MHT CET 2026 Chemistry question
The bond dissociation enthalpy of , and are 434, 242 and 431 kJ respectively. Calculate the enthalpy of formation of HCl
Answer: (A)
Solution
Formation reaction: . \sum\sum\tfrac{1}{2}\tfrac{1}{2}$ × 242) - 431 = (217 + 121) - 431 = 338 - 431
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