Question

The equilibrium constant for the formation of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) from the elements is extremely large and that for the formation of \(\mathrm{NO}(\mathrm{g})\) from its elements is very small. This implies that (1) \(\mathrm{H}_{2} \mathrm{O}\) has a tendency to decompose into its elements. (2) NO has low tendency to decompose into its elements. (3) NO has appreciable tendency to decompose into its elements. (4) NO cannot be produced from direct reaction between nitrogen and oxygen.

Short answer

Option 3: NO has appreciable tendency to decompose into its elements.

Step-by-step solution

- Understand the Equilibrium Constant

The equilibrium constant, represented by K, indicates the extent of a reaction; a very large K implies the reaction strongly favors product formation, while a very small K means the reaction favors the reactants.

- Analyzing Formation of \(\text{H}_2 \text{O}\)

Given that the equilibrium constant for the formation of \(\text{H}_2 \text{O}\) is extremely large, it suggests that the formation of water from its elements (hydrogen and oxygen) is highly favorable and water has very low tendency to decompose.

- Analyzing Formation of \(\text{NO}\)

Given that the equilibrium constant for the formation of \(\text{NO}\) is very small, it indicates that the formation of nitrogen monoxide from its elements (nitrogen and oxygen) is not favorable. This implies \(\text{NO}\) has a high tendency to decompose into nitrogen and oxygen.

- Review the Given Options

Compare each option against the analysis: 1. \(\text{H}_2 \text{O}\) has a tendency to decompose into its elements. (False, it does not decompose easily) 2. \(\text{NO}\) has low tendency to decompose into its elements. (False, it has a high tendency to decompose) 3. \(\text{NO}\) has appreciable tendency to decompose into its elements. (True, it decomposes easily) 4. \(\text{NO}\) cannot be produced from direct reaction between nitrogen and oxygen. (False, it can be produced directly but in very small amounts)

Key concepts

Reaction Tendency

The reaction tendency of a compound refers to its likelihood of forming or decomposing in a given set of conditions. In chemical reactions, the equilibrium constant (K) plays an essential role in determining reaction tendency. A large K means that the reaction favors the formation of products, indicating a high tendency for the reaction to proceed forward. In contrast, a small K value means the reaction favors the reactants and suggests that the products tend to decompose back into reactants.
Using the exercise as an example, the formation of \(\text{H}_2 \text{O}\) has an extremely large equilibrium constant. This indicates that \(\text{H}_2 \text{O}\) is highly stable and has a low tendency to decompose into \(\text{H}_2\) and \(\text{O}_2\). Conversely, the formation of \(\text{NO}\) has a very small equilibrium constant, suggesting that \(\text{NO}\) is not very stable and has a high tendency to decompose back into nitrogen and oxygen.

Formation Reactions

Formation reactions are those in which elements combine to form a compound. For instance, when hydrogen and oxygen gases react to form water (\(\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2 \text{O}\)), this is a formation reaction.
The equilibrium constant (K) tells us how favorable the formation reaction is. In our exercise, the extremely large K for \(\text{H}_2 \text{O}\) formation indicates that during the reaction, almost all hydrogen and oxygen are converted into water. This means that if we have \(\text{H}_2\) and \(\text{O}_2\) together, they will naturally tend to form water until almost none of the reactants are left.
On the other hand, a very small K for \(\text{NO}\) formation means only a small amount of \(\text{N}_2\) and \(\text{O}_2\) will convert into \(\text{NO}\). This is why synthesizing \(\text{NO}\) in appreciable quantities is challenging.

Chemical Decomposition

Chemical decomposition is the process where a compound breaks down into its elements or simpler compounds. The tendency of a compound to decompose depends on its stability and the equilibrium constant for its formation reaction.
For \(\text{H}_2 \text{O}\), because the formation reaction's equilibrium constant is extremely large, \(\text{H}_2 \text{O}\) is very stable and does not decompose easily into \(\text{H}_2\) and \(\text{O}_2\). This results in a very low tendency for decomposition.
In contrast, \(\text{NO}\) has a very small equilibrium constant for its formation, indicating that it is less stable and more likely to decompose back into nitrogen and oxygen. This means \(\text{NO}\) has a high tendency for chemical decomposition, aligning with the fact that it is only present in small amounts when produced directly from \(\text{N}_2\) and \(\text{O}_2\).