NEET2008ChemistryChemical EquilibriumActual
Bond dissociation enthalpy of H ₂, Cl ₂ and HCl Enthalpy of formation of HCl is
Options
- A93 ~kJ ~mol ⁻¹
- B-245 ~kJ ~mol ⁻¹
- C-93 ~kJ ~mol ⁻¹
- D245 ~kJ ~mol ⁻¹
Correct answer
C. -93 ~kJ ~mol ⁻¹
Step-by-step solution
Key Idea: H_ reaction = Bond energy of reactant - Bond energy of product array ll Here, & H _ H - H =434 ~kJ ~mol ⁻¹ & H _ Cl - Cl =242 ~kJ ~mol ⁻¹ & H _ H - Cl =431 ~kJ ~mol ⁻¹ & 1 2 H ₂+ 1 2 Cl ₂ HCl array aligned H_ reaction & = 1 2 H_ H - H + 1 2 H _ Cl - Cl - H _ H - Cl & = 1 2 434+ 1 2 242-431 & =217+121-431 & =-93 ~kJ ~mol ⁻¹ aligned