AIPMT MAIN2008Chemistry-Thermochemistry
AIPMT MAIN 2008 Chemistry Enthalpy of Formation MCQ Question
Type: MCQ-numerical-Medium-Class 11
Bond dissociation enthalpy of H₂, Cl₂ and HCl are 434, 242 and 431 kJmol⁻¹ respectively. Enthalpy of formation of HCl is
A
245 kJmol⁻¹
B
93 kJmol⁻¹
C
-245 kJmol⁻¹
D
-93 kJmol⁻¹
Correct Answer
Option D
Detailed Explanation
The enthalpy change for the formation of HCl can be calculated using the bond dissociation energies. The reaction is: H₂ + Cl₂ → 2HCl. The enthalpy change is calculated as: ΔH = (bond energy of H₂ + bond energy of Cl₂) - (2 × bond energy of HCl) = (434 + 242) - (2 × 431) = -93 kJmol⁻¹.
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