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?

Verified step by step guidance

1. Identify whether the reaction absorbs or releases heat by analyzing the given enthalpy change (∆H). A positive ∆H value indicates that the reaction absorbs heat, making it endothermic, while a negative ∆H value would indicate an exothermic reaction.
2. In this problem, the enthalpy change (∆H) is given as +69 kcal/mol (+285 kJ/mol). Since the value is positive, this means the reaction absorbs heat from the surroundings.
3. Conclude that the reaction is endothermic because it requires energy input to proceed, as indicated by the positive ∆H value.
4. Recall that endothermic reactions are typically associated with an increase in the energy of the system, as energy is absorbed to form the products.
5. Note that the equilibrium constant (K) being very small (2.68 x 10^-29) at 25 °C suggests that the reaction strongly favors the reactants (O2) under standard conditions, but this does not affect the classification of the reaction as endothermic.

Key Concepts

Endothermic and Exothermic Reactions

Endothermic reactions absorb heat from their surroundings, resulting in a positive change in enthalpy (∆H > 0). Conversely, exothermic reactions release heat, leading to a negative change in enthalpy (∆H < 0). In this case, since the reaction has a positive ∆H of +69 kcal/mol, it indicates that the reaction is endothermic.

Enthalpy Change (∆H)

Enthalpy change (∆H) is a measure of the heat content of a system at constant pressure. It indicates whether a reaction absorbs or releases energy. A positive ∆H value signifies that energy is absorbed during the reaction, while a negative value indicates energy release. For the given reaction, the positive ∆H confirms it is endothermic.

Equilibrium Constant (K)

The equilibrium constant (K) quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. A very small K value, such as 2.68 x 10^-29, suggests that at equilibrium, the reactants are favored over the products, indicating that the formation of ozone (O3) from oxygen (O2) is not favored under standard conditions.