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In school, while studying chemical kinetics, we were learning about the speed law and equation for different chemical reactions. We learned that in general, the exponents of the concentration of reactants in the speed equation must be determined experimentally. However, earlier this week, we learned that if we are talking about basic elementary reactions, the coefficients of the reactants become the exponents in The equation. This puzzled me. Can anyone explain why the coefficients would become exponents? Thanks in advance

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A reaction occurs when its reactives are in the same region of the space. If you have an elmental reaction of the form $a A + b B \rightarrow $, for the reaction takes place it is needed that $a$ molecules of $A$ and $b$ molecules of $B$ collide. How probably is this event? Well, if you only need one molecule of $A$ ($a=1$), the probability that an $A$ molecule be in this specific place (at some time) is proportional to the $A$ concentration. If $(a=2)$ we need another $A$ molecule, as the probability for a second molecule being in that space region (at some time, in particular at the same time) is independent, the probability of finding two $A$ molecules in the same space region (at the same time) is the product of the two individual probabilities (is like throwing a dice two times). In general, for the mixture the probability of the encounter is proportional to $[A]^a [B]^b$. The probability of encounter is proportional to the number of encounters in some time interval. And the later to the reaction velocity.

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  • $\begingroup$ This sort of made sense but you made it seem as though the concentration is the probabillity of finding a particule at that soace. Is this true? $\endgroup$ – Anthony B Jun 2 '16 at 2:51
  • $\begingroup$ It is not the probability, but it is proportional to the probability. Proportionality factor depends on temperature and volition radii. Also, once the volition takes place, there is another probability for the collision be reactive $\endgroup$ – user1420303 Jun 2 '16 at 2:59
  • $\begingroup$ Volition = collision after spellchecker $\endgroup$ – user1420303 Jun 2 '16 at 3:00
  • $\begingroup$ Ok, so say the coefficient of one of the reactants is 3 and at a particular moment in the reaction its concentration is 2 moles/ liter. I would think that since the reaction requires 3 of these molecules to collide at the same time, it would be less likely than if it only needed 2 and thus the reaction would take longer. However, the value of 2 cubed is larger than 2 squared wich leads to think the reaction would take less time, where is my logic flawed? $\endgroup$ – Anthony B Jun 2 '16 at 3:04
  • $\begingroup$ In the omission of proportionally coefficient, if it is small enough to make the product between it and concentration smaller than 1, then the cube will be smaller that the square $\endgroup$ – user1420303 Jun 2 '16 at 3:13

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