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Chemical equilibrium refers to the state of a chemical reaction when the concentrations of the reactants and products stop changing over time. At this stage, the forward and reverse reactions occur at equal rates, resulting in a stable ratio of reactants and products. For this reaction: \[ \mathrm{NO}(g) + \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g) + \mathrm{CO}(g) \] When the system reaches equilibrium, you'll see the constant concentrations of \(\mathrm{NO}\), \(\mathrm{CO}_{2}\), \(\mathrm{NO}_{2}\), and \(\mathrm{CO}\) within the container. Knowing this, we use the concentration measurements at equilibrium to find the equilibrium constant \(K_c\). This constant is important as it tells us the extent to which a reaction occurs. A large \(K_c\) indicates a greater concentration of products compared to reactants at equilibrium. This balance is crucial for understanding how chemical reactions behave under different conditions.

The stoichiometry of a reaction involves the quantitative relationships between reactants and products. It's based on the coefficients found in the balanced equation. In this example: \[ \mathrm{NO}(g) + \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g) + \mathrm{CO}(g) \] Each molecule of \(\mathrm{NO}\) reacts with one molecule of \(\mathrm{CO}_{2}\) to produce one molecule of \(\mathrm{NO}_{2}\) and one molecule of \(\mathrm{CO}\). This 1:1:1:1 ratio simplifies calculations because the change in moles affects each component equally. If \(0.77\) moles of \(\mathrm{CO}_{2}\) are consumed, the same change applies to \(\mathrm{NO}\), and \(\mathrm{NO}_{2}\) and \(\mathrm{CO}\) are formed in equal amounts. Understanding these proportional relationships is important because it allows us to calculate how much product will form based on how much reactant is used.

Moles are a fundamental unit in chemistry representing quantity at the molecular level. When reactions occur, they often involve substances in the unit of moles. In the exercise, you start with the initial moles of \(\mathrm{NO}\) and \(\mathrm{CO}_{2}\), and find the moles remaining at equilibrium. 1. Initial moles were 3.9 for \(\mathrm{NO}\) and 0.88 for \(\mathrm{CO}_{2}\). 2. At equilibrium, \(0.11\) moles of \(\mathrm{CO}_{2}\) remain. The difference tells us \(0.77\) moles underwent reaction. The calculation is simple: subtract the moles at equilibrium from the initial moles to find the change. From the reaction stoichiometry, this change applies equally to \(\mathrm{NO}\), while \(0.77\) moles is the amount formed for both \(\mathrm{NO}_{2}\) and \(\mathrm{CO}\). Understanding moles and how they relate to chemical reactions is vital for predicting how substances interact in a chemical process.

Concentration refers to the amount of a substance in a given volume. In chemical equilibrium, we often express concentrations in terms of molarity, moles per liter (M). Assuming the volume of the container remains constant throughout the reaction is a common simplification. This allows us to express equilibrium concentrations easily. The equilibrium concentrations become:- \(\mathrm{[NO]} = \frac{3.13}{V}\)- \(\mathrm{[CO_{2}]} = \frac{0.11}{V}\)- \(\mathrm{[NO_{2}]} = \frac{0.77}{V}\)- \(\mathrm{[CO]} = \frac{0.77}{V}\) These calculations are straightforward when assuming a constant volume \(V\), and this method greatly simplifies finding the equilibrium constant \(K_c\). This approach highlights how crucial concentration is in determining the equilibrium state of a reaction and its constant.