The ionization enthalpy of Na formation from Na is 495.8kJ mol , while the electron gain enthalpy of — Chemical Bonding Chemistry Question
Question
The ionization enthalpy of Na formation from Na is 495.8kJ mol , while the electron gain enthalpy of Br is -325.0 kJ mol . Given the lattice enthalpy of NaBr is -728.4 kJ mol . The energy for the formation of NaBr ionic solid is (-) _______ × 10 kJ mol . + (g) -1 -1 -1 -1 -1
💡 Solution & Explanation
**Step 1: Identify the Born-Landé cycle components** The formation of NaBr solid from gaseous atoms follows: Na(g) + Br(g) → NaBr(s) The energy required equals: - Ionization enthalpy of Na: +495.8 kJ/mol - Electron gain enthalpy of Br: −325.0 kJ/mol - Lattice enthalpy of NaBr: −728.4 kJ/mol **Step 2: Apply the thermochemical relationship** For ionic solid formation from gaseous atoms: ΔH_formation = IE(Na) + EGE(Br) + Lattice enthalpy **Step 3: Substitute the values** ΔH_formation = 495.8 + (−325.0) + (−728.4) ΔH_formation = 495.8 − 325.0 − 728.4 **Step 4: Calculate step-by-step** 495.8 − 325.0 = 170.8 kJ/mol 170.8 − 728.4 = −557.6 kJ/mol **Step 5: Express in the required format** The question asks for the answer in (−) _____ × 10¹ kJ/mol −557.6 = (−) 5576.0 × 10¹ kJ/mol **Therefore, the answer is 5576.00.**