Question

Determine the work potential of 1 lbmol of diesel fuel \(\left(\mathrm{C}_{12} \mathrm{H}_{26}\right)\) at \(77^{\circ} \mathrm{F}\) and 1 atm in an environment at the same state.

Short answer

Answer: The work potential of 1 lbmol of diesel fuel at given conditions is 137,276.795 J/mol.

Step-by-step solution

Convert temperature to Kelvin

First, we need to convert the given temperature in Fahrenheit (77°F) to Kelvin. The conversion formula is: $$T (K) = \frac{5}{9}(T(\text{°F}) - 32) + 273.15$$ Plugging in the values, we get: $$T (K) = \frac{5}{9}(77 - 32) + 273.15$$ $$T (K) = 298.15\ K$$

Determine enthalpy change, \(\Delta H\)

We now need to determine the enthalpy change for 1 lbmol of diesel fuel, C12H26. This is typically found in thermodynamics or chemistry tables and denoted as \(\Delta H^\circ_f\). For diesel fuel: $$\Delta H^\circ_f(\text{C}_{12}\text{H}_{26}) = -350.1\ \text{kJ/mol}$$

Determine entropy change, \(\Delta S\)

Similarly, we need to determine the entropy change for 1 lbmol of diesel fuel, also typically found in tables and denoted as \(\Delta S^\circ_f\). For diesel fuel: $$\Delta S^\circ_f(\text{C}_{12}\text{H}_{26}) = -713.7\ \text{J/mol K}$$ Note that the value of entropy is negative because the diesel fuel is in a more ordered state compared to the products it forms when combusted.

Calculate Gibbs free energy change, \(\Delta G\)

Now we can calculate the Gibbs free energy change using the formula introduced earlier: $$\Delta G = \Delta H - T \Delta S$$ Plugging in the values, we get: $$\Delta G = -350.1\ \text{kJ/mol} - (298.15\ \text{K})\times(-713.7\ \text{J/mol K})$$ $$\Delta G = -350100\ \text{J/mol} + (298.15\ \text{K})\times(713.7\ \text{J/mol K})$$ $$\Delta G = -350100\ \text{J/mol} + 212823.205\ \text{J/mol}$$ $$\Delta G = -137276.795\ \text{J/mol}$$

Determine work potential for 1 lbmol of diesel fuel

Finally, we can determine the work potential of 1 lbmol of diesel fuel using the Gibbs free energy change: $$\text{Work potential} = -\Delta G = 137276.795\ \text{J/mol}$$ The work potential of 1 lbmol of diesel fuel at given conditions is therefore 137276.795 J/mol.

Key concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a measure of the total heat content of a system and reflects the energy exchanged with the surroundings at constant pressure. It's a vital concept in thermodynamics because it helps predict how much energy a substance will absorb or release during a chemical reaction.

In the context of diesel fuel, the enthalpy change for the combustion of one pound-mole (lbmol) of diesel, with the chemical formula \( \text{C}_{12}\text{H}_{26} \), is significantly negative. This indicates that the combustion reaction is exothermic; it releases heat to the surroundings. The negative value, \( -350.1 \text{kJ/mol} \), represents a large release of energy and coincides with the intuitive understanding that fuels burn to release energy, providing work capacity for engines and turbines. Accurate enthalpy values are essential for engineers to design efficient energy systems.

Entropy Change

Entropy change, denoted as \( \Delta S \), provides insight into the disorder or randomness of a system's particles. It's a fundamental concept in thermodynamics linked to the second law, which states that the total entropy of an isolated system can never decrease over time.

For our diesel fuel \( \text{C}_{12}\text{H}_{26} \), the entropy change is reported as \( -713.7 \text{J/mol K} \). The negative sign indicates that the initial state of the diesel fuel has more order compared to the final state of the products after combustion. A more ordered initial state implies that upon combustion, the diesel becomes more disordered, increasing the entropy of the system. Recognizing the role of entropy in reactions enables the understanding of not only the energy aspects but also the direction and spontaneity of the processes.

Gibbs Free Energy

Gibbs free energy change, denoted as \( \Delta G \), combines the concepts of enthalpy and entropy into a single value that determines the spontaneity of a reaction at constant temperature and pressure. The formula to calculate it is \( \Delta G = \Delta H - T \Delta S \), where \( T \) is the temperature in Kelvin.

After calculating, we found that the Gibbs free energy change for the combustion of diesel is \( -137276.795 \text{J/mol} \). A negative \( \Delta G \) implies that the reaction is spontaneous under the given conditions. This information is crucial for designing processes that harness the energy change of a reaction effectively. In practical terms, a more negative Gibbs free energy change suggests that a pound-mole of diesel fuel has significant potential to do work, such as powering an engine or generating electricity. Understanding Gibbs free energy not only aids in predicting reaction spontaneity but also provides insights for engineers and scientists when assessing the maximal useful work that can be derived from chemical reactions.