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

Jul 5, 2017

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

#### Explanation:

The $\text{enthalpy of atomization}$ 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. $\text{atoms}$, from one mole of gaseous $\text{dihydrogen MOLECULES}$ (which of course is the standard state of ${H}_{2}$) , given the prior definition,

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

Agreed?