Question

If the equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at 750 is 49 , then the equilibrium constant for the reaction, \(\mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons 1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})\) at the same temperature will be: (a) \(1 / 49\) (b) 49 (c) \(1 / 7\) (d) \(49^{2}\)

Short answer

The equilibrium constant for the reverse reaction is \(\frac{1}{49}\). The answer is (a).

Step-by-step solution

Identify the Original Reaction

We are given the equilibrium constant for \[ \mathrm{N}_{2}( ext{g}) + 3\mathrm{H}_{2}( ext{g}) \rightleftharpoons 2\mathrm{NH}_{3}( ext{g}) \]with the equilibrium constant \(K_c = 49\).

Identify the Reverse Reaction

The desired reaction is the reverse of forming ammonia:\[ \mathrm{NH}_{3}( ext{g}) \rightleftharpoons \frac{1}{2}\mathrm{N}_{2}( ext{g}) + \frac{3}{2}\mathrm{H}_{2}( ext{g}) \]

Relate the Equilibrium Constants

The equilibrium constant of a reaction is the reciprocal of the constant for its reverse reaction.Thus, if the equilibrium constant for the formation of \(\mathrm{NH}_{3}\) is \(K_c\), then for the decomposition of \(\mathrm{NH}_{3}\) it is \(1/K_c\).

Calculate the Desired Equilibrium Constant

Since the equilibrium constant for the formation of \(\mathrm{NH}_{3}\) is 49, the equilibrium constant for its decomposition reaction is:\[ K_{c, ext{reverse}} = \frac{1}{49} \]

Check Against Given Options

The answer choices provided include:(a) \(\frac{1}{49}\), (b) 49, (c) \(\frac{1}{7}\), and (d) \(49^{2}\).The correct choice according to our calculation is \(\frac{1}{49}\), which corresponds to option (a).

Key concepts

Equilibrium Constant

The concept of an equilibrium constant is central to understanding chemical reactions at equilibrium. It represents the ratio of the concentration of products to reactants at the equilibrium state, each raised to their stoichiometric coefficients. For a general reaction:

• A + B \(\rightleftharpoons\) C + D

The equilibrium constant (\(K_c\)) is expressed as:\[K_c = \frac{[C][D]}{[A][B]}\] Here, the symbols in the brackets indicate the molar concentrations of the substances.

An important thing to note is that the equilibrium constant is temperature-dependent. This means that it will vary if the temperature changes. Hence, it is imperative to always specify the temperature when quoting an equilibrium constant value.

Lastly, a high \(K_c\) value means that, at equilibrium, the reaction mixture contains more products than reactants, while a low \(K_c\) signifies a greater concentration of reactants. This concept is vital for predicting which side of the reaction is favored at equilibrium.

Ammonia Synthesis Reaction

The ammonia synthesis reaction, also known as the Haber-Bosch process, is a critical industrial method for producing ammonia (\(\mathrm{NH}_3\)). The balanced chemical equation for the reaction is:

• \(\mathrm{N}_2(\mathrm{~g}) + 3\mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2\mathrm{NH}_3(\mathrm{~g})\)

This reaction is important in agriculture for producing fertilizers.

For this specific reaction, the equilibrium constant \(K_c\) is given as 49 at a temperature of 750. This suggests that the formation of ammonia is favored at this condition. For learners, it's fundamental to understand that the reaction is highly sensitive to temperature and pressure changes, which need to be carefully managed to optimize ammonia yields.

This synthesis is exothermic, meaning it releases heat. Therefore, increasing temperature favors the reverse reaction due to Le Chatelier's principle, as the system seeks to absorb added heat by shifting towards the reactants.

Reverse Reaction Equilibrium

In discussing reversible reactions, the concept of reverse reaction equilibrium becomes essential. It's fascinating to note that reversing a chemical reaction involves inverting its equilibrium constant. For example, consider the ammonia synthesis reaction:

• \(\mathrm{N}_2(\mathrm{~g}) + 3\mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2\mathrm{NH}_3(\mathrm{~g})\)

The equilibrium constant for this forward reaction at 750 is 49.

When the reaction is reversed, and ammonia decomposes back into nitrogen and hydrogen:

• \(\mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons \frac{1}{2}\mathrm{N}_2(\mathrm{~g}) + \frac{3}{2}\mathrm{H}_2(\mathrm{~g})\)

The equilibrium constant then becomes the reciprocal of the forward reaction's constant:\[K_{c, \text{reverse}} = \frac{1}{K_c} = \frac{1}{49}\]Thus, the equilibrium constant for this reverse reaction is \(\frac{1}{49}\), meaning at equilibrium, nitrogen and hydrogen are less favored compared to ammonia. Analyzing both forward and reverse reactions helps chemists and students understand the dynamic nature of chemical equilibria.