Jerryt12 Posted March 27, 2010 Share Posted March 27, 2010 For example: gasoline Is it because of their electron configuration? If so, how does it plays in the role of producing vast amount of energy? Link to comment Share on other sites More sharing options...
Horza2002 Posted March 27, 2010 Share Posted March 27, 2010 You need to consider the stability of the products. When you burn petrol, the products (carbon dioxide and water) are far more stable than the starting petrol. This energy that is released on forming the strong bonds is what is given out as heat. There are some reactions that actually get colder as they react because the products are less stable than the starting material (have a look for endothermic reactions) Link to comment Share on other sites More sharing options...
Jerryt12 Posted March 28, 2010 Author Share Posted March 28, 2010 If endothermic reactions is less stable, then it is easier for them to go the reverse way right? Link to comment Share on other sites More sharing options...
Horza2002 Posted March 28, 2010 Share Posted March 28, 2010 If you only consider enthalpy (thats changes in bond strength) then yes, endothermic reactions are erasy to reverse. HOwever, for a reaction to occur, the change in Gibbs free energy has to be negative. This depends on the change in enthalpy and entropy. [math]\Delta[/math]G=[math]\Delta[/math]H-T[math]\Delta[/math]S If the entalpy is positive (i.e. endothermic) then the change in enthalpy (S) has to be very positive to overcome this. So if a reaction is endothermic, there has to be a large increase in the disorder of the system to make it work. 1 Link to comment Share on other sites More sharing options...
UC Posted March 29, 2010 Share Posted March 29, 2010 If you only consider enthalpy (that's changes in bond strength) then yes, endothermic reactions are easy to reverse. However, for a reaction to occur, the change in Gibbs free energy has to be negative. This depends on the change in enthalpy and entropy. [math]\Delta[/math]G=[math]\Delta[/math]H-T[math]\Delta[/math]S If the enthalpy is positive (i.e. endothermic) then the change in entropy (S) has to be very positive to overcome this. So if a reaction is endothermic, there has to be a large increase in the disorder of the system to make it work. Fixed that for you. Link to comment Share on other sites More sharing options...
pioneer Posted March 29, 2010 Share Posted March 29, 2010 Another consideration is the activation energy hill. The climb up the front side, is the amount of energy needed to push the reactants to where they will react. Once at the top of the hill, the product slides down the hill to a lower energy level giving off energy. If the climb up the front of the hill is easy (low energy) and the slide down the back side of the hill is very long (lot of energy output), this provides the most explosive situation. It takes very little energy to get a chain reaction, and since the products are so far down the backside of the hill, little energy is wasted in reversed reactions. 1 Link to comment Share on other sites More sharing options...
Horza2002 Posted March 29, 2010 Share Posted March 29, 2010 Pioneer, the difference between the reactant and product is the change in Gibbs free energy I described above. The height of the activation barrier has no effect on the amount of energy that is released. The height of the barrier only controls the rate of reaction (i.e. the kinetics) not the thermodynamics. An run away explosion may occur if there is a large energy difference and a small barrier height, but you need both conditions for that to occur. Link to comment Share on other sites More sharing options...
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