Question
The bond dissociation enthalpy of H _2, Cl _2 and HCl are 434, 242 and 431 kJ mol ^ -1 respectively. Calculate the enthalpy of formation of HCl.
MHT CET
Chemistry
Thermodynamics (C)
2026
MHT CET 2026 (11 April Shift 2)
MHT CET Chemistry Question (2026) — Solution
Options
- A. -93 kJ mol ^ -1
- B. 245 kJ mol ^ -1
- C. 93 kJ mol ^ -1
- D. -245 kJ mol ^ -1
Answer
A. -93 kJ mol ^ -1
Step-by-step solution
The reaction for the formation of HCl is: 1 2 H _2(g) + 1 2 Cl _2(g) HCl (g) The enthalpy of formation is given by: _f H = BE ( reactants ) - BE ( products ) _f H = [ 1 2 BE ( H _2) + 1 2 BE ( Cl _2) ] - BE ( HCl ) Substituting the given values: _f H = [ 1 2 (434) + 1 2 (242) ] - 431 _f H = (217 + 121) - 431 _f H = 338 - 431 = -93 kJ mol ^ -1 Answer: -93 kJ mol ^ -1
Related: Chemistry — Thermodynamics (C) · All PYQ Banks