Enthalpy of combustion of ethane

The combustion of ethane goes:

C2H6 (g) + 3.502 (g) ------> 2C02 (g) + 3H20 (l)

Ethane= -83.6 carbon dioxide= -393 water= -285.5

Since combustion = reactants - products the answer should be -154.1 kj mol-1

But i did some research and the answer is -1560 kjmol-1? I'm so confused

Reply 1

TheSnail

Original post by CrocodileCrunch

Have you encountered Hess' cycles?

/\f H2 = 0 (as it's an element)
/\f O2 = 0
/\f H20 = -285.5
/\f CO2 = -393
/\f C2H6 = -83.6

C2H6 (g) + 3.502 (g) ------> 2C02 (g) + 3H20 (l)
/|\ | | |
| -83.6 | 3.5x0 | 2x(-393) | 3x(-285.5)
| | | |

imagine all those arrows point up (as it's formation)

right so, /\c = -(-83.6) + [(2x-393)+(3x-285.5)]
(it's - for the -83.6 as you are going against the arrow then + for the products as you are going with the arrow)

= -1558.9

which I guess is roughly the same as your researched answer

EDIT - i'll fix this in a sec