Problem 25 - Chemical Principles: The Quest for Insight

Question

For the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(400 \mathrm{~K}, K=41\). Find the value of \(\mathrm{K}\) for each of the following reactions at the same temperature. (a) \(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(2 \mathrm{~N}_{2}(\mathrm{~g})+6 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NH}_{3}(\mathrm{~g})\)

Step-by-Step Solution

Answer

The equilibrium constants for the reactions at 400 K are: (a) \(K^{\'}_a = \frac{1}{41}\), (b) \(K_b = \sqrt{41}\), and (c) \(K_c = 41^2\).

Step 1 - Identifying the Relationship between the Given and the New Reactions

Notice that reaction (a) is the reverse of the given reaction, while reactions (b) and (c) are obtained by dividing and multiplying the coefficients of the balanced given reaction by 2 and 1/2 respectively. The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. When coefficients of a balanced equation are multiplied or divided by a number, the equilibrium constant is raised to the power of that number.

Step 2 - Calculate the Equilibrium Constant for Reaction (a)

Since reaction (a) is the reverse of the given reaction, the equilibrium constant for reaction (a), which we'll call \(K^{\'}_a\), is the reciprocal of the given equilibrium constant \(K\). Hence, \[K^{\'}_a = \frac{1}{K} = \frac{1}{41}\]

Step 3 - Calculate the Equilibrium Constant for Reaction (b)

Reaction (b) is obtained by dividing all the coefficients of the given reaction by 2. The equilibrium constant for reaction (b), which we'll call \(K_b\), can be found by taking the square root of the given equilibrium constant (since 2 is the reciprocal of 1/2) \[K_b = K^{\frac{1}{2}} = 41^{\frac{1}{2}}\]

Step 4 - Calculate the Equilibrium Constant for Reaction (c)

Reaction (c) is obtained by doubling all the coefficients of the given reaction. The equilibrium constant for reaction (c), which we'll call \(K_c\), can be found by squaring the given equilibrium constant \[K_c = K^{2} = 41^2\]

Key Concepts

Chemical Equilibrium

Chemical equilibrium represents the state of a chemical reaction where the rates of the forward and reverse reactions are equal, so there's no net change in the concentrations of reactants and products over time. It is important to understand that at equilibrium, the reaction hasn't stopped, but continues dynamically with reactants turning into products and vice versa at an equal rate.

For the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\rightleftharpoons2 \mathrm{NH}_{3}(\mathrm{~g})\) at 400 K, the equilibrium constant (K) is given as 41. This means at 400 K, the ratio of the products to reactants, raised to the power of their stoichiometric coefficients, is constant at 41, assuming a closed system. The equilibrium constant is crucial as it can provide insight into the position of the equilibrium, helping predict the concentrations of substances present when the system is at equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle provides insight into how a system at equilibrium reacts to external changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This can include alterations in concentration, pressure, volume, or temperature of the system.

In the context of our reaction, if more \(\mathrm{N}_{2}\) or \(\mathrm{H}_{2}\) were added to the system at equilibrium, the reaction would shift to the right to produce more \(\mathrm{NH}_{3}\), and vice versa if \(\mathrm{NH}_{3}\) were added. Likewise, increasing pressure by decreasing volume favors the side with fewer moles of gas, which in this case is the production of \(\mathrm{NH}_{3}\), while an increase in temperature generally favors the endothermic direction of the reaction.

Reaction Quotient

The reaction quotient (Q) is a measure that tells us which direction a reaction will proceed to reach equilibrium. It is calculated in the same way as K, using the current concentrations instead of the equilibrium concentrations.

When Q is compared to K: if \(QK\), the reaction will go in reverse to yield more reactants; and if \(Q=K\), the system is already at equilibrium. Taking our given reaction at 400 K, if the amount of \(\mathrm{NH}_{3}\) in the reaction mixture is reduced, Q would decrease and the reaction would shift to the right to re-establish equilibrium by producing more \(\mathrm{NH}_{3}\). Monitoring Q in comparison to K allows chemists to predict the movement of the equilibrium position and the likely mixture of chemicals at any given time.