Let's break down the **step-by-step solution** for calculating the chemical exergy correction factor (φ) for each fuel using the provided formula: \[ \phi = 1.041 + .1728 \frac{\beta}{\alpha} + .0432 \frac{\delta}{\alpha} + .2169 \frac{\gamma}{\alpha} \left( 1 - 2.062 \frac{\beta}{\alpha} \right) \] where: - \(\alpha\) = Number of Carbon atoms (C) - \(\beta\) = Number of Hydrogen atoms (H) - \(\gamma\) = Number of Nitrogen atoms (N) - \(\delta\) = Number of Oxygen atoms (O) We will fill out the table with the chemical structure and the exergy correction factor for each fuel. --- ## **Step 1: List Molecular Formulas** | Fuel | Formula | α | β | γ | δ | |--------------|----------|---|---|---|---| | Methane | CH₄ | 1 | 4 | | | | Propane | C₃H₈ | 3 | 8 | | | | Butane | C₄H₁₀ | 4 | 10| | | | Octane | C₈H₁₈ | 8 | 18| | | | Benzene | C₆H₆ | 6 | 6 | | | | Naphthalene | C₁₀H₈ |10 | 8 | | | | Pyridine | C₅H₅N | 5 | 5 | 1 | | | Pyrrole | C₄H₅N | 4 | 5 | 1 | | | Imidazole | C₃H₄N₂ | 3 | 4 | 2 | | | Furan | C₄H₄O | 4 | 4 | | 1 | | Dioxin | C₄H₄O₂ | 4 | 4 | | 2 | --- ## **Step 2: Calculate φ for Each Fuel** Let's plug the values into the formula: ### **Example Calculation: Methane (CH₄)** - α = 1, β = 4, γ = , δ = \[ \phi_{\text{CH}_4} = 1.041 + .1728 \times \frac{4}{1} + .0432 \times \frac{}{1} + .2169 \times \frac{}{1} \times (1 - 2.062 \times \frac{4}{1}) \] \[ = 1.041 + .6912 + + = 1.7322 \] --- Now, calculate for each: | Fuel | Formula | α | β | γ | δ | φ (rounded 3 dp) | |-------------|------------|---|---|---|---|------------------| | **Methane** | CH₄ | 1 | 4 | | | 1.732 | | **Propane** | C₃H₈ | 3 | 8 | | | 1.500 | | **Butane** | C₄H₁₀ | 4 | 10| | | 1.472 | | **Octane** | C₈H₁₈ | 8 | 18| | | 1.439 | | **Benzene** | C₆H₆ | 6 | 6 | | | 1.213 | | **Naphthalene**| C₁₀H₈ |10 | 8 | | | 1.125 | | **Pyridine**| C₅H₅N | 5 | 5 | 1 | | 1.260 | | **Pyrrole** | C₄H₅N | 4 | 5 | 1 | | 1.295 | | **Imidazole**| C₃H₄N₂ | 3 | 4 | 2 | | 1.388 | | **Furan** | C₄H₄O | 4 | 4 | | 1 | 1.224 | | **Dioxin** | C₄H₄O₂ | 4 | 4 | | 2 | 1.267 | --- ## **Step 3: Table Summary** | Fuel | Structure | φ (chemical exergy) | |--------------|------------|---------------------| | Methane | CH₄ | 1.732 | | Propane | C₃H₈ | 1.500 | | Butane | C₄H₁₀ | 1.472 | | Octane | C₈H₁₈ | 1.439 | | Benzene | C₆H₆ | 1.213 | | Naphthalene | C₁₀H₈ | 1.125 | | Pyridine | C₅H₅N | 1.260 | | Pyrrole | C₄H₅N | 1.295 | | Imidazole | C₃H₄N₂ | 1.388 | | Furan | C₄H₄O | 1.224 | | Dioxin | C₄H₄O₂ | 1.267 | --- ## **Step 4: Answer the Questions** ### **a. Linear vs Aromatic Hydrocarbons** **Trend:** Linear hydrocarbons (alkanes) have higher chemical exergy values (\(\phi\)) than aromatic hydrocarbons. This is because the higher hydrogen-to-carbon ratio in alkanes increases \(\phi\), while aromatics have more stable structures and lower hydrogen content per carbon, decreasing \(\phi\). **Relationship:** - **Higher H/C ratio (Linear) → Higher \(\phi\)** - **Greater Aromaticity → Lower \(\phi\)** --- ### **b. Aromatics vs Heteroarenes** **Trend:** Heteroarenes (aromatic rings with N or O) generally have slightly higher \(\phi\) than pure aromatics, but lower than alkanes. Heteroatoms (N, O) increase the denominator in the formula, slightly raising the chemical exergy compared to pure aromatics. **Relationship:** - **Introducing N or O into aromatic rings slightly raises \(\phi\), but they remain below linear hydrocarbons.** - **The presence of heteroatoms counteracts some of the aromatic stability, increasing exergy.** --- ## **Summary Table** | Fuel | Formula | φ (chemical exergy) | |--------------|-----------|---------------------| | Methane | CH₄ | 1.732 | | Propane | C₃H₈ | 1.500 | | Butane | C₄H₁₀ | 1.472 | | Octane | C₈H₁₈ | 1.439 | | Benzene | C₆H₆ | 1.213 | | Naphthalene | C₁₀H₈ | 1.125 | | Pyridine | C₅H₅N | 1.260 | | Pyrrole | C₄H₅N | 1.295 | | Imidazole | C₃H₄N₂ | 1.388 | | Furan | C₄H₄O | 1.224 | | Dioxin | C₄H₄O₂ | 1.267 | --- If you need the calculations for each specific molecule, let me know!