Dec 14, 2009
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I am a little bit confusing this part. am I right on this?

0 order – related with K
First order – related with concentration
Second order – related with concentration

First order half life – independent of concentration for all first order reaction and inversely related with K

please correct me If I am wrong abut this
 

UndergradGuy7

10+ Year Member
Jun 23, 2007
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rate law is rate = k[concentration A][concentration B] etc...
that is if you have a A and B

for 0 order you do not take into account the concentrations or rate = k[conc]^0 anything ^0 = 1 so its concentration does not matter.

For 1 order rate = k[A]^1

for 2 order rate = k[A]^1 * ^1 or k=[A]^2
notice for 2nd order you can have it two different ways as I wrote above. What makes it second order the how the exponents on A and B add up. if A is 2 then its second order on A and overall. If A's exponent is 1 and B is 1 then the reaction is 1st order with respect to A, but overall it is 2 because the exponent of A + B, or 1+1=2.

So yes, you were right about this
0 order – related with K
First order – related with concentration
Second order – related with concentration

It would also be good to know the units of k for the different orders and how the graphs look.


I think this is true too...
First order half life – independent of concentration for all first order reaction and inversely related with K