Reaction,H_(2)(g)+I_(2)(g)rarr2HI(g)DeltaH=-12.40 kcal.According to th – InterviewSolution

1.	Reaction,H_(2)(g)+I_(2)(g)rarr2HI(g)DeltaH=-12.40 kcal.According to this, the heat of formation of HI will be
Answer» 12.4 KCAL `-12.4` kcal `-6.20` kcal 6.20 kcal Solution :HEAT of formation is for 1 mole. Hence `DeltaH_(F)^(@)(HI)=-12.40//2=-6.20 kcal`.

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