Question

divide eq 2 by 2:

2H(g)........> H2(g) delta H= -104.2Kcal
H(g) ........> 1/2 H2(g) delta H= -72.1Kcal

Now reverse the reaction:
1/2H2(g).......>H(g) delta H=+72.1Kcal
subtract this eq from eq 1
1/2H(g)+1/2I2(g).....>HI(g) delta H=-1.1kcal
+-1/2H2(g)......>+-H(g) delta H=+-72.1Kcal
= 1/2I2(g).....>HI(g)- H(g)
on rearranging:
H(g)+I(g).........>HI(g) delta H=-73.2Kcal

Answer 1

There is no question.