# Facing a problem in enthalpy calculation

If 1 mole of gaseous carbon atoms are converted into diamonds, calculate enthalpy change of process given that bond enthalpy of $$\ce{C-C}$$ bond is $$\pu{400 kJ mol-1}.$$

According to me, the answer should be that since C(diamond) has 4 bonds and C(gas) doesn't have any bonds, 4 bonds are being formed so enthalpy change should be

$$ΔH = 4\cdot (\pu{-400 kJ mol-1})$$

But in the correct solution it says each bond is counted twice, so answer will be

$$ΔH = \frac{4\cdot (\pu{-400 kJ mol-1})}{2}$$

I don't get why are we doing this. In other questions we don't consider the double counting of bonds. For example, another question I did had 1 mole phosphorus gas converting into $$\ce{P4}$$ molecule, and even though $$\ce{P4}$$ has 6 bonds, we did not consider what we did with diamond, so what is the reason for that?

• Consider the simpler question: If 1 mole of hydrogen atoms (H) form dihydrogen (H-H), how many bonds are formed? Using your logic (each hydrogen makes one bond), there should be one mole bonds. But only 1/2 mole of dihydrogen was made. That would be 1 mole of single bonds in 1/2 mole of dihydrogen, not possible. – Karsten Theis Jul 22 '19 at 13:36

$$\ce{P4}$$ forms a molecular solid, whereas diamond is a covalent-network solid (see eg definitions here).
When estimating the enthalpy change, you consider only covalent bonds, not weak intermolecular forces. Therefore, for $$\ce{P4}$$ you consider only the 6 P-P bonds, whereas in diamond you consider covalent bonds to nearest neighbours, and correct for the fact that when you rupture a bond to a neighbour, that neighbour also loses a bond, therefore the division by 2.
• @bscripts Apparently there are exotic phases of nitrogen, see en.wikipedia.org/wiki/Solid_nitrogen. I would not call solid phases of molecular nitrogen $\ce{N2}$ covalent network solids. – Buck Thorn Jul 22 '19 at 18:56