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In the reaction between hydrochloric acid and Magnesium the overall reaction is:

$$\ce{Mg + 2HCl -> MgCl2 + H2}$$

My proposed mechanism of reaction is that $\ce{HCl}$ dissociates in $\ce{H2O}$ like so: $$\ce{H2O + HCl <=> H3O+ + Cl- }$$

And then the $\ce{H3O+}$ oxidises the magnesium like so:

$$\ce{2H3O+ + Mg -> Mg^{2+} + H2O + H2}$$

This, in an experiment to determine the rate of the reaction, measuring the $\ce{H2}$ produced would measure this mechanism, because it is the step in which hydrogen gas is evolved. Presumably these Magnesium ions are then in solution. I would therefore deduce that the step: $$\ce{2H3O+ + Mg -> Mg^{2+} + H2O + H2}$$ Is the rate-determining step, because it is the step in which magnesium ions are formed and go into solution. My rate law, deduced from these mechanisms, would look like this:

$$\ce{rate = k * [H3O+] }$$ Because it is the concentration of the $\ce{H3O+}$ which determines the formation of the Mg ions. From there, I would assume, the $\ce{Mg^{2+}}$ and the $\ce{Cl-}$ would remain in solution, as the reaction is carried out in the acid, obviously consisting of water.

From this, can I say that the rate law for the reaction

$$\ce{Mg + 2HCl -> MgCl2 + H2}$$

is merely: $$\ce{rate = k * [H3O+] }$$ But $\ce{[H3O+]}$ is: $$\ce{[H3O+] = (K*[H2O][HCl])/[Cl^{-}] }$$ Where K is the equilibrium constant of the acid equilibrium.

So can we write our rate law like:

$$\ce{rate = k*(K[H2O][HCl])/[Cl^{-}] }$$ Where K is the equilibrium constant and k is the rate constant and might look like this (following the Arrhenius equation): $$\ce{k = A e^{-E_a/(R T)}}$$

To make our rate law like this: $$\ce{ rate = A e^{-E_a/(R T)} *((K*[H2O][HCl])/[Cl^{-}]) }$$

My question is: Is my rate law correct for this reaction? If not, where have I gone wrong?

Additionally: This was one of our practicals. We ended up concluding that rate goes up quadratically or exponentially with concentration. This rate law would seem to indicate a linear increase, not exponential. I assume in my rate law there is no contribution from the solid Mg. Obviously in a powder, more surface area is exposed, and a high rate is experienced. Is there a way to factor this into my rate law? Thanks for the help! :)

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Your rate law should be $$ r = k_\text{obs} \times [\ce{H3O+}]^2 = A_\text{obs} \exp\left(\frac{E_\text{a, obs}}{RT}\right) \times [\ce{H3O+}]^2\; ,$$ which nicely fits your experimental data of a quadratic dependence on the concentration for the rate.

The exponent is necessary because the reaction is seemingly second-order in the concentration of $\ce{H+}$.

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  • $\begingroup$ Why? What steps/mechanisms would be in the reaction if this was the rate law? :) Thanks $\endgroup$ – Swedish Architect Mar 7 '14 at 17:16
  • $\begingroup$ Is H3O order 2 because there are 2 molecules needed to oxidise every atom of Mg?? Thanks :) $\endgroup$ – Swedish Architect Mar 10 '14 at 17:13
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    $\begingroup$ Well, according to the mechanism you propose: yes. If it really happens concertedly or via an intermediary step ($\ce{H+ + Mg -> Mg+ + H^{.}}$) I don't know. $\endgroup$ – tschoppi Mar 10 '14 at 19:58
  • $\begingroup$ OK. Thanks for clarifying. I would think that it would happen with an intermediate, so Mg is first oxidized to Mg+ then to Mg+2 a second time. But I don;t know. :) $\endgroup$ – Swedish Architect Mar 10 '14 at 20:36
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    $\begingroup$ Yeah, it seems more reasonable. But hey, if it fits your data (and it's not valuable, to-be-published research), go ahead and propose the concerted mechanism ;) $\endgroup$ – tschoppi Mar 10 '14 at 20:56
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For the reaction $\ce{2H3O+ + Mg -> H2 + Mg^{-2}}$; the actual rate law is $r=k \times [\ce{H3O+}]^2 \times (\ce{Mg})^1.$

The most obvious error is that HCl is a strong acid, thus not in equilibrium. Cl- is merely a spectator ion.

A more accurate reaction step method is:

$\ce{H3O+ + Mg -> H2O + MgH}$

$\ce{H3O+ + MgH -> H2 + Mg^{-2}}$

The surface area of magnesium directly affects the k value, but magnesium is still a valid part of the reaction. The magnesium term may appear zeroth order if the magnesium is in far excess, so significantly lowering the initial input of magnesium will probably result in the correct rate law.

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  • $\begingroup$ I hope while editing I haven't messed up anything!! Feel free to rollback if i have messed up stuff :-) $\endgroup$ – Freddy Oct 30 '14 at 7:28
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    $\begingroup$ I didn't think solids could be considered in a rate law - [] indicates concentration, which can only happen in liquids/gases, not solid. Clearly the surface area of the Mg has an effect - typically taken into account in the A term of the Arrhenius equation $\endgroup$ – Swedish Architect Oct 31 '14 at 16:43

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