Question

2C8H18+25O2 ->16CO2+18H2O
I need help to work out the enthalpy for the combustion of octane....^^balanced equation is above. I keep doing something wrong which causes me to get really high numbers. Please help.....

Answer 1

2C8H18+25O2 ->16CO2+18H2O
Left
16=C
36=H
50=O
Right
36H
32+18=50 =O
16=C
it s all right :)

Answer 2

i know that the balancing is right. I need help to work out the enthalpy using these bond energies. C-H: 413, C-C: 347, O=O: 498, C=O: 799, H-O: 464.

Answer 3

yes find the bond energy of the right molecules using above formula
for example there X no of C-H bond in octane so X*413 = P
and there are Y no of C-C bonds so = Y*347 = Q
P+q will give you the energy of that molecule similarly you find for all
and then energy of right molecules - energy of left molecule;es you do u get ur results

Answer 4

did u get what i am trying to say you ?

Answer 5

@adriana_98 ??????

Answer 6

@adriana_98 are you there?

Answer 7

Yes i am here @chmvijay @Jesstho.-.

Answer 8

2C8H18+25O2 ->16CO2+18H2O
2[(18xC-H) + (7xC-C)] + 25(O=O) – 16(4xC=O) + 18(2xH-O)
2(7434 + 2429) +12450 – 16(3196) + 18(928)
(19 544+ 12 450) – (51 136 + 16 704)
31 994 – 67 840
= -35 846 kJ
The number is really big and when i divide it by two to get 1 mL of octane,i get -17 923 kJ. It just seems too big.

Answer 9

final answer is for 2 mole of octane combustion u dived it by 2 you get the combustion of octane per mole

Answer 10

sorry, i am a bit confused??

Answer 11

2C8H18+25O2 ->16CO2+18H2O
This is for 2 moles of C8H18 u have to find for 1 mole of C8H18 right?...

Answer 12

yes, is it correct that i need to divide my final answer by 2??

Answer 13

@chmvijay

Answer 14

:) then what is the problem ur facing then ?

Answer 15

the answer is really big and i am not sure if it is right. Is the working out correct?

Answer 16

And also another question, how do i find enthalpy for blends of fuels so i need to find it for 1 part octane and 9 parts ethanol. What process do i need to take to find that and is there a equation i can use aswell. I am not sure

Answer 17

its correct u jhave made one mistake in 16(4xC=O) + 18(2xH-O) not this
but 16(2xC=O) - 18(2xH-O)
i just did it
and got 10,096 if i didved by 2 i get 5048 KJ/mol
which can be compareble with real values as shown in this table
http://www.ausetute.com.au/heatcomb.html

Answer 18

2[(18xC-H) + (7xC-C)] + 25(O=O) – 16(4xC=O) + 18(2xH-O)
12450+14868+4858-(25568+16704)=32176-42272 = -10096 KJ

Answer 19

look at once again ur calcualtions :P

Answer 20

what bond energies did you use to do your calculations 2 messages earlier because i can't work out how you got those answers. Sorry for all the questions

Answer 21

the one you have mentioned above :)

Answer 22

Nevermind i got the same answer as you

Answer 23

really ?....

Answer 24

yes,i mucked around with it

Answer 25

now i am unsure how to do this for blends. Do you know??

Answer 26

thats good :)

Answer 27

any other information they have given for that ?

Answer 28

no, nothing

Answer 29

the blends are like 1 part octane, 9 parts ethanol.... 2 parts octane, 8 parts ethanol etc... until we get to 0 parts octane and 10 parts ethanol. If that helps....

Answer 30

u mean to find enthalpy for blends of fuels ;)

Answer 31

yes

Answer 32

i did a blend of 1 mL so for example for 1:9, i used 0.1mL of octane and 0.9mL of ethanol. I just need to know how to work out the enthalpy in mL instead of moles this time. If you know how to do that...

Answer 33

find our enthalapy of the combustion for one mole of ethanol first similarly as we did above:)