The equilibrium constant for the formation of \(H_2O\)(g) from the elements is extremely large and that for the formation of \(NO\)(g) from its elements is very small. This implies that

\(NO\) has appreciable tendency to decompose into its elements.

The correct answer is option 3. \(NO\) has appreciable tendency to decompose into its elements.

Let us delve into the concepts of equilibrium constants and their implications for the stability and decomposition tendencies of compounds.

Equilibrium Constants and Reaction Tendencies

The equilibrium constant (\(K\)) of a chemical reaction provides a measure of the extent to which reactants are converted to products at equilibrium. It is defined by the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. For a general reaction:

\(aA + bB \rightleftharpoons cC + dD \)

the equilibrium constant \(K\) is given by:

\(K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \)

Large \(K\) Value: Indicates that the equilibrium position is far to the right, favoring the formation of products.

Small \(K\) Value: Indicates that the equilibrium position is far to the left, favoring the reactants.

Given Reactions

Formation of \(H_2O(g)\): \(2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(g) \)

Large \(K\) Value: The equilibrium constant for this reaction is extremely large, indicating that, at equilibrium, the concentration of \(H_2O(g)\) is much higher than that of \(H_2(g)\) and \(O_2(g)\). This suggests that \(H_2O(g)\) is very stable and does not tend to decompose into \(H_2(g)\) and \(O_2(g)\).

Formation of \(NO(g)\):

\(N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \)

Small \(K\) Value: The equilibrium constant for this reaction is very small, indicating that, at equilibrium, the concentration of \(N_2(g)\) and \(O_2(g)\) is much higher than that of \(NO(g)\). This suggests that \(NO(g)\) is not very stable and has a significant tendency to decompose back into \(N_2(g)\) and \(O_2(g)\).

Analysis of Statements

(1) \(H_2O\) has a tendency to decompose into its elements:

A large equilibrium constant for the formation of \(H_2O(g)\) means that \(H_2O(g)\) is very stable and the reaction strongly favors the formation of water from hydrogen and oxygen. Therefore, \(H_2O(g)\) has a low tendency to decompose into \(H_2(g)\) and \(O_2(g)\). This statement is incorrect.

(2) \(NO\) has low tendency to decompose into its elements:

A small equilibrium constant for the formation of \(NO(g)\) means that the reaction does not favor the formation of \(NO(g)\); instead, it favors the reactants \(N_2(g)\) and \(O_2(g)\). This indicates that \(NO(g)\) is unstable and tends to decompose back into \(N_2(g)\) and \(O_2(g)\). This statement is incorrect.

(3) \(NO\) has appreciable tendency to decompose into its elements.

Given the small equilibrium constant for the formation of \(NO(g)\), the reaction favors the reactants \(N_2(g)\) and \(O_2(g)\) at equilibrium. This means \(NO(g)\) is unstable and has a significant tendency to decompose into its elements. This statement is correct.

(4) \(NO\) cannot be produced from direct reaction between nitrogen and oxygen.

A small equilibrium constant does not mean that \(NO(g)\) cannot be produced at all; it means that at equilibrium, very little \(NO(g)\) will be present because the reaction strongly favors the reactants. However, \(NO(g)\) can still be produced, particularly at high temperatures or under conditions where the reaction is driven away from equilibrium. This statement is incorrect.

Detailed Explanation of Correct Option: (3)\(NO\) has appreciable tendency to decompose into its elements.

Equilibrium Constant (\(K\)): For the reaction \( N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \), a small \(K\) value indicates that, at equilibrium, the concentration of \(NO(g)\) is much lower than that of \(N_2(g)\) and \(O_2(g)\). This is because the position of equilibrium lies far to the left, favoring the reactants.

Stability of \(NO(g)\): The small \(K\) value reflects the instability of \(NO(g)\), meaning it does not persist in significant amounts and tends to revert to \(N_2(g)\) and \(O_2(g)\).

Tendency to Decompose: The significant tendency for \(NO(g)\) to decompose is a direct consequence of the equilibrium favoring the reactants. In practical terms, this means that in any mixture of \(NO(g)\), \(N_2(g)\), and \(O_2(g)\), the \(NO(g)\) will predominantly break down into \(N_2(g)\) and \(O_2(g)\).

Based on the equilibrium constants and their implications:

Correct Statement**: (3) \(NO\) has appreciable tendency to decompose into its elements.

The small equilibrium constant for the formation of \(NO(g)\) from its elements indicates that \(NO(g)\) is unstable and tends to decompose back into \(N_2(g)\) and \(O_2(g)\).