Calculate the enthalpy change for the following reaction, using given bond energy (kJ/mol) (C-H = 414 , H-O = 463 , H-Cl = 431 , C-Cl = 326 and C-O = 335 ) CH ₃ OH _ (g) + HCl _ (g) CH ₃ Cl _ (g) + H ₂ O _ (g)

Step-by-step solution

The enthalpy of reaction is given by: H = BE(Reactants) - BE(Products) H = [3 BE(C-H) + BE(C-O) + BE(O-H) + BE(H-Cl) ] - [3 BE(C-H) + BE(C-Cl) + 2 BE(O-H) ] Canceling the common bond energies on both sides: H = BE(C-O) + BE(H-Cl) - BE(C-Cl) - BE(O-H) Substituting the given values: H = 335 + 431 - 326 - 463 H = 766 - 789 = -23 kJ mol ⁻¹ Answer: -23 kJmol ⁻¹