Question

The most stable oxides of nitrogen will be : (a) \(2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) ; \quad K=6.7 \times 10^{16} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+5 \mathrm{O}_{2}(g) ; \quad K=1.2 \times 10^{24} \mathrm{~mol}^{5} \mathrm{~L}^{-5}\) (c) \(2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)\) \(K=2.2 \times 10^{30}\) (d) \(2 \mathrm{~N}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) ; \quad K=3.5 \times 10^{33} \mathrm{~mol} \mathrm{~L}^{-1}\)

Short answer

\(\mathrm{NO}_{2}\) is the most stable oxide of nitrogen.

Step-by-step solution

Understanding the Problem

We are provided with the equilibrium constants (K) for four different decomposition reactions of nitrogen oxides. A larger K value means the reaction proceeds more favorably to the right, forming more products (in this case, nitrogen and oxygen gases) and implying that the oxide in the reaction is less stable (since it decomposes readily).

Comparing the K Values

We can compare the K values to determine which nitrogen oxide is the least stable. Larger K values indicate that the reaction heavily favors product formation, corresponding to a less stable oxide.

Identifying the Most Stable Oxide

By examining the K values, we can find the smallest K, which indicates the most stable oxide, because the reaction does not proceed as readily to the right, thus producing fewer products.

Conclusion

After comparing the given K values, we can see that (a) has the smallest K value (\(6.7 \times 10^{16} \text{mol L}^{-1}\)), indicating that \(\mathrm{NO}_{2}\) is the most stable oxide among the options given.

Key concepts

Equilibrium Constant

An equilibrium constant (\(K\text{c}\) for concentrations or \(K\text{p}\) for partial pressures) is a number that represents the ratio of the concentration of the products to the reactants at equilibrium, raised to their respective stoichiometric coefficients. It is indispensable in determining the extent to which a reaction occurs before reaching chemical equilibrium.

The magnitude of the equilibrium constant gives us information regarding the position of equilibrium. A large value indicates a reaction that favors product formation, assuming we are talking about a reaction moving towards the right, as is conventionally represented. Conversely, a small value suggests a reaction that hardly proceeds at all, skewing towards reactant formation.

Understanding the equilibrium constant is crucial for predicting the behavior of chemical reactions under different conditions. For instance, in the example given regarding nitrogen oxides stability, the stability can be inferred based on how large the equilibrium constant is. In chemical systems, as the temperature or pressure changes, so can the value of the equilibrium constant, affecting the balance between products and reactants. This forms the basis for Le Chatelier's Principle, which states that a system at equilibrium will shift its composition to counteract any applied change.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate, leading to the constant ratio of the concentration of reactants to products. This point of dynamic balance does not mean that the reactants and products are equal in concentration, but that their concentrations no longer change with time.

At equilibrium, the rate at which the reactants turn into products is equal to the rate at which the products revert into reactants. For the nitrogen oxides mentioned in the problem, the chemical equilibrium can be visualized as a balance between the intact nitrogen oxide molecules and their decomposition into nitrogen and oxygen gases.

A vital aspect of equilibrium is that it can be reached from either direction of the reaction, starting with the reactants or the products. This state can be influenced by changes in temperature, pressure, or concentration, which can shift the position of equilibrium according to Le Chatelier's Principle, as previously noted.

Decomposition Reactions

In decomposition reactions, a compound is broken down into smaller compounds or elements. These reactions are integral to the field of chemistry because they help in understanding the stability of compounds and the energy required to break the chemical bonds within a compound.

For the nitrogen oxides stability problem, we're examining the decomposition reactions of various nitrogen oxides into nitrogen and oxygen. Each type of nitrogen oxide has different stability, meaning some are more prone to decompose than others. The equilibrium constant values offer insight into how readily each compound decomposes.

When analyzing these reactions, it's important to note that a higher stability correlates with less tendency to decompose and thus a lower equilibrium constant value. It also implies the release of less energy upon decomposition, as more stable compounds generally have lower potential energy. Teaching students how to interpret equilibrium constants helps them gain a deeper understanding of the stability of chemical compounds.