Question

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), can be synthesized in two ways. The first method involves reduction of oxygen by hydrogen, $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(\ell) $$ The second method involves oxidation of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{2}(\ell) $$ Find the free energy of formation, \(\Delta \mathrm{G}^{\circ}\), for both processes and predict which process is more efficient for the commercial preparation of hydrogen peroxide.

Short answer

The free energy of formation for Process 1 is approximately \(48.1 \text{ kJ/mol}\) and for Process 2 is approximately \(-94.1 \text{ kJ/mol}\). Since Process 2 has a lower (more negative) ΔG° value, it is more exergonic and therefore more efficient for the commercial preparation of hydrogen peroxide.

Step-by-step solution

Locate standard enthalpy changes

Table: Standard enthalpy changes and standard entropy changes for the elements and molecule involved. Elements and molecule | ΔH° (kJ/mol) | ΔS° (J/mol K) -------------------|--------|--------- \(\mathrm{H}_{2} (\mathrm{~g})\) | 0 | 131 \(\mathrm{O}_{2} (\mathrm{~g})\) | 0 | 205 \(\mathrm{H}_{2}\mathrm{O}_{2} (\ell)\) | -188 | 110 Note: Hydrogen and oxygen in their standard states both have ΔH° = 0 kJ/mol and ΔS° = 0 kJ/mol K.

Calculate the standard enthalpy change for the reaction

Using the given reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(\ell)\), calculate the standard enthalpy change: $$ \Delta \mathrm{H}^{\circ}_{\text{rxn}}=\Delta \mathrm{H}^{\circ}_{\mathrm{H}_{2}\mathrm{O}_{2}} - \left(\Delta \mathrm{H}^{\circ}_{\mathrm{H}_{2}} + \Delta \mathrm{H}^{\circ}_{\mathrm{O}_{2}}\right)= -188 - (0+0) \text{ kJ/mol} = -188 \text{ kJ/mol} $$

Calculate the standard entropy change for the reaction

Calculate the standard entropy change for the reaction: $$ \Delta \mathrm{S}^{\circ}_{\text{rxn}}=\Delta \mathrm{S}^{\circ}_{\mathrm{H}_{2}\mathrm{O}_{2}} - \left(\Delta \mathrm{S}^{\circ}_{\mathrm{H}_{2}} + \Delta \mathrm{S}^{\circ}_{\mathrm{O}_{2}}\right)= 110 - (131+205) \text{ J/mol K} = -226 \text{ J/mol K} $$

Calculate the free energy of formation for Process 1

Using the free energy equation, ΔG° = ΔH° - TΔS°, and assuming standard conditions (T = 298 K), calculate ΔG° for Process 1: $$ \Delta \mathrm{G}^{\circ}_{\text{Process 1}} = -188 \text{ kJ/mol} - (298 \mathrm{K})( -226 \text{ J/mol K})(1 \text{ kJ/1000 J}) \approx 48.1 \text{ kJ/mol} $$ **Process 2**

Calculate the standard enthalpy change for the reaction

In process 2, the water molecules are involved in the reaction, so the stoichiometric coefficients should be considered. Calculate the standard enthalpy change for the reaction: $$ \Delta \mathrm{H}^{\circ}_{\text{rxn}}=2\Delta \mathrm{H}^{\circ}_{\mathrm{H}_{2}\mathrm{O}_{2}} - \left( 2\Delta \mathrm{H}^{\circ}_{\mathrm{H}_{2}\mathrm{O}} + \Delta \mathrm{H}^{\circ}_{\mathrm{O}_{2}}\right)= -376 - (0+0) \text{ kJ/mol} = -376 \text{ kJ/mol} $$

Calculate the standard entropy change for the reaction

Calculate the standard entropy change for the reaction: $$ \Delta \mathrm{S}^{\circ}_{\text{rxn}}=2\Delta \mathrm{S}^{\circ}_{\mathrm{H}_{2}\mathrm{O}_{2}} - \left( 2\Delta \mathrm{S}^{\circ}_{\mathrm{H}_{2}\mathrm{O}} + \Delta \mathrm{S}^{\circ}_{\mathrm{O}_{2}}\right)= 220 - (0+0) \text{ J/mol K} = 220 \text{ J/mol K} $$

Calculate the free energy of formation for Process 2

Using the free energy equation, ΔG° = ΔH° - TΔS°, and assuming standard conditions (T = 298 K), calculate ΔG° for Process 2: $$ \Delta \mathrm{G}^{\circ}_{\text{Process 2}} = -376 \text{ kJ/mol} - (298 \mathrm{K})( 220 \text{ J/mol K})(1 \text{ kJ/1000 J}) \approx -94.1 \text{ kJ/mol} $$

Compare the free energy of formation for both processes

Process 1 has a higher ΔG° value, which means that Process 2 is more exergonic and, therefore, more efficient for the commercial preparation of hydrogen peroxide. ΔG°(Process 1) ≈ 48.1 kJ/mol ΔG°(Process 2) ≈ -94.1 kJ/mol Process 2 is more efficient for the commercial preparation of hydrogen peroxide.

Key concepts

Free Energy of Formation

The free energy of formation, denoted as \(\Delta G^{\circ}\), is a fundamental concept in chemical thermodynamics that measures the change in Gibbs free energy when 1 mole of a compound is formed from its elements at standard state conditions (298 K, 1 atm). This value helps predict if a chemical process is spontaneous and feasible under those standard conditions.

For hydrogen peroxide synthesis, calculating \(\Delta G^{\circ}\) for each process allows us to determine which method is more energetically favorable. This calculation incorporates both enthalpy (\(\Delta H^{\circ}\)) and entropy (\(\Delta S^{\circ}\)) changes according to the equation \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), with T representing temperature in Kelvin. A negative \(\Delta G^{\circ}\) indicates the reaction occurs spontaneously, demonstrating higher synthesis efficiency.

Enthalpy Change Calculation

Enthalpy change, represented by \(\Delta H^{\circ}\), quantifies the heat exchange in a chemical reaction at constant pressure. It reflects the total energy change, combining the breaking and forming of chemical bonds.

To calculate \(\Delta H^{\circ}\) in a reaction, we subtract the sum of enthalpies of the reactants from that of the products. A negative \(\Delta H^{\circ}\) suggests the reaction is exothermic (releases energy), while a positive value indicates it's endothermic (absorbs energy). In the context of hydrogen peroxide synthesis, calculating the \(\Delta H^{\circ}\) for both methods helps us assess their thermodynamic feasibility and provides a partial insight into their efficiency.

Entropy Change Calculation

Entropy, with the symbol S, is a measure of the disorder or randomness of a system. The change in entropy, \(\Delta S^{\circ}\), during a chemical reaction indicates if the system is becoming more or less ordered.

To calculate \(\Delta S^{\circ}\) in reactions, we take the difference between the entropy values of the products and the reactants. An increase in the number of moles of gas, or a change from a more structured to a less structured state, typically increases the system's entropy. In hydrogen peroxide synthesis, analyzing the entropy change is crucial to determine the second component of the Gibbs free energy equation and, therefore, the overall spontaneity of each synthesis method.

Chemical Thermodynamics

Chemical thermodynamics is the branch of chemistry that deals with the relationship between chemical reactions and energy changes involving heat, work, temperature, and entropy. It allows us to predict the direction and extent of chemical reactions and the conditions under which they occur.

In the case of hydrogen peroxide synthesis, thermodynamics principles guide us to comprehend how the changes in enthalpy and entropy of the reaction influence the free energy, which in turn informs about the spontaneity and equilibrium conditions. This field lays the foundation for understanding the efficiency and feasibility of various synthesis processes in the chemical industry.

Synthesis Efficiency

Synthesis efficiency refers to how effectively a chemical process converts reactants into desired products. This efficiency can be influenced by several factors, including reaction kinetics, the amount of energy consumed, and the yield of the reaction.

When evaluating the two methods of hydrogen peroxide synthesis, efficiency is gauged by considering the thermodynamic favorability (\(\Delta G^{\circ}\)) and the practical aspects such as reaction rates and costs. A lower \(\Delta G^{\circ}\) value for a given process usually corresponds to higher synthesis efficiency as it indicates a greater tendency for the reaction to proceed spontaneously, contributing to a more cost-effective industrial production.