Enthalpy of Combustion from mean bond enthalpies

So first you have to form an equation for the combustion of ethane:

2(C2H6) + 7(O2) = 4(CO2) + 6(H2O)

Then you have to break all the bonds in ethane and oxygen:

12 x C-H (which requires 12 x 413 kJ mol-1)
2 x C-C (which requires 2 x 348 kJ mol-1)
7 x O=O (which requires 7 x 498 kJ mol-1)

which requires in total (12x413)+(2x348)+(7x498)= 9138 kJmol-1
so you have used up 9138 kJ mol-1


Now you have to work out how much energy is released when bonds are made making CO2 + H2O:

O=C=O H-O-H

8 x C=O (8 x 748)
12 x O-H (12 x 468)

so in total, making bonds has released 11600 kJmol-1

so 11600-9138= 2462 kJ mol-1

So 2462 kJ mol-1 has been released, so the reaction is exothermic, so the enthalpy of combustion is negative-

so (delta)HC = -2462 kJ mol-1

Hope this helps. You should draw a Hess' law diagram on your paper- it will make it easier to calculate and will allow the examiner to see that you have calculated it correctly.

Good luck!!

My answer is correct, as long as the values for the mean bond enthalpies you quoted are right (I didn't check them).

Wouldn't it be C2H6 + 3 and a half O2 -----> 2(CO2) + 3H2O??

even when I tried this I got a different answer to the book.....ahhhh

even when I tried this I got a different answer to the book.....ahhhh

So first you have to form an equation for the combustion of ethane:

2(C2H6) + 7(O2) = 4(CO2) + 6(H2O)

Then you have to break all the bonds in ethane and oxygen:

12 x C-H (which requires 12 x 413 kJ mol-1)
2 x C-C (which requires 2 x 348 kJ mol-1)
7 x O=O (which requires 7 x 498 kJ mol-1)

which requires in total (12x413)+(2x348)+(7x498)= 9138 kJmol-1
so you have used up 9138 kJ mol-1


Now you have to work out how much energy is released when bonds are made making CO2 + H2O:

O=C=O H-O-H

8 x C=O (8 x 748)
12 x O-H (12 x 468)

so in total, making bonds has released 11600 kJmol-1

so 11600-9138= 2462 kJ mol-1

So 2462 kJ mol-1 has been released, so the reaction is exothermic, so the enthalpy of combustion is negative-

so (delta)HC = -2462 kJ mol-1

Hope this helps. You should draw a Hess' law diagram on your paper- it will make it easier to calculate and will allow the examiner to see that you have calculated it correctly.

Good luck!!

I just did this and got the answer that the book gave.

C-C = 348
(6) C-H = 2478
(3,1/2) O=O = 1734
TOTAL = 4569

(4) C=O = 2972
(6) H-O = 2778
TOTAL = 5750

4569 - 5750 = -1181

It;s not 4569 its 4560, you totalled it up wrong

so the answer is then -1190

This is the final answer.

C-C = 348
(6) C-H = 2478
(3,1/2) O=O = 1743
TOTAL = 4569

(4) C=O = 2972
(6) H-O = 2778
TOTAL = 5750

4569 - 5750 = -1181

This. I forgot to half my answer.

Anyone tempted to workout combustion of ethanol?

What should the answer be and I'll post if I got the same.