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The bond dissociation energy of gaseous `H_(2),Cl_(2)` and `HCl` are `104,58` and `103 kcal mol^(-1)` respecitvely. Calculate the enthalpy of formation for `HCl` gas.A. `-22` kcalB. `+22` kcalC. `+184` kcalD. `-184` kcal |
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Answer» Correct Answer - A `(1)/(2)H_(2)+(1)/(2)Cl_(2) rarr HCl` `DeltaH=(1)/(2)xx104+(1)/(2)xx58-103=-"22 kcal"` |
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