From the activation energy, How do I know what mechanism control the reaction rate?

Question

I did the experiment about oxide dissolution in acid. I found that the activation energy is about 9-11 kcal/mol. From these value, can I define the mechanism which control the reaction rate? Diffusion, chemical or mixed control?

Answer 1

Mixed control.

Answer 2

Lancenigo di Villorba (TV), Italy

Wow! You made a very interesting work!


YOUR EXPERIMENT
As you wrote, you dipped some METAL OXIDE's pieces in a Strong Acidic Aqueous Solution : I think you chosen H2SO4's aqueous solutions since it results able to dissolve many chemical oxides

MxOy(s) + y H2SO4(aq) ---> Mx(SO4)y(aq) + y H2O(aq)

Surely, you recorded the Dissoluted Oxide in the acidic mixture against the Time Coordinate ; when you collected several data, you re-start the studies working on a different Temperature value.
In this fashion, you obtained several "Isothermal Data's Enquiries".


DATA's REGRESSION
Speaking on any Isothermal, you calculate the Reaction's Rate

Reaction's Rate = 0 - d[Dissoluted Oxide] / d[Time]

so you obtain several Rate's Data fo any Isothermal ; so you get picture of "Reaction's Rate vs. Dissoluted Oxide" by drawing the related graph (e.g. I suggest you LOGARITMIC GRAPHIC SCALE).
Now, these study gives you the Observed value of the Kinetic Constant

k.exp = d[ LN(Reaction's Rate) ] / d[ LN(Dissoluted Oxide) ]

Finally, you execute the ultimate Differential Enquiriment

E.att = R * d[ LN(k.exp) ] / d[ 1 / T ]


DISCUSSION
On the basis of

E.att = R * d[ LN(k.exp) ] / d[ 1 / T ]

where Chemical Textbooks delucidated the role played by Kinetic Constant "k" and Mass Transfer's One as ratio "D / s"

k.exp = k * D / (k * s + D)

where "k" is the Kinetic Constant on the Oxide's Surface while "D" is the Mass's Diffusivity in liquid phase while "s" is the "Lewis-Whitman" 's film thickness which acids have to permeate to join on the Oxide's Surface.
Executing the Differential Work you will be able to show that

E.att = (E.diff + E.real) / 2

that are respectively the DIFFUSION CONTRIBUTE (e.g. related to "D" and "s")

E.diff = R * d[ LN(D) ] / d[ 1 / T ]

and the CHEMICAL REACTION's CONTRIBUTE

E.real = R * d[ LN(k) ] / d[ 1 / T ]


CONCLUSIONs
As I said, it is "E.att = (E.diff + E.real) / 2".
As you know, DIFFUSIVE EVENTS RESULT CHARACTERIZED FROM SMALL "E.diff" VALUES.
AT COLD TEMPERATURE, "E.real" RESULTS VERY GREATER THAN "E.diff", SO IT IS "E.exper = E.real / 2".
AT HIGH TEMPERATURE, "E.real" FALLS DOWN, SO "E.real" JOINS NEIGHBOUR TO "E.diff".
In your case, you gave "E.att LIKE 9 or 11 kcal/mol" and I thought to KINETIC CONTROL.

I hope this could be clear.

Answer 3

The reactivity relies upon on many components. The length of the molecule, the section on the molecule of the halogen, the type of halogen, or perhaps which halogen the tertiary ought to reason the molecule to be extra polar and subsequently extra reactive.