Question
Calculate the enthalpy change of the reaction, H _2 (g) + Cl _2 (g) 2 HCl(g) if bond energies (kJ mol^ -1 ): H–H = 436, Cl–Cl = 242, H–Cl = 431
MHT CET
Chemistry
Thermodynamics (C)
2026
MHT CET 2026 (13 April Shift 1)
MHT CET Chemistry Question (2026) — Solution
Options
- A. -184 kJ/mol
- B. -246 kJ/mol
- C. -242 kJ/mol
- D. -431 kJ/mol
Answer
A. -184 kJ/mol
Step-by-step solution
The enthalpy change of the reaction is given by the difference between the sum of bond energies of reactants and the sum of bond energies of products. H = BE(Reactants) - BE(Products) For the given reaction: H _2 (g) + Cl _2 (g) 2 HCl(g) H = [ BE(H-H) + BE(Cl-Cl) ] - [2 BE(H-Cl) ] Substituting the given values: H = (436 + 242) - (2 431) H = 678 - 862 H = -184 kJ mol ^ -1 Answer: -184 kJ/mol
Related: Chemistry — Thermodynamics (C) · All PYQ Banks