# Question 460c8

Jul 23, 2017

$\frac{352 k J}{m o l}$

The tabulated/measured value deviates from this by 10% or so, but for the purposes of this question, that would be the answer you're probably looking for.

#### Explanation:

Firstly, ${H}_{2}$, and ${N}_{2}$ actually have a bond enthalpy of the foregoing enthalpies with the opposite sign, not an enthalpy of formation. Thus, we aren't particularly concerned with enthalpy of formation here (I may be wrong!).

3/2H_2(g)+1/2N_2(g) to NH_3" "DeltaH_f^°[NH_3] = (-46.0kJ)/(mol)

Anyways, we have the equation of one mole of the molecule of interest above, and now we must apply the concept of bond enthalpies which is the energy it takes to form/break a bond in a molecule. Moreover, we're looking to solve the equation relating bonds being broken and formed in the foregoing reaction.

DeltaH_(rxn)^° = SigmaDeltaH_R - SigmaDeltaH_R

That's a fancy way of saying since it takes energy to break bonds and energy is emitted when bonds are formed, we will realize the net enthalpy of the reaction in that way.

$\frac{- 46.0 k J}{m o l} = \left[\frac{3}{2} \cdot \frac{436.0 k J}{m o l} + \frac{1}{2} \cdot \frac{712.0 k J}{m o l}\right] - N {H}_{3}$

Doing some algebra, and

$N {H}_{3} = \frac{1056.0 k J}{m o l}$

And, if you'd like to find the average bond enthalpy of an N—H# bond in the molecule of interest, you'd simply divide our answer by 3 to realize (since there are three of these bonds in the structure of $N {H}_{3}$ molecules):

$\frac{352 k J}{m o l}$