8
$\begingroup$

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 $1.5$ with reference to $[\ce{H+}]$ and hence the rate equation is

$$\mathrm{rate} = k[\ce{H+}]^{1.5}.\tag{1}$$

How does this fractional order correlate with 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:

$$\ce{A ->[$x$] B ->[$y$] C}$$ $$x + y = n$$ $$\mathrm{rate} = k[\ce{A}]^n.\tag{2}$$

Is this right? 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?

$\endgroup$
2
  • 2
    $\begingroup$ $x + y = n$ 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. $\endgroup$
    – MaxW
    Apr 7, 2016 at 3:29
  • $\begingroup$ A reaction mechanism is (generally) a very complicated thing which you can't hope to deduce from orders alone. One may say a fractional order indicates that the actual mechanism is "not simple" (that is, more complicated than "all molecules simply meet and turn to products at once, just like the equation says"). But then again, it is pretty much never simple. $\endgroup$ Nov 3, 2023 at 20:25

1 Answer 1

13
$\begingroup$

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.

$\endgroup$
2
  • 5
    $\begingroup$ 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. $\endgroup$
    – Lighthart
    Apr 7, 2016 at 3:09
  • $\begingroup$ So Can I claim for an elementary reaction, the order can never be fractional or negative ( if it involves no intermediates ) Also what does it mean to sy if the order with respect to an reactant is negative ? $\endgroup$
    – Shashaank
    Oct 25, 2022 at 4:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.