Given #H_2+Delta rarr 2dotH# #;DeltaH=104*kcal*mol^-1#, what is #DeltaH_f^@# for #dotH(g)#?

1 Answer
Jul 5, 2017

Answer:

#DeltaH_"atomization"^@("Hydrogen")=52*kcal*mol^-1#.

Explanation:

The #"enthalpy of atomization"# is the energy required to produce one mole of ATOMS from an element in its standard state under standard conditions.......

We were given data for the reaction........

#H_2(g) +Deltararr2dotH; DeltaH^@=104*kcals*mol^-1 #

Since this reaction results in the formation of 2 mol of hydrogen radicals, i.e. #"atoms"#, from one mole of gaseous #"dihydrogen MOLECULES"# (which of course is the standard state of #H_2#) , given the prior definition,

#DeltaH_"atomization"^@("Hydrogen")=52*kcal*mol^-1#.

Agreed?