Question

A student warned his friends not to swim in a river close to an electric plant. He claimed that the ozone produced by the plant turned the river water to hydrogen peroxide, which would bleach hair. The reaction is $$ \mathrm{O}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{O}_{2}(g) $$ Assuming that the river water is at \(25^{\circ} \mathrm{C}\) and all species are at standard concentrations, show by calculation whether his claim is plausible. Take \(\Delta G_{i}^{\circ} \mathrm{O}_{3}(g)\) at \(25^{\circ} \mathrm{C}\) to be \(+163.2 \mathrm{~kJ} / \mathrm{mol}\) $$ \text { and } \Delta G_{f}^{\circ} \mathrm{H}_{2} \mathrm{O}_{2}(a q)=-134 \mathrm{~kJ} / \mathrm{mol} \text { . } $$

Short answer

Based on the given information and calculations, the student's claim that ozone reacts with water vapor to thermodynamically favor the formation of hydrogen peroxide is plausible. This is because the overall change in Gibbs Free Energy for the reaction is negative, indicating that the reaction is spontaneous under standard conditions at 25°C.

Step-by-step solution

Write down the given information

We are given the following information: - ΔGᵢ° O₃(g) = +163.2 kJ/mol - ΔGᶠ° H₂O₂(aq) = -134 kJ/mol

Calculate the change in Gibbs Free Energy for the other species involved in the reaction

Since, both water and oxygen are in their standard state, ΔG = 0 for both the species: - ΔGᵢ° H₂O(l) = 0 kJ/mol - ΔGᶠ° O₂(g) = 0 kJ/mol

Calculate the overall change in Gibbs Free Energy for the reaction using the given data

The overall change in Gibbs Free Energy, ΔG°, is the sum of the products' Gibbs Free Energies minus the sum of the reactants' Gibbs Free Energies: ΔG° = [ΔGᶠ°(H₂O₂) + ΔGᶠ°(O₂)] - [ΔGᵢ°(H₂O) + ΔGᵢ°(O₃)] Substitute the given values of each species and solve for ΔG° : ΔG° = [-134 kJ/mol + 0 kJ/mol] - [0 kJ/mol + 163.2 kJ/mol] = -134 kJ/mol - 163.2 kJ/mol = -(297.2 kJ/mol)

Determine the plausibility of the student's claim

Since the overall change in Gibbs Free Energy, ΔG°, is negative (-297.2 kJ/mol), the reaction is spontaneous at 25°C and standard conditions. This means the student's claim could be plausible, as the production of hydrogen peroxide is thermodynamically favored under these conditions.

Key concepts

Chemical Spontaneity

Understanding chemical spontaneity is crucial when evaluating whether a reaction will occur under specific conditions without external energy input. Spontaneity is dictated by a value known as Gibbs Free Energy, often symbolized as \( \Delta G \). In essence, if a reaction has a negative \( \Delta G \), it is considered spontaneous, meaning it is capable of proceeding forward on its own. Now, spontaneity doesn't necessarily mean that a reaction will happen quickly; it might still be slow due to kinetic barriers.

For example, the claim about the river turning into hydrogen peroxide can be evaluated for spontaneity by calculating the Gibbs Free Energy of the reaction. In our exercise, the \( \Delta G^{\circ} \) for the production of hydrogen peroxide and oxygen from ozone and water was found to be negative, suggesting that under standard conditions at 25°C, this transformation is spontaneous. Hence, according to thermodynamics, the student's claim could indeed have merit. However, keep in mind that factors other than \( \Delta G \) might affect the actual occurrence of this reaction in the environment.

Thermodynamic Favorability

The thermodynamic favorability of a reaction is intricately tied to its Gibbs Free Energy change, as mentioned in the context of spontaneity. To determine if a particular reaction is thermodynamically favorable, we look at the sign and magnitude of the \( \Delta G \) value. A reaction with a \( \Delta G \) less than zero is thermodynamically favorable, implying that it can release free energy, often in the form of work or heat, into its surroundings.

In the exercise, by finding that \( \Delta G^{\circ} \) was -297.2 kJ/mol, it tells us that not only is the reaction spontaneous, but it also favors the formation of products (hydrogen peroxide and oxygen in this case) from the reactants (ozone and water). This thermodynamic favorability contributes to our understanding of why certain reactions occur in nature, lending credibility to potential alterations in the environment, such as the formation of hydrogen peroxide in a river nearby a power plant.

Chemical Reaction Calculations

Performing chemical reaction calculations allows us to predict the behavior of a reaction quantitatively. In the context of Gibbs Free Energy, we use the values of \( \Delta G^{\circ} \) for each reactant and product to calculate the overall change in free energy for the reaction. The formula \( \Delta G^{\circ} = \sum \Delta G^{\circ}_{\text{products}} - \sum \Delta G^{\circ}_{\text{reactants}} \) is fundamental in these calculations.

Let's delve into our exercise: we calculated the \( \Delta G^{\circ} \) for the overall reaction by subtracting the sum of \( \Delta G^{\circ} \) for reactants from that of the products. This process involves analyzing the standard Gibbs Free Energy of formation \( (\Delta G^{\circ}_f) \) or ionization \( (\Delta G^{\circ}_i) \) for each species involved. In scenarios where water and oxygen are in standard states (as in our problem), their \( \Delta G^{\circ} \) values are zero. These calculations form the basis for predicting whether reactions can happen and under what conditions, which is a fundamental aspect of chemistry studies.