MHT CET202611 April 2026Evening ShiftChemistryThermodynamics (C)Actual
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.
Options
- A-93 kJ mol ⁻¹
- B245 kJ mol ⁻¹
- C93 kJ mol ⁻¹
- D-245 kJ mol ⁻¹
Correct 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