Problem 79 - Chemical Principles: The Quest for Insight

Question

At \(500^{\circ} \mathrm{C}, K_{c}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})\). If analysis shows that the composition is \(3.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}, 2.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\) and \(0.500 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\), is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Step-by-Step Solution

Answer

The reaction is not at equilibrium because the reaction quotient (\(Q_c\)) is less than the equilibrium constant (\(K_c\)). The reaction will proceed to the right to produce more \(\mathrm{NH}_3\) and reach equilibrium.

Step 1: Write the expression for the equilibrium constant (\(K_c\))

Based on the reaction provided, the equilibrium constant expression in terms of concentrations at temperature 500°C is given by: \[K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3}\] where [X] represents the molar concentration of substance X at equilibrium.

Step 2: Plug in the given concentrations into the the equilibrium constant expression

Using the provided concentrations, the reaction quotient (\(Q_c\)) is calculated by substituting the concentrations into the equilibrium expression: \[Q_c = \frac{(0.500)^2}{(3.00)(2.00)^3}\]

Step 3: Calculate the reaction quotient (\(Q_c\))

Calculate the reaction quotient by evaluating the expression from Step 2: \[Q_c = \frac{0.25}{3.00 \times 8.00} = \frac{0.25}{24.00} = 0.0104\]

Step 4: Compare the reaction quotient (\(Q_c\)) with the equilibrium constant (\(K_c\))

The value of the reaction quotient indicates if the reaction is at equilibrium (if \(Q_c = K_c\)), or which direction it will proceed to reach equilibrium (\(Q_c < K_c\) indicates the reaction will proceed to the right, forming more products; \(Q_c > K_c\) indicates the reaction will proceed to the left, forming more reactants).

Step 5: Determine the reaction's direction toward equilibrium

Since \(Q_c = 0.0104\) is less than \(K_c = 0.061\), the reaction will proceed to the right (toward the production of more \(\mathrm{NH}_3\) gas) to reach equilibrium.

Key Concepts

Understanding the Equilibrium Constant (Kc)

The equilibrium constant, represented as K_c, plays a pivotal role in the study of chemical reactions. It is a value that indicates the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction at constant temperature.

When we look at the reaction \( \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \), the equilibrium constant expression is
\[ K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \]
Here, the concentrations are raised to the power of their stoichiometric coefficients in the balanced chemical equation. Importantly, the concentration of solid and liquid compounds do not appear in this expression as they are constant and do not affect the balance of the reaction.

The value of K_c, in this case, is given as 0.061 at 500°C. This constant value is crucial because it signifies the extent of the reaction: a larger K_c (much greater than 1) means that, at equilibrium, the reaction favors the products, whereas a smaller K_c (much less than 1) shows a reaction that favors the reactants. It's important to note that K_c does not give any information about how fast the reaction achieves equilibrium, only about the position of equilibrium.

Using the Reaction Quotient (Qc) to Predict Reaction Direction

The reaction quotient, Q_c, is calculated similarly to the equilibrium constant but it applies to any set of given concentrations, not necessarily at equilibrium. For the reaction \( \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \), the expression is
\[ Q_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \]
Using the concentrations at any point in time, we can determine the status of the reaction.

If Q_c is calculated (as seen in Step 3 of the solution) and found to be different from K_c, the reaction is not at equilibrium and will shift to restore balance. If Q_c is less than K_c, the reaction will proceed to produce more products. If Q_c is greater than K_c, the reaction will proceed to produce more reactants. This is instrumental in predicting the direction in which the reaction will proceed to achieve equilibrium.

Le Chatelier's Principle and Its Effects on Equilibrium

Le Chatelier's principle is a foundational concept in chemistry that describes how a system at equilibrium responds to external changes.

According to this principle, if a system at equilibrium experiences a change in concentration, temperature, or pressure, the system will adjust accordingly to counteract the change and reattain a new equilibrium state. For the given reaction involving N₂, H₂, and NH₃, here are a few scenarios to consider:

• Increase in N₂ or H₂ concentration: The system will shift to the right, producing more NH₃.
• Decrease in NH₃ concentration: The system will also shift to the right to produce more NH₃.
• Increase in temperature (for an exothermic reaction): The system will shift to the left, reducing the amount of NH₃.
• Increase in pressure: The system will shift to the side with fewer gas molecules, which, in this case, would be a shift to the right, favoring the formation of NH₃.

Understanding Le Chatelier's principle helps in controlling reactions to desired outcomes, such as increasing product yield in industrial processes. It underscores the dynamic nature of equilibrium, highlighting that it is not static but adaptable to the conditions imposed on the system.