Questions 43-46 refer to the following information. 2 NO2(g) ⇄ N2O4(g) dark brown colorless The dime — Equilibrium Chemistry Question

STEPS:

1. Identify the equilibrium reaction and species colors:
The given dimerization reaction is represented as:
$$2\ \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$$
* The reactant, $\text{NO}_2(g)$, is a dark brown gas.
* The product, $\text{N}_2\text{O}_4(g)$, is a colorless gas.

2. Incorporate the thermodynamic nature of the reaction:
The reaction is described as an exothermic process. Because an exothermic reaction releases heat, we can treat heat as a product on the right side of the chemical equation:
$$2\ \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) + \mathbf{heat}$$

3. Apply Le Chatelier's principle to a temperature decrease:
* The gas mixture is initially at dynamic equilibrium inside a rigid reaction vessel.
* When the temperature of the mixture is decreased, heat is removed from the system, which stresses the equilibrium.
* To relieve this stress and produce more thermal energy, the equilibrium must shift in the exothermic direction (to the right, favoring the products).

4. Determine concentration changes resulting from the shift:
* Because the system shifts to the right, reactant molecules ($\text{NO}_2$) are consumed and product molecules ($\text{N}_2\text{O}_4$) are formed.
* Consequently, the concentration of dinitrogen tetroxide, $[\text{N}_2\text{O}_4]$, must increase, while the concentration of nitrogen dioxide, $[\text{NO}_2]$, must decrease.

5. Relate concentration changes to color changes:
* As the concentration of the dark brown $\text{NO}_2$ reactant decreases and the concentration of the colorless $\text{N}_2\text{O}_4$ product increases, the dark brown color will fade.
* Therefore, the mixture will turn a lighter brown, which makes Option B the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This option claims that $[\text{NO}_2]$ will increase and the mixture will turn a darker brown. This describes a shift to the left (favoring reactants), which would only happen if the temperature of this exothermic system were *increased* (adding heat).
• Option C is incorrect: Although this option correctly notes that the mixture will turn a lighter brown, it incorrectly states that $[\text{N}_2\text{O}_4]$ will decrease. Shifting to the right must increase the concentration of the product $\text{N}_2\text{O}_4$.
• Option D is incorrect: This option states that no change will be observed. Changing the temperature of a chemical equilibrium alters the value of the equilibrium constant ($K_c$) and shifts the position of the equilibrium, which will visually change the color of the gaseous mixture.