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The bond dissociation energy of gaseous H_(2),Cl_(2) and HCl are 104, 58 and 103 "kcal/mole" respectively. Calculate the enthalpy of formation of HCl gas. |
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Answer» SOLUTION :For the reaction `H_(2)(g)+Cl_(2)(g) to 2HCl(g)` For reactants Bond energy OF1 mole of H-H bond `=104` kcal Bond energy of 1 mole of Cl-Cl bond `=58` kcal For products : Energy of formation of 2 MOLES of H-Cl bond `=-2xx103` kcal Thus `DeltaH` of the above reaction `=104+58-206=-44` kcal SINCE, for `H_(2)+Cl_(2) to 2HCl` , `DeltaH=-44kcal` then for, `(1)/(2)H_(2)+(1)/(2)Cl_(2) to HCl` , `DeltaH=-22kcal` |
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