Problem 88 - Chemistry The Science in Context

Question

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

Step-by-Step Solution

Answer

Answer: The equilibrium concentration of N2O4 in the air is 7.93 × 10^(-12) M.

Step 1: Understand the given reaction

The given equilibrium reaction is: $$ N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g) $$ This tells us that 1 mole of N2O4 is in equilibrium with 2 moles of NO2.

Step 2: Write the equilibrium constant (Kc) expression

The equilibrium constant expression for the given reaction is: $$ K_{c} = \frac{[NO_{2}]^{2}}{[N_{2}O_{4}]} $$ Here, \([NO_{2}]\) represents the equilibrium concentration of NO2 and \([N_{2}O_{4}]\) represents the equilibrium concentration of N2O4. We are given \(K_{c}=6.1 \times 10^{-3}\) and \([NO_{2}] = 2.2 \times 10^{-7} M\).

Step 3: Substitute the given values into the Kc expression

Substitute the given values \([NO_{2}]\) and \(K_{c}\) into the Kc expression: $$ 6.1 \times 10^{-3} = \frac{(2.2 \times 10^{-7})^{2}}{[N_{2}O_{4}]} $$

Step 4: Solve the equation for \([N_{2}O_{4}]\)

Now, solve the equation for \([N_{2}O_{4}]\): $$ [N_{2}O_{4}] = \frac{(2.2 \times 10^{-7})^{2}}{6.1 \times 10^{-3}} $$ $$ [N_{2}O_{4}] = \frac{4.84 \times 10^{-14}}{6.1 \times 10^{-3}} $$ $$ [N_{2}O_{4}] = 7.93 \times 10^{-12} \thinspace M $$ Thus, the equilibrium concentration of N2O4 in the air is \(7.93 \times 10^{-12} M\).

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a crucial concept in understanding chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants at equilibrium for a particular reaction at a given temperature. For the reaction \( N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g) \), \( K_c \) is expressed as: \[ K_c = \frac{[NO_{2}]^{2}}{[N_{2}O_{4}]} \]Here, the concentrations are raised to the power of their respective stoichiometric coefficients in the balanced chemical equation.

• A large \( K_c \) value indicates a higher concentration of products at equilibrium, signifying that the reaction favors the formation of products.
• A small \( K_c \) value suggests that the reactants are favored at equilibrium.

In our example, \( K_c = 6.1 \times 10^{-3} \), indicating that even though NO2 is more prevalent, N2O4 is still significantly present. This understanding is critical when interpreting the dynamics of reversible reactions.

Concentration Calculation

Calculating concentrations in chemical reactions involves understanding how the equilibrium constant formula can be rearranged to find unknown concentrations. In our exercise, the concentration of NO2 was given as \( 2.2 \times 10^{-7} M \), and we needed to find the concentration of N2O4. To do this, we use the equilibrium expression:\[ K_c = \frac{[NO_{2}]^{2}}{[N_{2}O_{4}]} \]Rearrange the formula to solve for \([N_{2}O_{4}]\):\[ [N_{2}O_{4}] = \frac{(2.2 \times 10^{-7})^{2}}{6.1 \times 10^{-3}} \]This calculation shows how small the concentration of N2O4 is, illustrating how far the equilibrium lies towards NO2 under the given conditions. Understanding this aspect helps predict and control product yields in various chemical processes.

Reaction Quotient

The reaction quotient, \( Q \), serves as a valuable tool in predicting the direction in which a reaction will proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant, but with the concentrations of reactants and products at any moment, not just at equilibrium.

• If \( Q = K_c \), the system is at equilibrium.
• If \( Q < K_c \), the reaction will proceed forward to form more products, shifting to the right.
• If \( Q > K_c \), the reaction will proceed backward, forming more reactants, shifting to the left.

While \( Q \) wasn’t directly discussed in the solution, understanding it complements the exploration of \( K_c \) by contextualizing how reactions strive to reach equilibrium based on initial conditions. Knowing both \( K_c \) and \( Q \) provides a full picture of a reaction's status and potential adjustments needed to achieve equilibrium.