Question

Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at \(85^{\circ} \mathrm{C} .\) (a) Use data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of \(\Delta G\) for the reaction as the temperature increases? (c) What is the significance of the change in the magnitude of \(\Delta G\) with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

Short answer

In summary, the calculated values for the reaction of hydrogen gas and oxygen gas to form water are ∆H° = -571.6 kJ/mol and ∆S° = -237.8 J/mol • K. As the temperature increases, the magnitude of ΔG decreases, which indicates a decrease in the efficiency of hydrogen as a fuel. Fuel cell technology, which operates at lower temperatures, is more efficient than combustion processes for generating electrical energy from hydrogen.

Step-by-step solution

Find the standard enthalpy of formation of H2O and entropies of species involved.

Search Appendix C for standard enthalpy of formation (∆H°f) for H2O and standard entropy values (S°) for H2, O2 and H2O. The values in Appendix C are: - ∆H°f (H2O) = \(-285.8 \ \mathrm{kJ \ mol^{-1}}\) - S° (H2) = \(130.60 \ \mathrm{J \ mol^{-1} \ K^{-1}}\) - S° (O2) = \(205.10 \ \mathrm{J \ mol^{-1} \ K^{-1}}\) - S° (H2O) = \(69.90 \ \mathrm{J \ mol^{-1} \ K^{-1}}\)

Calculate ∆H° for the reaction.

Use the stoichiometry of the reaction and the standard enthalpies of formation to calculate ∆H° as follows: ∆H° = [2 * ∆H°f (H2O)] - [2 * ∆H°f (H2) + ∆H°f (O2)] The enthalpies of formation of H2 and O2 are both 0 as they are diatomic gases in their standard states. Therefore, ∆H° = [2 * (-285.8 kJ/mol)] = -571.6 kJ/mol

Calculate ∆S° for the reaction.

Use the stoichiometry of the reaction and the standard entropies of the substances to calculate ∆S°: ∆S° = [2 * S° (H2O)] - [2 * S° (H2) + S° (O2)] ∆S° = [2 * (69.9 J/mol • K)] - [2 * (130.6 J/mol • K) + (205.1 J/mol • K)] = -237.8 J/mol • K #b) Determine the trend for ΔG with increasing temperature#

Calculate ΔG under standard conditions at 85°C.

We can find ΔG using the relationship: ΔG = ΔH - TΔS First, convert the temperature 85°C to Kelvin: 85 + 273.15 = 358.15 K ΔG = -571.6 kJ/mol - (358.15 K * (-237.8 J/mol • K)) ΔG = -326.2 kJ/mol

Trend for ΔG with increasing temperature.

As T increases, the negative (-TΔS) term will become less negative, and therefore, ΔG will also become less negative. In summary, as temperature increases, the magnitude of ΔG will decrease. #c) Analyze the significance of the change in ΔG with temperature with respect to the utility of hydrogen as a fuel#

Consider the relationship between ΔG and the maximum work attainable.

A less negative ΔG indicates a reduction in the amount of energy that can be transformed into useful work, which means the efficiency decreases when the temperature increases. #d) Determine which method to generate electrical energy from hydrogen is more efficient#

Combustion vs. Fuel Cells.

Fuel cells operate at a lower temperature than combustion processes, which means that the ΔG values remain relatively lower for fuel cells. Since a lower ΔG value indicates a higher energy conversion efficiency, fuel cell technology is more efficient than combustion for generating electrical energy from hydrogen.

Key concepts

Enthalpy and Entropy

When discussing the potential of hydrogen as a fuel, it's crucial to understand two fundamental thermodynamic concepts: enthalpy and entropy. Enthalpy (\( \Delta H \) ) represents the total energy change in a chemical reaction. It includes internal energy changes and pressures and volumes at constant pressure. Entropy (\( \Delta S \) ) describes the degree of disorder or randomness in a system.

For the reaction of hydrogen gas with oxygen to form water, the change in enthalpy was calculated to be -571.6 kJ/mol. This negative value indicates an exothermic reaction, meaning it releases energy in the form of heat. The entropy change was calculated as -237.8 J/mol · K, showing a decrease in disorder as gas molecules form liquid water. Together, these values help us predict the feasibility and spontaneity of the reaction, which becomes more insightful when considered alongside Gibbs Free Energy.

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G \) ) is an important factor for assessing the feasibility of a reaction to perform work. It combines enthalpy and entropy changes in a system. The equation \( \Delta G = \Delta H - T\Delta S \) expresses this relationship.

For hydrogen combustion, \( \Delta G \) at 85°C is -326.2 kJ/mol, suggesting the reaction is spontaneous, providing potential for use as fuel. As temperature increases, \( T\Delta S \) becomes less negative, making \( \Delta G \) less negative.
This implies the reaction releases less free energy available for work, i.e., its efficiency decreases. Understanding how \( \Delta G \) changes with temperature helps us evaluate hydrogen fuel's utility under various conditions.

Fuel Cells

Fuel cells offer an innovative way to harness hydrogen gas for energy. They function similarly to batteries but are continuously supplied with reactants. They combine hydrogen and oxygen to produce water, electricity, and heat.

The efficiency of fuel cells is higher compared to traditional combustion because they can directly convert chemical energy into electrical energy without going through an intermediate thermal stage. They operate with lower \( \Delta G \) values, maintaining higher energy efficiency. This makes fuel cells more advantageous, especially for applications where minimizing energy loss is crucial.- Continuous operation with consistent fuel supply- High efficiency compared to combustion- Reduced emissions, as water is the primary by-product

Combustion

Combustion of hydrogen involves burning hydrogen gas in the presence of oxygen to produce heat, water, and energy. This method is similar to traditional methods using fossil fuels to drive turbines for electricity.

While combustion can be less efficient than fuel cells due to energy lost as heat, the exothermic nature of hydrogen combustion still makes it an effective method.
However, the potential reduction in efficiency due to \( \Delta G \) becoming less negative at higher temperatures means more energy is lost. Despite this, combustion remains useful for large-scale energy production when fuel cells are not viable.

Energy Efficiency

Energy efficiency plays a paramount role in evaluating hydrogen’s utility as a fuel. It measures how much useful energy is successfully converted and utilized compared to the total energy input. In the context of generating electricity from hydrogen, efficiency comes down to how \( \Delta G \) and other thermodynamic properties are managed.

Fuel cells exhibit high efficiency due to minimal energy loss. They effectively use the reaction’s free energy, leading to better conversion rates. On the other hand, combustion typically involves greater energy losses as heat.
- Fuel cells: Directly convert chemical to electrical energy- Combustion: Involves energy loss through heat- Hydrogen's potential lies in maximizing its efficiency for cleaner and sustainable energy solutions