aommaster Posted December 7, 2003 Posted December 7, 2003 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?
aommaster Posted December 7, 2003 Author Posted December 7, 2003 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
YT2095 Posted December 7, 2003 Posted December 7, 2003 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? 1
aommaster Posted December 7, 2003 Author Posted December 7, 2003 so is the balanced equation: c4h9oh + 8o2 ------------> 4co2 + 5h20 ?????
aommaster Posted December 7, 2003 Author Posted December 7, 2003 shouldnt it be 3 o2 as in ur equation there are 9 on the left and 7 on the right???
YT2095 Posted December 7, 2003 Posted December 7, 2003 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).
aommaster Posted December 7, 2003 Author Posted December 7, 2003 yeah, cause, looking at the equation: c4h9oh + 4o2 ------------> 4co2 + 5h2o | 3O2 is needed to correct it
YT2095 Posted December 7, 2003 Posted December 7, 2003 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
aommaster Posted December 7, 2003 Author Posted December 7, 2003 OH sorry sorry i didnt count the 4 in fornt of the carbon ok thanx
aommaster Posted December 7, 2003 Author Posted December 7, 2003 umm... i wroked it out, BUT I STILL GOT AN ENDOTHERMIC REACTION!!!! its driving me mad!!! but i now get 13153.2 -------------->11080
aommaster Posted December 7, 2003 Author Posted December 7, 2003 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
YT2095 Posted December 7, 2003 Posted December 7, 2003 well that`s not right, coz 4O2 will just do the Carbon. where did the Hydrogen get its 2 and half O2 molecules from?
aommaster Posted December 7, 2003 Author Posted December 7, 2003 man, im more confused then u r!!!! I confused myself!!! ok, what do we have to do now to correct the equation above?
YT2095 Posted December 7, 2003 Posted December 7, 2003 add 5 more Oxygen ATOMS or 2.5 O2 molecules, don`t forget, you have a single spare O atom in your fuel
aommaster Posted December 7, 2003 Author Posted December 7, 2003 so... would this be correct and balanced? 2C4H9OH + 12O2 -----------------> 8CO2 + 10H2O
YT2095 Posted December 7, 2003 Posted December 7, 2003 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
aommaster Posted December 7, 2003 Author Posted December 7, 2003 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?
YT2095 Posted December 7, 2003 Posted December 7, 2003 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
aommaster Posted December 7, 2003 Author Posted December 7, 2003 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
YT2095 Posted December 7, 2003 Posted December 7, 2003 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 1
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