Oxygen can be converted into ozone by the action of lightning or electric sparks: 3 O 2 (g) ⇌ 2 O 3 (g) For this reaction, ∆ H = +69kcal/mol (+285 kj/mol) and K = 2.68 × 10 -29 at 25 °C. b. Are the reactants or the products favored at equilibrium?

Hello. The following conversion of molecular bromine to atomic bombing has an equilibrium constant of 1.1 times 10 to minus three at 1080 degrees Celsius were asked to determine whether the reactant or the product is favored at equilibrium. Let's begin by writing our equilibrium constant expression. Recall that it's our product concentration. So that would be bro main and the power comes from the coefficient in the balanced reaction equation. So it would be squared divided by the concentration of our reactant. So molecular browning, we're told this is equal to 1.1 times 10 to minus three. This is less than one. So that tells us then that our numerator is small and our denominator is large, we have the reactant in the denominator. So if this is large, it says then that the reaction favors reactant an equilibrium. This then corresponds to answer a thanks for watching. Hope this help.

Ch.7 Chemical Reactions: Energy, Rate and Equilibrium

McMurry - Fundamentals of GOB 8th Edition

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McMurry 8th Edition

Ch.7 Chemical Reactions: Energy, Rate and Equilibrium

Problem 59b

Chapter 7, Problem 59b

Oxygen can be converted into ozone by the action of lightning or electric sparks:
3 O₂(g) ⇌ 2 O₃(g)
For this reaction, ∆H = +69kcal/mol (+285 kj/mol) and K = 2.68 × 10^-29 at 25 °C.
b. Are the reactants or the products favored at equilibrium?

Verified step by step guidance

Step 1: Analyze the equilibrium constant (K). The equilibrium constant (K) indicates the ratio of the concentrations of products to reactants at equilibrium. A very small K value (e.g., 2.68 x 10^-29) suggests that the concentration of products is much smaller than the concentration of reactants at equilibrium.

Step 2: Interpret the given K value. Since K is extremely small, the reaction heavily favors the reactants at equilibrium, meaning that the formation of ozone (O3) is not favored under these conditions.

Step 3: Consider the enthalpy change (∆H). The reaction has a positive ∆H value (+69 kcal/mol), indicating that it is endothermic. This means the reaction requires energy input (e.g., from lightning or electric sparks) to proceed toward the formation of ozone.

Step 4: Relate ∆H to temperature. For an endothermic reaction, increasing the temperature would favor the formation of products (O3) because the system absorbs heat. However, at 25 °C, the reaction still strongly favors the reactants due to the very small K value.

Step 5: Conclude the equilibrium position. At 25 °C, the reactants (O2) are heavily favored at equilibrium because the equilibrium constant (K) is extremely small, indicating that only a negligible amount of ozone (O3) is formed under these conditions.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (K)

The equilibrium constant (K) quantifies the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. A small K value, such as 2.68 x 10^-29, indicates that at equilibrium, the concentration of reactants is much greater than that of products, suggesting that the reaction favors the reactants.

Gibbs Free Energy (∆G)

Gibbs Free Energy (∆G) is a thermodynamic potential that helps predict the direction of a chemical reaction. It is related to the enthalpy change (∆H) and entropy change (∆S) of the system. A positive ∆H, as given in the question, typically indicates that the reaction is endothermic, which can also suggest that the formation of products is less favorable under standard conditions.

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust to counteract the change and restore a new equilibrium. In this context, understanding how changes in temperature, pressure, or concentration affect the position of equilibrium can help predict whether reactants or products will be favored.

Related Practice

Textbook Question

Use your answer from Problem 7.53 to calculate the following:

a. [N₂O₄] at equilibrium when [NO₂] = 0.0250 mol/L

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Textbook Question

Use your answer from Problem 7.54 to calculate the following:

a. [O₂] at equilibrium when [CO₂] = 0.18 mol/L and [CO] = 0.0200 mol/L

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Textbook Question

Oxygen can be converted into ozone by the action of lightning or electric sparks:

3 O₂(g) ⇌ 2 O₃(g)

For this reaction, ∆H = +69kcal/mol (+285 kj/mol) and K = 2.68 × 10^-29 at 25 °C.

a. Is the reaction exothermic or endothermic?

2515

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Textbook Question

Hydrogen chloride can be made from the reaction of chlorine and hydrogen:

Cl₂(g) + H₂(g) → 2 HCl(g)

For this reaction, K = 26 × 10³³ and ∆H = -44 kcal/mol(-184 kJ/mol) at 25 °C.

a. Is the reaction endothermic or exothermic?

2350

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Textbook Question

When the following equilibria are disturbed by increasing the pressure, does the concentration of reaction products increase, decrease, or remain the same?

a. 2 CO₂(g) ⇌ 2 CO(g) + O₂(g)

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Textbook Question

For the following equilibria, use Le Châtelier's principle to predict the direction of the reaction when the pressure is increased by decreasing the volume of the equilibrium mixture.

a. C(s) + H₂O(g) ⇌ CO(g) + H₂(g)

1751

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Oxygen can be converted into ozone by the action of lightning or electric sparks: 3 O2(g) ⇌ 2 O3(g) For this reaction, ∆H = +69kcal/mol (+285 kj/mol) and K = 2.68 × 10-29 at 25 °C. b. Are the reactants or the products favored at equilibrium?

Verified video answer for a similar problem:

Key Concepts

Equilibrium Constant (K)

Gibbs Free Energy (∆G)

Le Chatelier's Principle

Oxygen can be converted into ozone by the action of lightning or electric sparks:
3 O₂(g) ⇌ 2 O₃(g)
For this reaction, ∆H = +69kcal/mol (+285 kj/mol) and K = 2.68 × 10^-29 at 25 °C.
b. Are the reactants or the products favored at equilibrium?