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?

Step by step solution

01

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) = \(69.90 \ \mathrm{J \ mol^{-1} \ K^{-1}}\)

02

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

03

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#

04

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

05

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#

06

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#

07

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.

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