ΔHhyd??= ΔHlatt??+ ΔHsol?
ΔHlatt??= ΔHhyd??- ΔHsol?
Answer
Step 1:?Draw the energy cycle of KCl
ΔHhyd??= (ΔHlatt?[KCl]) + (ΔHsol?[KCl])
(ΔHhyd?[K+]) + (ΔHhyd?[Cl-])?? =? (ΔHlatt?[KCl]) + (ΔHsol?[KCl])
(ΔHhyd?[Cl-])?? = (ΔHlatt?[KCl]) + (ΔHsol?[KCl]) - (ΔHhyd?[K+])
ΔHhyd??[Cl-]?? = (-711) + (+26) - (-322) = -363 kJ mol-1
Answer
ΔHhyd??= (ΔHlatt?[MgCl2]) + (ΔHsol??[MgCl2])
(ΔHhyd?[Mg2+]) + (2ΔHhyd??[Cl-])?? = (ΔHlatt??[MgCl2]) + (ΔHsol??[MgCl2])
(ΔHhyd?[Mg2+])?? = (ΔHlatt?[MgCl2]) + (ΔHsol?[MgCl2]) - (2ΔHhyd?[Cl-])
ΔHhyd?[Mg2+]? = (-2592) + (-55) - (2 x -363) = -1921 kJ mol-1
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