The standard heat of formation (∆H ) of ethane (in kJ/mol), if the heat of combustion of ethane, hyd — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Write the combustion equations** - Ethane: C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l); ∆H_c = -1560 kJ/mol - Carbon: C(s) + O₂(g) → CO₂(g); ∆H_c = -393.5 kJ/mol - Hydrogen: H₂(g) + 1/2 O₂(g) → H₂O(l); ∆H_c = -286 kJ/mol **Step 2: Apply Hess's Law** The formation of ethane from elements is: 2C(s) + 3H₂(g) → C₂H₆(g); ∆H_f = ? **Step 3: Construct the target equation using Hess's Law** Multiply the carbon combustion equation by 2: 2C(s) + 2O₂(g) → 2CO₂(g); ∆H = 2(-393.5) = -787 kJ Multiply the hydrogen combustion equation by 3: 3H₂(g) + 3/2 O₂(g) → 3H₂O(l); ∆H = 3(-286) = -858 kJ Reverse the ethane combustion equation: 2CO₂(g) + 3H₂O(l) → C₂H₆(g) + 7/2 O₂(g); ∆H = +1560 kJ **Step 4: Add all three equations** ∆H_f = -787 + (-858) + 1560 ∆H_f = -1645 + 1560 ∆H_f = -85 kJ/mol **Step 5: Correct calculation** ∆H_f = 2(-393.5) + 3(-286) - (-1560) ∆H_f = -787 - 858 + 1