Examples of Exploring Reaction Quotient and Equilibrium

Exploring the Relationship Between Reaction Quotient and Equilibrium

Understanding the relationship between the reaction quotient (Q) and the equilibrium constant (K) is crucial in predicting the direction of chemical reactions. Here, we explore three diverse examples that illustrate this concept in real-world contexts.

Example 1: The Formation of Ammonia

Context

The Haber process is a well-known chemical reaction used to synthesize ammonia from nitrogen and hydrogen gases. It provides a practical example of how Q relates to K in industrial applications.

In this reaction, the equilibrium constant (K) is defined as:

\[K = \frac{[NH_3]^2}{[N_2][H_2]^3}\]

Example

Suppose we have a reaction mixture at equilibrium with the following concentrations:

• [NH₃] = 0.5 M
• [N₂] = 0.1 M
• [H₂] = 0.3 M

First, we can calculate K:

• \[K = \frac{(0.5)^2}{(0.1)(0.3)^3} = \frac{0.25}{0.0027} \approx 92.59\]

Now, let’s calculate Q using the same concentrations:

• \[Q = \frac{(0.5)^2}{(0.1)(0.3)^3} = 92.59\]

Notes

Since Q = K, the system is at equilibrium. If we were to decrease the concentration of ammonia, Q would decrease, shifting the reaction to the right to produce more ammonia until equilibrium is re-established.

Example 2: The Dissociation of Acetic Acid

Context

Acetic acid (CH₃COOH) is a weak acid that partially dissociates in water. This example illustrates how we can use Q and K to understand the extent of dissociation under varying conditions.

The equilibrium expression for the dissociation of acetic acid is:

\[K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}\]

Example

Consider a solution of acetic acid where:

• [H⁺] = 0.01 M
• [CH₃COO⁻] = 0.01 M
• [CH₃COOH] = 0.1 M

Calculating Kₐ:

• \[K_a = \frac{(0.01)(0.01)}{0.1} = 0.001\]

Now, let’s check Q:

• \[Q = \frac{(0.01)(0.01)}{0.1} = 0.001\]

Notes

Again, since Q = K, the system is at equilibrium. If we were to increase the concentration of acetic acid, Q would decrease, favoring the formation of more ions until equilibrium is restored.

Example 3: The Synthesis of Ethanol

Context

The production of ethanol from ethylene and water is an important industrial process. Understanding the relationship between Q and K can help optimize the production conditions.

The equilibrium expression for this reaction is:

\[K = \frac{[C_2H_5OH]}{[C_2H_4][H_2O]}\]

Example

Assume the following concentrations at a given moment:

• [C₂H₅OH] = 0.8 M
• [C₂H₄] = 0.3 M
• [H₂O] = 0.4 M

Calculating K:

• \[K = \frac{0.8}{(0.3)(0.4)} = \frac{0.8}{0.12} \approx 6.67\]

Now, let’s calculate Q:

• \[Q = \frac{0.8}{(0.3)(0.4)} = 6.67\]

Notes

Since Q = K, the reaction is at equilibrium. If the temperature were increased, it might shift the equilibrium position, affecting the concentrations and thus Q. Adjusting the conditions can help in optimizing the production of ethanol.

These examples illustrate the practical applications of exploring the relationship between reaction quotient and equilibrium. By understanding these concepts, it becomes easier to predict how changes in concentration and conditions will affect chemical reactions.