculation is a key step in finding the equilibrium constant and understanding how far a reaction will proceed. In this exercise, the equilibrium concentrations of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are given, from which we need to calculate \(K_{e}\).
The equilibrium constant expression for the reaction \(2 \mathrm{NO}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}\) is:
\[K_{e} = \frac{[\mathrm{N}_{2}\mathrm{O}_{4}]_{eq}}{[\mathrm{NO}_{2}]_{eq}^2}\]
To calculate \(K_{e}\), plug the equilibrium concentrations into the expression:

• \([\mathrm{N}_{2}\mathrm{O}_{4}] = 2.9 \times 10^{-3}\, \mathrm{M}\)
• \([\mathrm{NO}_{2}] = 4.2 \times 10^{-3}\, \mathrm{M}\)

Using these values:\[K_{e} = \frac{2.9 \times 10^{-3}}{(4.2 \times 10^{-3})^2} = 0.164\]
This calculation provides the equilibrium constant, indicating how the positions of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) are balanced at a specific temperature during the reaction.