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

1 Answer
anor277
Jul 5, 2017

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?

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