Question

13.58 Separate streams of hydrogen \(\left(\mathrm{H}_{2}\right)\) and oxygen \(\left(\mathrm{O}_{2}\right)\) at \(25^{\circ} \mathrm{C}, 1\) atm enter a fuel cell operating at steady state, and liquid water exits at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). The hydrogen flow rate is \(2 \times 10^{-4}\) \(\mathrm{kmol} / \mathrm{s}\). If the fuel cell operates isothermally at \(25^{\circ} \mathrm{C}\), determine the maximum theoretical power it can develop and the accompanying rate of heat transfer, each in \(\mathrm{kW}\). Kinetic and potentially energy effects are negligible.

Short answer

Maximum theoretical power: -94.88 kW.Rate of heat transfer: 94.88 kW.

Step-by-step solution

Write the balanced chemical reaction

The balanced chemical reaction for the formation of water from hydrogen and oxygen is: \[ 2 H_2 + O_2 \rightarrow 2 H_2O \]

Calculate the moles of reactants

Determine the amount of oxygen required for the reaction. Given that the flow rate of hydrogen is \(2 \times 10^{-4}\) kmol/s, use the stoichiometry of the reaction: From the balanced equation: \[ 2 \text{ mol } H_2 : 1 \text{ mol } O_2 \] Therefore, the flow rate of oxygen required is: \(\dot{n}_{O_2} = \frac{1}{2} \times \dot{n}_{H_2} = \frac{1}{2} \times 2 \times 10^{-4} = 1 \times 10^{-4} \text{ kmol/s}\)

Find the Gibbs free energy change

For the reaction at standard conditions (25°C, 1 atm), the Gibbs free energy change, \(\Delta G^0_f\), for the formation of liquid water (H2O) is: \(\Delta G^0_f = -237.2 \text{ kJ/mol}\) For the reaction: \[ 2H_2 + O_2 \rightarrow 2H_2O \] So, \(\Delta G^0 = 2 \times (-237.2) = -474.4 \text{ kJ/mol}\)

Calculate the maximum theoretical power

Maximum theoretical power is given by the total Gibbs free energy change per second. Using the flow rate of hydrogen: \(\dot{n}_{H_2} = 2 \times 10^{-4} \text{ kmol/s}\). The corresponding Gibbs free energy change per second is: \(\dot{W}_{max} = \dot{n}_{H_2} \times (-474.4 \text{ kJ/mol}) = 2 \times 10^{-4} \text{ kmol/s} \times (-474.4 \text{ kJ/mol}) \times 1000 \text{ (to convert kmol to mol)} = -94.88 \text{ kW}\)

Calculate the rate of heat transfer

Using the first law of thermodynamics for a steady-state, isothermal process: \(0 = \dot{Q} - \dot{W}_{max}\) Thus, the rate of heat transfer: \(\dot{Q} = \dot{W}_{max} \). Therefore, \(\dot{Q} = 94.88 \text{ kW}\) (heat transfer is in the same magnitude as the maximum theoretical power but considering sign convention, it’s positive value)

Key concepts

Chemical Reaction Stoichiometry

In our exercise, hydrogen \(\text{H}_2\) and oxygen \(\text{O}_2\) combine to form water \(\text{H}_2\text{O}\). Stoichiometry helps us understand the proportions needed for this reaction. By using the balanced chemical equation \[ 2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} \], we see that 2 moles of hydrogen will react with 1 mole of oxygen to produce 2 moles of water.

Gibbs Free Energy Change

The Gibbs Free Energy Change (ΔG) indicates the energy available to do work during a chemical reaction. For our reaction, the standard Gibbs free energy change is \(\text{-237.2 kJ/mol}\) for the formation of liquid water from hydrogen and oxygen at 25°C and 1 atm. In the balanced equation \[ 2\text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} \], the total Gibbs free energy change is \(-474.4 \text{kJ/mol}\), since two moles of water are formed.

Maximum Theoretical Power

To find the maximum theoretical power, we use the Gibbs free energy change per second. Given the hydrogen flow rate \(2 \times 10^{-4} \text{ kmol/s}\), the power \(W_{\text{max}}\) can be calculated as follows: \[ \text{W}_{\text{max}} = (\text{Hydrogen Flow Rate}) \times (\text{Gibbs Free Energy Change}) \]. Substituting the values, we get: \[ \text{W}_{\text{max}} = 2 \times 10^{-4} \times (-474.4 \text{ kJ/mol}) \times 1000 = -94.88 \text{ kW} \]

Rate of Heat Transfer

In a steady-state, isothermal process, the rate of heat transfer ( Q ) relates to the maximum theoretical power using the First Law of Thermodynamics: \[ 0 = \text{Q} - \text{W}_{\text{max}}\]. This simplifies to \(\text{Q} = \text{W}_{\text{max}}\). Therefore, the rate of heat transfer is \(\text{94.88 kW}\). It is positive, indicating that the system absorbs this much heat from the surroundings.