# What does a fractional order of reaction mean for the mechanism?

I'm doing a high school/sixth form college investigation of the kinetics between magnesium ribbon and hydrochloric acid.

I have obtained a rate order was 1.5 w.r.t [H+] and hence the rate equation is rate = k[H+]1.5.

What would that mean about the mechanism?

I remember my teacher once telling me that a fractional rate order just means that there is an intermediate stage and each step has its own rate order which contributes to the rate equation - is this right?

Below is a pictorial representation of what my teacher told me:

Also, what would this mean for my rate-determining step? If the above is true, what would I choose as the RDS if they both contribute a different amount to the rate order?

• $\mathrm{x + y = n}$ certainly won't hold for all cases. For instance in this case all of the A could be quickly converted to B, but C forms very slowly. – MaxW Apr 7 '16 at 3:29

Here's an example.

Let's consider a reaction of the generic form:

$$\ce{3A -> 2C}$$

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

Step one is fast and reversible.

$$\ce{A <=>[k_1][k_{-1}] 2B}$$

Step 2 is slow and irreversible.

$$\ce{A + B ->[k_2] C}$$

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.

• This is well explained, but it should be commented that the only real conclusion that can be drawn from a fractional order coefficient is that there are multiple steps involved in the reaction. The actual orders are not explicitly informative. – Lighthart Apr 7 '16 at 3:09