Problem 71 - Chemistry An Atoms First Approach

Question

Calculate a value for the equilibrium constant for the reaction $$\mathbf{O}_{2}(g)+\mathbf{O}(g) \rightleftharpoons \mathbf{O}_{3}(g)$$.given $$\begin{aligned}& \mathrm{NO}_{2}(g) \stackrel{h v}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) & & K=6.8 \times 10^{-49} \\\\\mathrm{O}_{3}(g)+\mathrm{NO}(g) & \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & & K=5.8 \times 10^{-34}\end{aligned}$$.(Hint: When reactions are added together, the equilibrium expressions are multiplied.) (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

Step-by-Step Solution

Answer

The equilibrium constant for the reaction O₂(g) + O(g) ⇄ O₃(g) is approximately 8.52 × 10¹⁵.

Step 1: Analyze given reactions

We are given the following reactions: 1. NO₂(g) ⇄ NO(g) + O(g) with K₁ = 6.8 × 10⁻⁴⁹ 2. O₃(g) + NO(g) ⇄ NO₂(g) + O₂(g) with K₂ = 5.8 × 10⁻³⁴ We want to manipulate these reactions in a way that results in the desired reaction: O₂(g) + O(g) ⇄ O₃(g).

Step 2: Determine a suitable combination of given reactions

ToAdd the given reactions together in a suitable manner, we can reverse reaction 1 and add it to reaction 2. This will result in the following reaction: 1. -(NO₂(g) ⇄ NO(g) + O(g)) 2. O₃(g) + NO(g) ⇄ NO₂(g) + O₂(g) -------------------------- Result: O₂(g) + O(g) ⇄ O₃(g)

Step 3: Calculate the new equilibrium constants for the manipulated reaction

Since we have reversed reaction 1, the equilibrium constant for this manipulated reaction will be the inverse of the initial equilibrium constant: K'₁ = 1 / K₁ = 1 / (6.8 × 10⁻⁴⁹)

Step 4: Multiply the equilibrium constants for the combined reactions

Now, we can multiply K'₁ by K₂ to obtain the equilibrium constant for the desired reaction: K₃ = K'₁ × K₂ = (1 / (6.8 × 10⁻⁴⁹)) × (5.8 × 10⁻³⁴)

Step 5: Calculate the value of the equilibrium constant

Finally, we can calculate the value of K₃: K₃ = (1 / 6.8 × 10⁻⁴⁹) × (5.8 × 10⁻³⁴) K₃ = (1 / 6.8) × (5.8 × 10¹⁵) K₃ = 0.147 × 5.8 × 10¹⁵ K₃ ≈ 8.52 × 10¹⁵ Thus, the equilibrium constant for the reaction O₂(g) + O(g) ⇄ O₃(g) is approximately 8.52 × 10¹⁵.

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is fundamental in chemical reactions, including synthesis and decomposition. It is the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. In this balanced state, no net change occurs in the concentrations of reactants and products. However, both processes continue to happen at equal rates, maintaining the system's stability.

When a system reaches equilibrium, it is often represented by an equilibrium constant, denoted as K, which is a numerical value that expresses the ratio of product concentrations to reactant concentrations, each raised to the power of their respective coefficients from the balanced equation. For the reaction \( A + B \rightleftharpoons C + D \), the equilibrium constant K is given by the expression: \[ K = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \] where the square brackets indicate concentrations, and a, b, c, d are the stoichiometric coefficients. This constant is crucial for predicting the direction of the reaction and understanding the reaction's extent under given conditions.

Le Chatelier's Principle

Le Chatelier's principle provides insight into the behavior of a system when changes are introduced. It postulates that if a dynamic equilibrium is disturbed by changing conditions, such as concentration, pressure, or temperature, the equilibrium will shift to oppose the change and establish a new equilibrium.

This principle is utilized to predict the effects of such changes. For instance, if the concentration of a reactant is increased, the system will respond by shifting the equilibrium to the right, producing more products to counteract the change. Conversely, decreasing the concentration of a product causes the equilibrium to shift to the left, producing more reactants.

Application in Exercises

Applying Le Chatelier's principle in calculations involves understanding how alterations affect K. When reactions are added together to form a new one, as shown in the step-by-step solution, the principle helps understand why and how the equilibrium constants of individual reactions relate to the new reaction's K value.

Reaction Quotient

The reaction quotient, Q, plays a pivotal role in predicting the direction of equilibrium shifts. It is calculated using the same formula as the equilibrium constant K, but with the initial concentrations instead of the equilibrium concentrations.

The equation for Q is: \[ Q = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \], and it reflects the state of a reaction mixture at any point in time, unlike K, which only represents the mixture at equilibrium. By comparing the values of Q and K, we can determine the reaction's progress:

• If \( Q < K \), the forward reaction is favored, and more products will be formed.
• If \( Q > K \), the reverse reaction is favored, and more reactants will be formed.
• If \( Q = K \), the system is at equilibrium, and no net change occurs.

Through the use of the reaction quotient, students can anticipate how a reaction will proceed and adjust the conditions to achieve a desired outcome or state of equilibrium.