For the reaction the reaction enthalpy (Round off to the Nearest Integer). [Given: Bond enthalpies i — Thermodynamics and Thermochemistry Chemistry Question
Question
For the reaction the reaction enthalpy (Round off to the Nearest Integer). [Given: Bond enthalpies in kJ mol : C – C : 347, C = C : 611; C – H : 414, H – H : 436] –1
💡 Solution & Explanation
# Solution: Reaction Enthalpy Calculation **Step 1: Identify the reaction** The reaction is: C₂H₄ + H₂ → C₂H₆ (Hydrogenation of ethene to ethane) **Step 2: Apply bond enthalpy formula** ΔH = Σ(bonds broken) - Σ(bonds formed) **Step 3: Calculate bonds broken (reactants)** C₂H₄: 1 C=C bond + 4 C-H bonds = 611 + 4(414) = 611 + 1656 = 2267 kJ/mol H₂: 1 H-H bond = 436 kJ/mol **Total bonds broken = 2267 + 436 = 2703 kJ/mol** **Step 4: Calculate bonds formed (products)** C₂H₆: 1 C-C bond + 6 C-H bonds = 347 + 6(414) = 347 + 2484 = 2831 kJ/mol **Total bonds formed = 2831 kJ/mol** **Step 5: Calculate reaction enthalpy** ΔH = 2703 - 2831 = -128 kJ/mol The negative sign indicates an exothermic reaction. The magnitude is 128 kJ/mol. Therefore, the answer is 128.00.