Question

13.61 An inventor has developed a device that at steady state takes in liquid water at ${25}^{\circ }\mathrm{C},1\mathrm{atm}$ with a mass flow rate of $4\mathrm{kg}/\mathrm{h}$ and produces separate streams of hydrogen $\left({\mathrm{H}}_2\right)$ and oxygen $\left({\mathrm{O}}_2\right)$, each at ${25}^{\circ }\mathrm{C},1$ atm. The inventor claims that the device requires an electrical power input of $14.6\mathrm{kW}$ when operating isothermally at ${25}^{\circ }\mathrm{C}$. Heat transfer with the surroundings occurs, but kinetic and potential energy effects can be ignored. Evaluate the inventor's claim.

Short answer

The inventor's claim of requiring 14.6 kW is realistic.

Step-by-step solution

Physical Understanding of the Process

The problem describes a device that takes in liquid water at steady state and separates it into hydrogen and oxygen gas. The process occurs isothermally at 25°C, with electrical power input and heat transfer with surroundings.

Determine the Chemical Reaction

The chemical reaction for the decomposition of water is: $$2H_{2}O(l)\to 2H_2(g)+O_2(g)$$

Calculate the Required Electrical Work

Use the standard Gibbs free energy change for the reaction at 25°C to calculate the required work. The Gibbs free energy change for water decomposition is: $\mathrm{△}G=237.2\text{ kJ/mol}$. Convert flow rate to moles per second: - Water flow rate: 4 kg/h - Molar mass of water: 18 g/mol Flow rate in moles: $$\text{Flow rate}=\frac{4\text{ kg/h}\times 1000\text{ g/kg}}{18\text{ g/mol}}=222.22\text{ mol/h}$$ Convert to seconds: $$=\frac{222.22}{3600}\text{ mol/s}=0.0617\text{ mol/s}$$

Calculate the Electrical Power Requirement

Using Gibbs free energy, calculate the required power: $$\text{Power}=0.0617\text{ mol/s}\times 237.2\text{ kJ/mol}=14.63\text{ kW}$$

Evaluate the Inventor's Claim

Compare the calculated electrical power requirement to the inventor’s claim: $14.6\text{ kW}\text{ (claimed)}\approx 14.63\text{ kW}\text{ (calculated)}$. This indicates that the inventor's claim is realistic within the precision of the provided data.

Key concepts

Gibbs Free Energy

Gibbs free energy is a crucial concept in thermodynamics that helps us understand the energy changes in a chemical reaction. It is denoted as $G$ and combines enthalpy (total heat content) and entropy (degree of disorder). The formula is: $$\mathrm{△}G=\mathrm{△}H-T\mathrm{△}S$$ where $\mathrm{△}H$ is the change in enthalpy, $T$ is the temperature in Kelvin, and $\mathrm{△}S$ is the change in entropy.
When $\mathrm{△}G$ is negative, the reaction occurs spontaneously. In this exercise, the Gibbs free energy for the decomposition of water is given as $237.2$ kJ/mol. This positive $\mathrm{△}G$ value indicates that input energy is required to drive the reaction, making it non-spontaneous.
This value plays a critical role in calculating the electrical work needed to decompose water into hydrogen and oxygen.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products. In this exercise, the chemical process of interest is the decomposition of water: $$2H_{2}O(l)\to 2H_2(g)+O_2(g)$$ This reaction illustrates how liquid water is split into hydrogen and oxygen gases.< ul>

Hydrogen gas ($H_2$) is highly flammable and has multiple industrial applications.

Oxygen gas ($O_2$) is essential for life and industrial processes.

Decomposing water is a standard reaction in various scientific fields, including energy production and hydrogen fuel technologies.
Understanding the stoichiometry is crucial. For every 2 moles of water decomposed, 2 moles of hydrogen and 1 mole of oxygen are produced.

Electrical Power Input

Electrical power input is the energy supplied to a system in the form of electrical work. In this exercise, the inventor claims that the device requires $14.6$ kW to operate at a steady state and isothermally. This power input is essential to drive the non-spontaneous reaction of water decomposition.< ul>

The calculation of power begins by determining the mass flow rate of water (\text{4 kg/h}).

Next, convert this to moles per second.

Using the Gibbs free energy change, calculate the electrical work required per unit of time.

In our calculations, the required power is $14.63$ kW. This figure is consistent with the inventor’s claim, indicating that the proposed device is theoretically sound.
Electrical power input is critical in various applications, from industrial chemical processes to everyday appliances.

Isothermal Process

An isothermal process occurs at a constant temperature. In this exercise, the device operates isothermally at ${25}^{\text{°}}C$. Maintaining constant temperature is key to ensuring that the Gibbs free energy change remains accurate for the calculation.< ul>

Isothermal processes are often idealized and assume perfect thermal equilibrium with surroundings.

In real-world applications, achieving perfect isothermality can be challenging.

Heat transfer with the surroundings might be necessary to maintain constant temperature.

In this scenario, ignoring kinetic and potential energy effects simplifies the calculations. It focuses solely on the thermal and chemical energy aspects.
Overall, isothermal processes are essential in thermodynamics for simplifying and understanding energy transformations.