Chemistry CW – Calculating the theoretical Energy transfer of 6 alcohols per mole of fuel?
Snap-crackle-pop asked:
For methanol, ethanol, propanol. butanolico, pentanolo and esanolo that I must find the energy transfer theoretical if they are burnt in the air (oxygen). This requires balancing of equations that I have already done but when it comes to actual calculations, my results have not model and aren 't all negative numbers (as it should be as they are esotermici.) I know are wrong but i can & # 039; t see where I'm making mistakes. If anyone could explain, facciami an example for, I would be so grateful, here 's what you need to know: The equation: Alcohol + oxygen —> Carbon dioxide + water Values slaves (Kj / mol) H-C – 413 Cc – 347 OC – 336 O = O – 498 C = O – 805 Oh – 464 for example balanced: 2CH3OH + 3O2 —> 2CO2 + 4H20 for all equations: O2 = double bond CO2 = C = 0 link double x2 all other residues / elements contain the individual links Thanks so much to anyone who can help, I are struggling with this for so long now. other equations balanced if necessary: Ethanol: C2H5OH + 3O2 —> 2CO2 + 3H20 Propanolo: 2C3H7OH + 9O2 —> 8H2O + 6CO2 Butanolico: C4H9OH + 6O2 —> 4CO2 + 5 H2O Pentanolo: 2C5H11OH + 15O2 —> 10CO2 +12 H2O Esanolo: C6H13OH + 9O2 —> 7H2O + 6CO2
For methanol, ethanol, propanol. butanolico, pentanolo and esanolo that I must find the energy transfer theoretical if they are burnt in the air (oxygen). This requires balancing of equations that I have already done but when it comes to actual calculations, my results have not model and aren 't all negative numbers (as it should be as they are esotermici.) I know are wrong but i can & # 039; t see where I'm making mistakes. If anyone could explain, facciami an example for, I would be so grateful, here 's what you need to know: The equation: Alcohol + oxygen —> Carbon dioxide + water Values slaves (Kj / mol) H-C – 413 Cc – 347 OC – 336 O = O – 498 C = O – 805 Oh – 464 for example balanced: 2CH3OH + 3O2 —> 2CO2 + 4H20 for all equations: O2 = double bond CO2 = C = 0 link double x2 all other residues / elements contain the individual links Thanks so much to anyone who can help, I are struggling with this for so long now. other equations balanced if necessary: Ethanol: C2H5OH + 3O2 —> 2CO2 + 3H20 Propanolo: 2C3H7OH + 9O2 —> 8H2O + 6CO2 Butanolico: C4H9OH + 6O2 —> 4CO2 + 5 H2O Pentanolo: 2C5H11OH + 15O2 —> 10CO2 +12 H2O Esanolo: C6H13OH + 9O2 —> 7H2O + 6CO2
Tags: Alcohol, Alcohols, Balanced Equations, Balancing Equations, Bond Values, Carbon Dioxide, Chemistry, Compounds, Double Bond, Elements, Energy Transfer, Fuel Ethanol, H2o, Methanol, Mole, Negative Numbers, Oxygen, Pentanol, Single Bonds, Water Bond
September 9th, 2008 at 8:23 am
In your example, bonds broken =
6 x C-H
2 x C-O
2 x O-H
3 x O=O
Bonds formed =
4 x C=O
8 x O-H
It should work! Remember to compare like with like, so you would halve this value because you started with 2 moles of methanol.