Predicting the rate law

It may be possible to distinguish between plausible mechanisms by comparing the experimental rate law with the overall rate law that the mechanism would predict. If they are the same, that mechanism could be operating.
 
For example, consider the reaction:  CH3Br + OH CH3OH + :Br
two step mechanism 
step 1: CH3Br +CH3 + :Br
(C–Br bond breaks - slow)
step 2: H3C+ + :OH CH3–OH
(C–O bond forms - fast)
single step mechanism
CH3Br + :OH   CH3OH + :Br
(simultaneous C–Br bond breaking and C–O bond formation)


How do I predict the experimenal rate law that would be observed for a given mechanism?
Each step in a reaction mechanism is an elementary reaction, that is one that could occurs on a single collision between reacting species, provided the collision occurred with sufficient energy.

The rate of an elementary reaction depends on the concentrations of the reactant species. The rate law for an elementary reaction can be written from the balanced equation for the reaction.  Therefore the predicted rate law for any step in a proposed mechanism has the rate constant k multiplied by the product of its reactant concentrations raised to their coefficient in the balanced equation.  See the examples below.
 
For Step 1 of the two-step mechanism:  rate = k[CH3Br]
For the single-step mechanism:  rate = k[CH3Br][OH

The slow step determines the rate of the overall process.  Therefore the rate law for the reaction depends on the rate law for this step. 

If the rate-determining step is first, the predicted rate law will only involve species that are reactants in the overall reaction and can be compared directly with the experimental one. If rate law is the same as the experimentally determined one, the mechanism that predicts that rate law may be operating.