91Ó°ÊÓ

Dinitrogen trioxide, a chemical compound with the formula \( \text{N}_2\text{O}_3 \), is an interesting substance in the realm of chemical equilibria. Dinitrogen trioxide is a deep blue liquid at low temperatures, but it becomes a colorless gas as the temperature rises. It is relatively unstable, particularly at higher temperatures, such as the 25.0°C in the given problem, where it decomposes readily.
In the provided exercise, we start with 4.00 moles of this compound, which are placed in a flask. As it decomposes, it breaks down into nitrogen dioxide (\( \text{NO}_2 \)) and nitric oxide (\( \text{NO} \)). This decomposition is reversible, meaning that reactions can proceed in both forward and backward directions, a hallmark of equilibrium reactions.
Understanding how dinitrogen trioxide interacts and transforms under certain conditions is key to predicting the behavior of similar chemical systems and solving related problems. Its decomposition exemplifies the dynamic balance that characterizes chemical equilibria.

The equilibrium constant (\( K \)) is a crucial concept in understanding chemical equilibria. It provides a quantitative measure of the ratio of products to reactants at equilibrium. Each reaction has its specific \( K \) value, reflecting how far the reaction will proceed before reaching equilibrium.
In the case of dinitrogen trioxide decomposing into nitrogen dioxide and nitric oxide, the equilibrium constant expression can be written using the concentrations of the gases involved:\[ K = \frac{[\text{NO}_2][\text{NO}]}{[\text{N}_2\text{O}_3]} \] Because we know the moles of each reactant and product at equilibrium, we can substitute these into the equilibrium expression. This ratio will give us the \( K \) value, showing us how the various species in the reaction are distributed at equilibrium.
The equilibrium constant helps predict how changes in concentration, temperature, or pressure can shift the position of equilibrium. A higher \( K \) value indicates a greater concentration of products compared to reactants at equilibrium, signifying a reaction that largely proceeds to the right.

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. Understanding stoichiometry allows us to predict how much product will form from a given quantity of reactants as is required in the exercise with dinitrogen trioxide.
For the reaction \( \text{N}_2\text{O}_3(g) \rightleftharpoons \text{NO}_2(g) + \text{NO}(g) \), the stoichiometry indicates that one mole of \( \text{N}_2\text{O}_3 \) decomposes to form one mole each of \( \text{NO}_2 \) and \( \text{NO} \).
In the problem, we know that 1.20 moles of \( \text{NO}_2 \) are formed at equilibrium. According to stoichiometry, this means 1.20 moles of \( \text{N}_2\text{O}_3 \) must have decomposed. Consequently, there will also be 1.20 moles of \( \text{NO} \) at equilibrium.
The initial number of moles of \( \text{N}_2\text{O}_3 \) was 4.00 moles, thus the remaining moles of \( \text{N}_2\text{O}_3 \) at equilibrium are calculated as 4.00 minus the 1.20 moles that decomposed, resulting in 2.80 moles left.
Reaction stoichiometry is a fundamental tool in solving chemical equilibrium problems, providing the necessary peripheral calculations for determining the amounts of substances present at equilibrium.