I. H rxn for the reaction below was determined using bond dissociation energies
ID: 1068594 • Letter: I
Question
I. Hrxn for the reaction below was determined using bond dissociation energies (BDEs) to be 57 kJ. Using the table of BDE below, find the bond energy for HBr.
Bond C-C C=C C-H C-Br
BDE (kJ/mol) 348 614 413 276
II. Use heats of formation to find the standard heat of reaction (H°rxn) for the reaction shown above. C2H4 and HBr are both gases and H°f for CH3CH2Br () is 85.3 kJ/mol. Use T16-4-1 to find any additional values of H°f needed.
*** Answers should come out to:
I. BDE(H—Br) = 366 kJ
II. H°rxn= 101.16 kJ
Explanation / Answer
SO we have the folowing reaction
C2H4 + HBr ----------> C2H5Br
Now part I , BDE of reaction = -57 kJ
Now Bond energy of reaction = Bond energy of products - Bond energy of reactants
BDEreaction = BDE p - BDEr = BDE(C2H4) +BDE (HBr)-BDE(C2H5Br)
Now , look at bonds in each
C2H5Br has 1 C-C , 5 C-H and 1 C-Br bond
C2H4 has1 C=C and 4 C-H bond
HBr has 1 H-Br bond
Now ,BDEreaction = BDE r - BDEp =BDE(C2H4) +BDE (HBr)- BDE(C2H5Br)
= ( 1* H-Br)+(1 *C=C + 4* C-H ) -(1 * C-C + 5* C-H + 1* C-Br)
-57 = BDE(H-Br) + (614 +4 * 413) -(348 +5*413 + 276)
BDE(H-Br) = 423-57 = 366 kJ
Part 2 Heat of formation
We define the heat of formation as the ethalpy for formation of 1 mole of a substance at standard state from its pure elements
Here we know Heatf(rxn) = Hf(prod) -Hf(reactant) = H(C2H5Br) - Hf(C2H4) -Hf(HBr) = 85.3 -52.4 - (-36.29)
= -101.41 kJ