chemistry enthalpy related question

Question

If bond energy of   C-H bond is 99 k cal and given that

1.  2C (S)+  2H₂(g) →    C₂H₄ (g) ,  ΔH = 12.5 k cal
2.  2C (S)+  2C (g) , ΔH (g) ,      ΔH =  2 x 170.5  k cal
3.  2H2(g)   →   4H (g)     ,               ΔH = 4 x 52.1  k cal

 

   THE C=C  bond energy  is ? 

1.     141.7 kcal
2.    537.7  kcal
3.   396.0 kcal
4.    208.4 kcal
    

   i have used the formula ΔH = ∑ bonds broken -  ∑ bonds formed 

   what i got was 29.6 k cal ....which is no near to the answers above

    

  

 

 

Answer 1

The reaction involving the formation of ethylene (C₂H₄) from carbon and hydrogen involves the following steps:

1) convert solid carbon to gaseous carbon atoms:  ΔH_vap

2) break H-H bond to form hydrogen atoms:    ΔH bond dissociation  (bond dissociation energy)

3) combine hydrogen atoms with carbon atoms to form ethylene:  bond energy for C=C and bond energy for 4 C-H bonds

The overall heat of reaction can be written as a function of the parts above:

ΔH_rxn = 2ΔH_vap for C +2BDE_H2 - BE_C=C - 4BE_C-H     note:  BDE for H₂ can be thought of as 4 times the bond energy of H-H

12.5 kcal = 2(170.5 kcal) +2(104 kcal) - BE_C=C - 4(99 kcal)  -->

BE_C=C = -12.5 +341 +208 - 396 = 140.5 kcal/mol

#1 is the answer

note that bond energy enthalpies are negative (energy given off as bonds form)