i have got these bond energies:

Butanol 5580 (total)

Water 928

Carbon Dioxide 1210

oxygen 498.3

 

When i work out the total bond energies on both sides, it shows me that the complete combustion of Butanol is ENDOTHERMIC!!!! Whats wrong with my calculation

 

C4H9OH + 6O2 ----------> 4CO2 + 5H2O

 

14149.8 11080

 

See?

butanol as in Butan-1-ol CH3(CH2)2CH2OH ?

well, honestly to tell you, i havent a clue what you are talking about!!! But all i know, is that the butanol we are using has a formula of C4H9OH

 

P.S i read ur question on the smallest movement that is detected and i have a question, Why do u ask impossible-to-answer questions lol

LOL, well in reverse order, it`s not an impossible question, there has to be some threshold for this movement idea.

 

Next, lets have a look at your formula....

C4 H9 OH well we can re-shuffle those a little as there`s 2 piles with H in it

 

C4 H10 O1

 

and we have as many Oxygens as we can use in our air too

 

we can make 4 lots of CO2 using 8 O from the air

 

and 5 lots of H2O using 4 lots of O from the air and remaining Oxygen in our fuel.

 

total air usage to combust 1 molecule of butanol would be 8 oxygen from the Air.

 

leaving 4 parts CO2 and 5 parts water

and completely combusted... can you work from there?

so is the balanced equation:

 

c4h9oh + 8o2 ------------> 4co2 + 5h20

 

?????

4O2 MOLECULES or 8 individual Oxygen ATOMS

and thats it

shouldnt it be 3 o2 as in ur equation there are 9 on the left and 7 on the right???

well unless I`ve made a grave mistake, yup, that`s it

that`s how I was taught anyway

post #4 should demonstrate it all (at least in principal).

 

yeah, cause, looking at the equation:

 

c4h9oh + 4o2 ------------> 4co2 + 5h2o

|

3O2 is needed to correct it

aommaster said in post #5 :

so is the balanced equation:

 

c4h9oh + 8o2 ------------> 4co2 + 5h20

 

?????

 

4x CO2 (that`s 4xO2)

5x H2O (that`s 5xO... or 2xO2 and just a single O)

 

don`t forget the fuel contains ONE of the O atoms

 

I count 13 O ATOMS from the air, or 6 O2 molecules and 1 ATOM.

and one used from the fuel making it 7O2

 

OH sorry

 

sorry

i didnt count the 4 in fornt of the carbon

 

ok

thanx

no worries

umm...

 

i wroked it out, BUT I STILL GOT AN ENDOTHERMIC REACTION!!!!

its driving me mad!!!

 

but i now get

13153.2 -------------->11080

PASS? you`ve really lost me there dude?

well, ill write out the whole reaction so u can see

 

C4H9OH + 4O2 --------------->4CO2 + 5H2O

 

(2 x 5580) + (4 x 498.3) ----------> (4 x 1610) + (5 x 928)

 

13153.2-------------------> 11080

well that`s not right, coz 4O2 will just do the Carbon.

where did the Hydrogen get its 2 and half O2 molecules from?

man, im more confused then u r!!!!

I confused myself!!!

 

ok, what do we have to do now to correct the equation above?

add 5 more Oxygen ATOMS or 2.5 O2 molecules, don`t forget, you have a single spare O atom in your fuel

so... would this be correct and balanced?

 

2C4H9OH + 12O2 -----------------> 8CO2 + 10H2O

use the method in post #4, re-shuffle into neat little piles of 3 Different elements and work from there

 

then you`ll be able to answer that

yeah. I wroked it out, its correct. Now the numbers before the chemicals are 2, 12 8 and 10, can i divide them by 2?

there`s 2 schools of thought about that in all honesty. for instance Phosphourous Pentoxide is often writen as P2O5 when in fact the molecule is P4O10.

so that`s entirely up to you, whichever works best and remains factual at the same time, is great

ok, so then, i'll divide it by 2 since my chemistry teacher tells be to simplify it. Thanx, and in case u were wondering y i was a total idiot, it was because im only in year 11, and balancing formulas as complex as this, i have not practiced

 

lol

 

I wondered no such thing dude, as I said that`s what we`re here for, just hope some the methods discussed were of benefit for future projects also.

All the Best