Mechanisms

The mechanism for a chemical reaction is described by the rate law.  Mechanisms describe the collisional events that allow a reaction to occur and must obey 3 rules.

1. A mechanism must be composed of elementary steps/reactions.
2. The elementary steps of a mechanism, when added together, must yield the overall reaction
3. The observed rate law for the overall reaction must be consistent with the rate law for the slowest step

Elementary steps are the collisions that occur in a reaction.  Thinking about the probability of a collision, we can simplify the picture a little bit because it is extremely unlikely that more than 2 particles will collide with the proper orientation and energy to react.  This means that all elementary steps are either unimolecular or bimolecular.  {Yes, termolecular elementary steps are possible, but they're rarely significant contributors so we will ignore them for now.}
Once we have this series of elementary steps, what do we do with them?  Because elementary steps describe collisions at the molecular level, we can write rate laws for each of the elementary steps based only upon the balanced reaction of the elementary step.  Since doubling the number of any molecule will double the probability of a collision (the rate), the elementary steps are first order with respect to each reacting molecule.
Rates are determined by the activation energy of a step or an overall process; the higher the activation energy, the slower the rate.  For a series of steps, whichever step has the highest activation energy will determine (or limit) the rate for the entire process, so it is known as the Rate Determining (or Limiting) Step, the RDS (or RLS).  With a little algebra, we can wrassle the rate law of the RDS into a form that looks like the observed rate law.  The rate law expression for the overall process shown below, A + B + D→E, is

Rate_obs = k_obs[A]₀^x[B]₀^y[D]₀^z

If the first step is RDS, we can write the rate law expression for the first step:

Rate₁=k₁[A]₀¹[B]₀^y

This rate law, looks just like the observed rate law if the reaction is first order with respect to [A] and [B], so if the first step is RDS, this is a pretty straight-forward problem with minimal algebraic wrasslin'.

For the second step RDS, we have to use a steady state approximation to make the rate law "consistent with" the observed rate law.  The rate law for the second step is:

Rate₂=k₂[C]₀¹[D]₀¹

But [C] does not appear in the overall observed rate law expression, so we have to figure out a way to make this expression "consistent with" the observed rate law for the overall process.  If most of "C+D" has quite a bit of energy, but not quite enough to get over the C+D→E hump, it probably has enough energy to go backwards and return to "A+B".  This means we can define another rate law, this time for the reverse of step 1, C+D→A+B:

Rate_-1=k_-1[C]₀¹[D]₀¹

As the rxn A+B→C+D proceeds, eventually we will build up a concentration of "C+D" that will remain essentially constant throughout the reaction.  This is the "steady state".  When we reach this steady state, the rate of step 1 going forward (Rate₁) will be equal to the rate of step 1 in reverse (Rate_-1):

Rate₁ = Rate_-1

Which means that:

k₁[A]₀¹[B]₀¹ = k_-1[C]₀¹[D]₀¹

Now we can solve for  [C]₀¹ :

[C]₀¹ = {k₁/k_-1}[A]₀¹[B]₀¹/[D]₀¹

Now, we can plug the expression for  [C]₀¹ into the rate law expression for step 2:

Rate₂ =  k₂({k₁/k_-1}[A]₀¹[B]₀¹/[D]₀¹)[D]₀¹ = k_combined[A]₀¹[B]₀¹  

Note: Since "k₁", "k_-1", and "k₂" are constants, we can just lump them together into a single constant, in this case labelled "k_combined".

So if the second step is slow, the observed rate law should be first order w.r.t. [A] and [B], zero order w.r.t. [D].  Note that this is the same as the observed rate law expression if the first step is slow, so how can we distinguish between these two mechanisms?  Typically there would have to be something else observable in the reaction.  If step 2 is slow, then during the course of the reaction, we might expect to see a measurable concentration of "C" appear when the steady state is established and then disappear when the reaction is complete.  We might also be able to analyse the value of "k_observed" to see a difference, but that's often a bit more challenging than detecting an intermediate in the reactions.

Yikes, that one got a little long.  Have a good weekend.