Homework Help: Rate law confusion.

hi mate,

very confused by something in my text book explaining rate law and would be very grateful if someone could clarify it for me. here goes...

text book excerpt:

text book
rate laws for elementary step

elementary step ----------------------- molecularity----------- rate law

.......(1) a -> product......................... unimolecular............................. k[A]

.......(2) 2A -> product ........................ bimolecular ........................... k[A]^2

.......(3) A + B -> product........................ bimolecular ........................... k [A]


why concentrations are multiplied in rate law?


collision theory



collision theory and rate law

textbook tells us: "if particles must collide to react, the laws of probability tell us why rate depends on product of reactant concentrations not sum."

one implication of this is that in the case of bimolecular elementary step reactions (see table above) if concentrations are doubled rate of reaction goes four time faster.

this seems pretty obvious for forumla 3 (see above table) and the text book shows a neat little diagram of how the number of possible collisions quadruples when you double the number particles in a vessel from 2 to 4 SEE DIAGRAM below (by the way ignore the full stops they are the only way i could create distance between the reactant particles the damn post kept condensing the spaces between characters)



VESSEL 1



-----------------------------
A

..................... B

------------------------------

VESSEL 2



-----------------------------
A ................... B


A ................. B

-----------------------------


clearly in the first instance there is only one possible collision and when u double the concentration of both reactants in the second you can draw four possible collisions between the particles thus explaining why the reaction proceeds four times quicker.


so far so good. the trouble is if you draw this same diagram for forumula (2) i.e.( four B particles in a vessel) in the table above you can draw six possible collisions in the second instance and yet judging by the formula [a]^2 there should only be a quadrupling of the rate. this disparity between n.o of possible collisions with increasing concentration and predicted rate increase gets worse with higher numbers. anyway my question is shouldn't the increase in rate for reaction (2) with increasing concentrations be higher than that for (3) and if it shouldn't why not????


if anyone could clear this up it would be great


thanks alot

john

 

sorry but i don't understand what you're saying. Would you mind elaborating further. i still cant wrap my head around the following example:

imagine two A particles in a vessel and in another vessel one A and one B particle.

see below
vessel 1
--------------------
A


. . . . . . . . . . A
--------------------

and
vessel 2
--------------------

A . . . . . . . . B


-------------------

clearly in both cases there is only one possible collision. now lets see what happens when you double concentration of reactants:

vessel 1
--------------------
A . . . . . . . . A



A . . . . . . . . A
---------------------

and
vessel 2
----------------------

A . . . . . . . . . B


A . . . . . . . . . B

----------------------

now in vessel one there are 6 possible collisions (6 times as many as were possible before) whilst in contrast in vessel two there are 4 possible collisions ( 4 times as many as before). according to the rate law for elementary steps the reaction rate in both cases should quadruple with a doubling in the concentration of the reactants. the explanation given is due to the quadrupling of the number of possible collisions, but in the case of the first vessel there is SIX times as many possible collisions!!!! by that logic in this particular example the rate in vessel one should be six times faster when the concentration is doubled but the forumla is [A]^2 which means it quadruples. whats going on?