What does a fractional order of reaction mean for the mechanism?
Here's an example.

Let's consider a reaction of the generic form:

$$\ce{3A -&gt; 2C}$$ 

Now, let's say we have some knowledge that this reaction proceeds by the following two-step mechanism:


  Step one is fast and reversible.


$$\ce{A &lt;=&gt;[k_1][k_{-1}] 2B}$$


  Step 2 is slow and irreversible.


$$\ce{A + B -&gt;[k_2] C}$$

The rate of the reaction (expressed as the change in concentration of the product per unit time) can be expressed in terms of the concentrations of A and B.

$$\mathrm{rate}=\dfrac{\Delta [\ce{C}]}{\Delta t}=k_2[\ce{A}][\ce{B}]$$

However, we would prefer to not use intermediates in our rate law. Since the first step is fast and reversible, we can use the law of mass action to create a relationship between the concentrations of A and B:

$$K_1=\dfrac{k_{1}}{k_{-1}}=\dfrac{[\ce{B}]^2}{[\ce{A}]}$$
$$[\ce{B}]^2=K_1[\ce{A}]$$
$$[\ce{B}]=\sqrt{K_1[\ce{A}]}$$

If we substitute into the  rate law:

$$\mathrm{rate}=\dfrac{\Delta [\ce{C}]}{\Delta t}=k_2[\ce{A}]\sqrt{K_1[\ce{A}]} =k_2 K_1^{1/2}[\ce{A}]^{3/2}$$

The term $k_2 K_1^{1/2} =\dfrac{k_2 k_1^{1/2}}{k_{-1}^{1/2}}$ becomes $k_{\mathrm{obs}}$, the observed rate constant of the overall reaction. The fractional order comes from the second step being rate-determining, but the first step being an equilibrium with stoichiometry that generates the fractional tern.