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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: chaneliman on June 20, 2008, 07:58:45 AM

Title: Rate of Reactions
Post by: chaneliman on June 20, 2008, 07:58:45 AM
Can you just clear some things up for me;

Natural gas begins to burn when lit with a match. Why does it continue to burn when the match is taken away? (i think it has something to do with activation energy?)

Why must the energy absorbed, when the bonds break, be less than the energy released when new bonds form in the combustion of methane; which results in an overall release of thermal energy?

Why does a small increase in temperature have a greater impact on the rate than an increase at higher temperatures?
Title: Re: Rate of Reactions
Post by: Astrokel on June 20, 2008, 12:26:49 PM
1. Yes the lit provides an external energy for the reaction to reach its activation energy, however, there must be another source of energy to keep it from burning.(hint: look at the main composition of natural gas and part 2)

2. That is because products of combustion of methane(CO2 &H2O) are more energetically stable than the reactants. So in a sense, reactants are relatively less stable, so the bonds are easier to be break, therefore energy absorbed is lesser. Whereas, products are relatively more stable, so their bonds are easier to formed, therefore more energy is released. Hence, overall is always an exothermic process.

3. This is probably due to the nature of exponetial graph. Do you know about Arrhenius' equation?