Dr. Boyle's Rate Law Exposition
(C) - Copyright, 1991 F.W. Boyle, Jr., Ph.D.

Rate Law of a Reaction:
Rate Law (or rate) = k·[Reactant 1]·...·[Reactant n]
so for
aA + bB + cC ---> products
Rate Law (rate) = k·[A]··[B]y·[C]z
where x, y, and z are the reaction orders for each reactant.
Rate of Reaction:
Rate of Reaction (rate) = abs(change in [component])
--------------------------
change in time
The component can be either a reactant of a product but
1. A reactant always decreases in concentration so a negative sign (-) must be placed in front of the equation.
2. A product always increases in concentration so the sign in front of the equation is positive (+).
Example:
aA + bB ---> cC + dD
rate (NOT RATE LAW) = 

               - 1 d[A]  =  - 1 d[B]  =  1 d[C]  =  1 d[D]
                 - ----       - ----     - ----     - ----
                 a  dt        b  dt      c  dt      d  dt
The "d" prefix means the change in concentration or time.
Remember the reaction cannot proceed any faster than the slowest portion, including the decomposition of a reactant.

Order of Component:
The order of a component is the power to which the concentration of the component is raised in the rate law equation. The order tells us the extent to which the reaction depends on the concentration of a particular component.
Calculating simple orders:
(rateh/ratel) = order for 1st order and some
(conch/concl) second order reactions
A general equation to calculate orders:
We know the the ratio of the rates is related to the ratio of the concentrations by:
(rateh/ratel) = (conch/concl)N
where N = the order for the particular component.
taking logarithms:
log(rateh/ratel) = N·log(conch/concl)
solving for N:
N = log(rateh/ratel)
log(conch/concl)
This equation will work for all orders of reaction including zeroth orders.
Examples:
1. H+(aq) + OH-(aq) ---> H2O(l)

[H+] M rate M min-1
1.0 1.0 x 10 -4
2.0 1.0 x 10 -4

Using equation above:
N = log{(1.0 x 10-4)/(1.0 x 10-4)}
log(2.0/1.0)

                  = log(1) =   0    = 0  so order of reaction wrt
                    log(2)   0.301       [H+] is zeroth.
2. H2O2(l) ---> H2O(l) + 1/2 O2(g)

[O2] M rate M min-1
0.5 1.0 x 10 -6
1.0 2.0 x 10 -6

N = log{(2 x 10-6)/(1 x 10-6)}
log(1.0)/0.5)

                  = log(2) = 0.301 = 1  so order of reaction wrt
                    log(2)   0.301      [O2(g)] is first.
3. NOBr(g) ---> 2NO(g) + Br2(g)

[NOBr] M rate M min-1
0.20 0.80
0.80 12.80

N = log(12.80/0.80)
log(0.80/0.20) 


                  = log(16) = 1.204 = 2 so order of reaction wrt
                     log(4)   0.602     [NOBr(g)] is second.
Using the Rate Law
Calculate the initial rate given the following:
1. The rate law for the reaction.
2. Initial (or starting) concentrations,
usually designated [ ]o.
3. A value for k
Example:
rate (rate law) = 1.3 x 10-3 M-2 min-1·[NO(g)]2·[Br2]
Calculate the initial rate, that is the rate at time = 0,
with these concentrations: [NO]o = [Br2]o = 3.0 x 10-4 M.
From this equation we can see that k = 1.3 x 10-3 M-2 min-1
and that the calcualtion of the initial rate is second order wrt [NO] and first order wrt [Br2].
Let's expand the equation and add the references to indicate that we are calculating an initial rate:
Rateo = 1.3 x 10-3 M-2 min-1·[NO]o·[NO]o·[Br2]o
We can see that the initial concentration of [N0] must be used twice or is squared and that the initial concentration of [Br2] is only used once. Substituting into the equation we get:
Rateo =
= 1.3 x 10-3 M-2 min-1·(3.0 x 10-4 M)·(3.0 x 104 M)·(3.0 x 10-4 M)
= 35.1 x 10-15 M min-1 = 3.51 x 10-14 M min-1
Therefore the initial rate of this reaction when [NO]o = [Br2]o= 3.0 x 10-4 M will be:
3.51 x 10-14 M min-1