Question

Using bond energy data calculate heat of formation of isoprene

Answer 1

Considering bonds broken and formed ONLY, what is the enthalpy change for the following reaction:

C40H82 ---> C16H34 + 2C12H24

Solution:

Comment: this is a bit of a trick question. Why? I'll let that evolve during the discussion.

1) Hess' Law for bond enthalpies is:

ΔH = Σ Ereactant bonds broken minus Σ Eproduct bonds broken

2) Let's consider the total bonds in one molecule of the reactant, C40H82:

C−C ⇒ 39 
C−H ⇒ 82

3) Let's consider the total bonds in the three product molecules:

one C16H34:

C−C ⇒ 15 
C−H ⇒ 34two C12H24:C−C ⇒ 24 
C−H ⇒ 48total:C−C ⇒ 15 + 24 = 39 
C−H ⇒ 34 + 48 = 82Comment: knowing the C−C bonds in the C12H24 molecule is, to the ChemTeam, the key. Note that its formula is of the form CnH2n. That indicates either (a) one double bond between two carbons or (b) a cyclic structure with only single bonds between carbons. I took the second assumption in solving this problem, because this is the assumption that makes this a trick question.

4) Calculate the ΔH:

Since there are an equal number of C−C and C−H bonds on each side, the ΔH equals zero.

Comment: suppose you were to assume that double bonds were to form. In that case, you would have this:

reactant

C−C ⇒ 39 
C−H ⇒ 82productC−C ⇒ 15 + 20 = 35 
C=C ⇒ 2 
C−H ⇒ 34 + 48 = 82Net result:reactant ⇒ four C−C broken 
product ⇒ two C=C made