Question

13.52 Carbon monoxide (CO) at \(25^{\circ} \mathrm{C}, 1\) atm enters an insulated reactor operating at steady state and reacts completely with the theoretical amount of air entering in a separate stream at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). The products of combustion exit as a mixture at \(1 \mathrm{~atm}\). For the reactor, determine the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per kmol of CO entering. Neglect kinetic and potential energy effects.

Short answer

0.0866 kJ/K per kmol of CO.

Step-by-step solution

Write the balanced reaction equation

The reaction of carbon monoxide (CO) with the theoretical amount of air can be given by:\[2 \text{CO}(g) + \text{O}_2(g) + 2 \text{N}_2(g) \rightarrow 2 \text{CO}_2(g) + 2 \text{N}_2(g)\]Here, air is assumed to consist of 21% oxygen and 79% nitrogen by volume.

Evaluate moles of reactants and products

For 1 kmol of CO, the moles of reactants and products are:Reactants: \[1 \text{kmol CO} \]\[0.5 \text{kmol O}_2 \]\[1.88 \text{kmol N}_2 \] (since air is 21% oxygen, so 79/21 times = 3.76/2 = 1.88 times more N_2 than O_2)Products: \[1 \text{kmol CO}_2 \]\[1.88 \text{kmol N}_2 \]

Calculate the entropy change of the reactants and products

The change in entropy can be evaluated as:\[ \text{ΔS}_{reactants} = \text{ΔS}_{CO} + \text{ΔS}_{O_2} + \text{ΔS}_{N_2} \]\[ \text{ΔS}_{products} = \text{ΔS}_{CO_2} + \text{ΔS}_{N_2} \]Given standard entropy values (at 298 K, 1 atm):\(\text{S}_{CO} = 197.7 \text{ J/mol·K}\)\(\text{S}_{O_2} = 205.0 \text{ J/mol·K}\)\(\text{S}_{N_2} = 191.5 \text{ J/mol·K}\)\(\text{S}_{CO_2} = 213.6 \text{ J/mol·K}\)So, we have:\[ \text{ΔS}_{reactants} = (1 \times 197.7) + (0.5 \times 205.0) + (1.88 \times 191.5) \]\[ \text{ΔS}_{reactants} = 197.7 + 102.5 + 359.02 = 659.22 \text{ J/K} \]\[ \text{ΔS}_{products} = (1 \times 213.6) + (1.88 \times 191.5) \]\[ \text{ΔS}_{products} = 213.6 + 359.02 = 572.62 \text{ J/K} \]

Calculate the change in entropy for the system

The change in entropy of the system is given by: \[ \text{ΔS}_{system} = \text{ΔS}_{products} - \text{ΔS}_{reactants} \]So, \[ \text{ΔS}_{system} = 572.62 - 659.22 = -86.6 \text{ J/K} \]

Calculate the entropy production rate

Since the process occurs in an insulated reactor (adiabatic), the entropy production must equal the change in entropy of the surroundings (which is zero, since there's no heat exchange):So, the rate of entropy production per kmol of CO entering is:\[\text{production rate} = - \text{ΔS}_{system} = 86.6 \text{ J/K}\]Convert to kJ/K: = 0.0866 \text{ kJ/K} per kmol of CO.

Key concepts

Thermodynamic entropy

Thermodynamic entropy is a measure of the disorder or randomness in a system. It is a cornerstone concept in thermodynamics and helps predict the direction of spontaneous processes. In the context of combustion reactions, entropy changes indicate how reactants (like carbon monoxide and oxygen) transform into products (like carbon dioxide and nitrogen). The total entropy of an isolated system can never decrease over time, which is an expression of the second law of thermodynamics. This law is crucial for understanding why certain reactions occur, as well as their efficiency. Notably, when analyzing a combustion process, we often calculate the entropy changes for both the reactants and the products. In a steady-state process, these changes help determine the production rate of entropy—a key factor in evaluating the performance of the system.
Entropy values are specific for different substances and states (e.g., gas, liquid). These values are usually given in units like J/mol·K or kJ/mol·K, and are required to solve entropy-related problems in thermodynamics.

Combustion reactions

Combustion reactions are chemical processes in which a fuel reacts with an oxidant, releasing heat and forming products. These reactions are exothermic, meaning they release energy in the form of heat. In the given exercise, the combustion of carbon monoxide (CO) with oxygen (O_2) is considered. This reaction can be simplified as:

• 2 CO(g) + O2(g) + 2 N2(g) → 2 CO2(g) + 2 N2(g)

Here, we assume the reaction takes place with the theoretical amount of air, consisting of 21% oxygen and 79% nitrogen. Combustion reactions are essential in many industrial processes and power generation. Understanding them helps in optimizing fuel efficiency and controlling pollutant emissions. Additionally, calculating the entropy change for such reactions assists in evaluating the system's thermodynamic performance. For example, knowing the standard entropy values of CO, O2, N2, and CO2 allows us to compute the total entropy changes during the reaction. This helps in understanding how much disorder increases or decreases in the system.

Steady-state process

A steady-state process is a condition where all properties of the system (like pressure, temperature, and chemical composition) remain constant over time, despite ongoing mass and energy transfer. In the context of the exercise, the combustion reaction occurs in an insulated reactor operating at steady state. This means all entering and exiting streams have consistent flow rates and properties. Steady-state conditions are crucial for simplifying the analysis of thermodynamic systems. They allow us to focus on reaction processes without worrying about temporal changes or oscillations. For example, when calculating entropy production, the steady-state assumption enables us to use constant entropy values for reactants and products, simplifying the calculations. Moreover, in real-world applications, maintaining a steady-state condition is often desirable for consistent product quality and efficient operation.

Adiabatic process

An adiabatic process is one in which no heat is exchanged with the surroundings. This is achieved through insulation or rapid compression/expansion. In the given exercise, the reactor is well-insulated, making the combustion an adiabatic process. This means that all the energy changes are due to work done and not due to heat transfer.

Adiabatic processes play a significant role in thermodynamics, especially in understanding entropy changes. In such systems, the only entropy change comes from the reactions themselves, as there is no heat flow to increase or decrease entropy. For example, in the step-by-step solution, the adiabatic condition implies that the entropy production must match the change in system entropy, as there's zero heat exchange with the environment. Moreover, understanding adiabatic processes helps engineers design efficient reactors and engines, as insulation prevents energy losses, making the processes more thermodynamically efficient.