how does fire work chemically?

[Danny]

ive never had a satisfactory description of how fire works. im talking about like wood, or alcohol and the sort.

[Horza2002]

To anwser this, I am assuming you mean the act of combustion and not the flame itself.

 

Burnign is, at the simplest level, a chemical reaction; one set of chemical bonds are replaced by another set (normally stronger). For a reaction to occur, the change in Gibbs free energy (G) should be negative. Since G=H-TS, for a given temperature, an exothermic reaction with an increase in the number of molecules will be very favourable.

 

The majority of fuels have lots of carbon-carbon and carbon-hydrogen bonds (e.g. petrol, wax, wood, etc.) in them. These bonds are relatively strong, however, carbon-oxygen bonds are stronger. When the fuel is burnt, these bonds are replaced with carbon-oxygen double bonds and hydrogen-oxygen bonds. These bonds are stronger than the starting bonds and so there is an enthalpic gain when they are replaced.

 

Carbon-Carbon = 347 kJmol^-1

Carbon-hydrogen = 435 kJmol^-1

Oxygen-carbon double = 805 kJmol^-1

Oxygen-hydrogen = 464kJmol^-1

Oxygen-oxygen = 498kJmol^-1

 

So for example, burning pentane, the equation is

C₅H₁₂ + 8O₂ ==> 5CO₂ + 6H₂O

 

The enthalpy of the starting material is:

5(C-C) + 12(C-H) + 8(O=O)

5(347) + 12(435) + 8(498) = 10939kJmol^-1

 

The enthalpy of the products:

10(C=O) + 12(O-H)

5(805) + 12(464) = 9593kJmol^-1

 

Therefore the change in enthalpy of this reaction is 9593-10939 = -1346kJmol^-1 (i.e. its exothermic and favourable).

 

Another favourable aspect to consider is that there is an increase in entropy. There are 9 starting material molecules and 11 product molecules...so overal, burning a fuel has a negative change in Gibbs energy and so is favourable.

 

Thats probably a little more indepth than you wanted but o well!

Edited March 15, 2011 by Horza2002

[mississippichem]

ive never had a satisfactory description of how fire works. im talking about like wood, or alcohol and the sort.

 

Short answer: the rapid oxidation of organic compounds. Oxygen steals electrons from said organic compounds. The final products (if combustion is complete) are [ce]H_{2}O[/ce] and [ce]CO_{2}[/ce].

 

These reactions give off heat so are said to be exothermic. Another point of interest is that these reactions happen extremely fast, surpassed only by the speed of acid base reactions. Acid base reactions are faster only because protons can tunnel between solvent molecules.

Edited March 16, 2011 by mississippichem

[Danny]

To anwser this, I am assuming you mean the act of combustion and not the flame itself.

 

Burnign is, at the simplest level, a chemical reaction; one set of chemical bonds are replaced by another set (normally stronger). For a reaction to occur, the change in Gibbs free energy (G) should be negative. Since G=H-TS, for a given temperature, an exothermic reaction with an increase in the number of molecules will be very favourable.

 

The majority of fuels have lots of carbon-carbon and carbon-hydrogen bonds (e.g. petrol, wax, wood, etc.) in them. These bonds are relatively strong, however, carbon-oxygen bonds are stronger. When the fuel is burnt, these bonds are replaced with carbon-oxygen double bonds and hydrogen-oxygen bonds. These bonds are stronger than the starting bonds and so there is an enthalpic gain when they are replaced.

 

Carbon-Carbon = 347 kJmol^-1

Carbon-hydrogen = 435 kJmol^-1

Oxygen-carbon double = 805 kJmol^-1

Oxygen-hydrogen = 464kJmol^-1

Oxygen-oxygen = 498kJmol^-1

 

So for example, burning pentane, the equation is

C₅H₁₂ + 8O₂ ==> 5CO₂ + 6H₂O

 

The enthalpy of the starting material is:

5(C-C) + 12(C-H) + 8(O=O)

5(347) + 12(435) + 8(498) = 10939kJmol^-1

 

The enthalpy of the products:

10(C=O) + 12(O-H)

5(805) + 12(464) = 9593kJmol^-1

 

Therefore the change in enthalpy of this reaction is 9593-10939 = -1346kJmol^-1 (i.e. its exothermic and favourable).

 

Another favourable aspect to consider is that there is an increase in entropy. There are 9 starting material molecules and 11 product molecules...so overal, burning a fuel has a negative change in Gibbs energy and so is favourable.

 

Thats probably a little more indepth than you wanted but o well!

 

 

I think you could have been a little bit more indepth actually, but i understand.... its quite expensive to ship a textbook!

 

 

 

thank you all very much. good answers, i learned a bit.

[Horza2002]

If you'd like a little more detail, then we can discuss the mechanism of combutions.

 

Combustion is generally accepted to be radical chain reaction mechanism; although it is not completely understood. The initiation is forming an oxygen radical from molecular oxygen which then abstracts a hydrogen to leave a hydrogen peroxide radical. This radical abstracts a second hydrogen from the fuel to give hydrogen peroxide, which then homolyses to give two hydroxyl radicals. These are very reactive species and then go onto abstract further hydrogens (to give water) and oxides the carbon atoms (to give carbon dioxide).

 

The obvious question here is where does the initial oxygen radical come from? The most accepted idea is that molecular oxygen (a triplet in its ground state) is converted to singlet oxygen (by flipping the spin of one of the unpaired electrons in molecular oxygen). To do this, a relatively large amount of energy is required because singlet oxygen is not very stable. This energy normally comes from whatever ignites the fuel (a spark, another flame, etc).