Problem 96 - Chemistry: An Atoms-Focused Approach

Question

Urban Air On a very smoggy day, the cquilibrium concentration of \(\mathrm{NO}_{2}\) in the air over an urban arca reaches \(2.2 \times 10^{-7} M .\) If the temperature of the air is \(25^{\circ} \mathrm{C},\) what is the concentration of the dimer \(\mathrm{N}_{2} \mathrm{O}_{4}\) in the air? $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad K_{e}=6.1 \times 10^{-3} $$

Step-by-Step Solution

Answer

Answer: The concentration of N2O4 in the air is approximately \(7.97 \times 10^{-12} M\).

Step 1: Write down the given information.

The given information is: 1. Equilibrium concentration of NO2: \(2.2 \times 10^{-7} M\) 2. Temperature: \(25^{\circ}C\) 3. Equilibrium constant, \(K_{e}\): \(6.1 \times 10^{-3}\) The balanced chemical equation for the reaction is: $$N_{2}O_{4}(g) \rightleftharpoons 2 NO_{2}(g)$$ We are given the concentration of NO2 and need to find the concentration of the dimer, N2O4.

Step 2: Write the expression for the equilibrium constant.\(tag_content#For the given reaction, the equilibrium constant \)K_{eq}$ can be expressed as follows: $$ K_{e} = \frac{[NO_{2}]^2}{[N_{2}O_{4}]} $$ Where: - \([NO_{2}]\) is the molar concentration of NO2 gas, in mol/L - \([N_{2}O_{4}]\) is the molar concentration of N2O4 gas, in mol/L

Step 3: Substitute the given values and solve for the dimer concentration.\(tag_content#We are given the concentration of NO2 and the value of \)K_{e}$. We can substitute these values into the expression from Step 2 to obtain: $$ 6.1 \times 10^{-3} = \frac{(2.2 \times 10^{-7})^2}{[N_{2}O_{4}]} $$ Now, we will solve for \([N_{2}O_{4}]\): \[ [N_{2}O_{4}] = \frac{(2.2 \times 10^{-7})^2}{6.1 \times 10^{-3}} \] \[ [N_{2}O_{4}] \approx 7.97 \times 10^{-12} \] Therefore, the concentration of N2O4 in the air is approximately \(7.97 \times 10^{-12} M\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a crucial concept in chemistry that describes a state where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of the reactants and products remain constant over time, although they are not necessarily equal. This equilibrium occurs in a closed system where no substances can enter or exit.
In the example of the reaction between 2O4 and NO2, a state of equilibrium is achieved when the rate at which 2O4 breaks down into 2 NO2 equals the rate at which NO2 recombines to form 2O4. This dynamic balance ensures that although both reactions continue to occur, the overall concentrations of each species do not change.
It's important to note that achieving equilibrium does not imply that the reactions have stopped. Instead, both reactions proceed at equal rates, maintaining concentration levels in the system.

Equilibrium Constant

The equilibrium constant, denoted as \( K_{eq} \), is a number that provides insight into the relative amounts of reactants and products at equilibrium for a given chemical reaction at a specific temperature.
This constant is calculated using the concentrations of the products raised to the power of their coefficients, divided by the concentrations of the reactants raised to the power of their coefficients in the balanced chemical equation.
For the reaction \( 2O4(g) \rightleftharpoons 2 NO2(g) \), the equilibrium constant expression is given by:

• \( K_{eq} = \frac{[NO_2]^2}{[N2O4]} \)

The value of \( K_{eq} \) indicates whether the equilibrium position favors the reactants or products:

• If \( K_{eq} >> 1 \), the reaction favors the formation of products.
• If \( K_{eq} << 1 \), the reaction favors the formation of reactants.

For our specific reaction involving 2O4 and NO2, \( K_{eq} = 6.1 \times 10^{-3} \), suggesting that the equilibrium does not strongly favor one side completely over the other but somewhat leans towards the reactants due to the relatively small value.

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a function similar to the equilibrium constant \( K_{eq} \), but it is applicable at any point in the reaction, not just at equilibrium. It helps in predicting the direction the reaction will go to achieve equilibrium.
The expression for \( Q \) is similar to that of \( K_{eq} \), but it uses the initial concentrations rather than those at equilibrium.
If the concentrations of the reactants and products for this reaction are not at equilibrium, we use the reaction quotient to compare it with the equilibrium constant:

• \( Q = \frac{[NO_2]^2}{[N2O4]} \)

By comparing \( Q \) and \( K_{eq} \):

• If \( Q < K_{eq} \), the forward reaction is favored to reach equilibrium.
• If \( Q > K_{eq} \), the reverse reaction is favored to reach equilibrium.
• If \( Q = K_{eq} \), the system is already at equilibrium.

In practice, calculating \( Q \) allows chemists to make predictions and adjustments needed to direct reactions toward a desired outcome, ensuring a thorough understanding of chemical processes.