[Big-Daddy]

Estimate the B-B single bond dissociation enthalpy in B₂Cl₄(g) using the following information:

Bond, Bond Dissociation Enthalpy (kJ/mol)
B–Cl: 443
Cl–Cl: 242

Compound, ΔH_f° (kJ/mol)
BCl3(g): –403
B2Cl4(g): –489

I wrote out the equations for which we have enthalpy changes, Cl2(g) -> 2 Cl (g), 2 B (s) + 2 Cl2 (g) -> B2Cl4 (g), etc. The only combination needed was the bond dissociation enthalpy of Cl-Cl and enthalpy of formation of B2Cl4. Then this was the enthalpy change for B2Cl4 -> 2 B + 4 Cl which involves breaking 4 B-Cl bonds and 1 B-B bond so I could work out the B-B bond strength from there, without needing BCl3, or so I thought - the answer I got was -799 kJ/mol, completely wrong clearly, whereas the actual answer was apparently +327 kJ/mol. How was this answer arrived at?

[Big-Daddy]

Can I have some clarification with this? I found it odd to be fooled by a Hess' law problem, particularly when it seems to be a non-trivial mistake I made.

[blaisem]

I am not sure why you can't go from the ground up to B₂Cl₄ as you did.  Is that a rule that the heat of formation of a molecule is just the sum of all its individual bonds?

In any case, this route doesn't get you to the right answer here.  If, however, you assume that whatever difference is present that is making your numbers wrong is also contained in BCl₃, then you can start from BCl₃ and go to B₂Cl₄ and assume the difference will be carried along in the process.

Here's the overall reaction, which might help you.

2BCl₃  B₂Cl₄ + Cl₂

Starting with the reactant, consider what bonds need to be broken for it to dimerize.  The final difference, as you would expect, is the heat of B-B.

[Big-Daddy]

I've seen the difficulty and managed to solve the problem now.

The bond dissociation enthalpies only apply to the bonds in the gaseous phase molecules, whereas the enthalpies of formation use B (s) rather than B (g). Therein is the problem.