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?

Short Answer

Expert verified
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 by step solution

01

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.
02

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: \[Q_c = \frac{(0.500)^2}{(3.00)(2.00)^3}\]
03

Calculate the reaction quotient (\(Q_c\))

Calculate the reaction quotient by evaluating the expression from Step 2: \[Q_c = \frac{0.25}{3.00 \times 8.00} = \frac{0.25}{24.00} = 0.0104\]
04

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).
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Equilibrium Constant (Kc)
The equilibrium constant, represented as Kc, 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 Kc, 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 Kc (much greater than 1) means that, at equilibrium, the reaction favors the products, whereas a smaller Kc (much less than 1) shows a reaction that favors the reactants. It's important to note that Kc 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, Qc, 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 Qc is calculated (as seen in Step 3 of the solution) and found to be different from Kc, the reaction is not at equilibrium and will shift to restore balance. If Qc is less than Kc, the reaction will proceed to produce more products. If Qc is greater than Kc, 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.

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Most popular questions from this chapter

State whether reactants or products will be favored by compression in each of the following equilibria. If no change occurs, explain why that is so. (a) \(2 \mathrm{O}_{3}(\mathrm{~g})=3 \mathrm{O}_{2}(\mathrm{~g})\) (b) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}(\mathrm{g})\) (c) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(2 \mathrm{HD}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2}(\mathrm{~g})\) (e) \(\mathrm{Cl}_{2}(\mathrm{~g})=2 \mathrm{Cl}(\mathrm{g})\)

Given that \(K_{c}=61\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(500 \mathrm{~K}\), calculate whether more ammonia will tend to form when a mixture of composition \(2.23 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(1.24 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(1.12 \times 10^{-4} \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\) is present in a container at \(500 \mathrm{~K}\).

A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

Determine \(K_{c}\) from the value of \(K\) for cach of the following equilibria. (a) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.4\) at \(1000 \mathrm{~K}\) (b) \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s})=\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}), K=9.4 \times 10^{-2}\) at \(24^{\circ} \mathrm{C}\)

Explain why the following cquilibria are heterogeneous and write the reaction quotient \(Q\) for each one. (a) \(2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{FcCl}_{3}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) (c) \(2 \mathrm{KNO}_{3}\) (s) \(\rightleftharpoons 2 \mathrm{KNO}_{2}\) (s) \(+\mathrm{O}_{2}\) (g)

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