Question
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)
MHT CET
Chemistry
Thermodynamics (C)
2026
MHT CET 2026 (11 April Shift 1)
MHT CET Chemistry Question (2026) — Solution
Options
- A. -23 kJmol ^ -1
- B. -42 kJmol ^ -1
- C. -59 kJmol ^ -1
- D. -51 kJmol ^ -1
Answer
A. -23 kJmol ^ -1
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
Related: Chemistry — Thermodynamics (C) · All PYQ Banks