# Multiple rate-determining steps and rate expression

I am trying to understand the concept of the rate laws governing hydrogen evolution of platinum. More specifically, I am trying to understand the multiple occurrence of a rds to the overall rate law of a reaction.

In the lecture, we talked about how there are three possible steps; the Volmer-step, the Heyrovsky step and the Tafel step:

For simplicity, lets talk about the Volmer-Tafel mechanism. We suppose the Volmer step to be very slow and rate-determining, and the Tafel step to be really fast; therefore, the Volmer reaction has to happen twice in order for the Tafel step to take place.

My professor derived the following rate expression:

This rate expression seems to agree with literature. However, I am not able to understand how neither the reaction order in protons nor the "Tafel slope" (exponential term, similar to Arrhenius) is not affected by the fact that the rate-determining step has to happen twice. I guess this is also a question in more general terms.

Why would multiple occurrence of an RDS not alter the rate expression?

In case I you only consider the electrochemical hydrogen reduction and its inverse reaction

$$\ce{H^+ + M + e <=>[k_1][k_{-1}] MH} \hspace{3ex}[1]$$

The rate of reaction [1] is given by

$$\ce{ v_1 = k_1 c_{H^+} (1 -\theta_H)}$$

where $$\ce{c_{H^+}}$$ is the surface concentration of hydrogen, and $$\theta_H$$ is the surface coverage of adsorbed hydrogen.

Since reaction [1] is an electron tranfer reaction on an electrode surface, it rate depends on the electrode potential (E) - or overpotential ($$\ce{\eta\,=\,E -E_0}$$). If the oxidation step at the applied potential, $$\ce{k_1(\eta)}$$ is given by

$$\rm{k_1 = k^0 e^{-\alpha f \eta} }$$

where $$\rm{\alpha}$$ is the symmetry factor and $$\rm{f = RT/F}$$. It is natural that this expression reminds you of the Arrhenius equation. Actually, you can interpret as describing the effect of the electrode potential of the height of the reaction barrier.

Thus, $$\ce{ v_1 = k^0 e^{-\alpha f \eta} c_{H^+} (1 -\theta_H)}$$

For low coverage this expression simplifies to

$$\ce{ v_1 \approx k^0 e^{-\alpha f \eta} c_{H^+} }$$

Since reaction [1] is the RDS, the overall rate depends on the rate of this step, and

$$\ce{ v_{HER} = \frac{v_1}{\color{red}{2}} = \frac{k^0}{2} e^{-\alpha f \eta} c_{H^+}}$$

Guess the 2 factor is subsumed into the experimental $$\ce{k^0_{exp}}$$. Maybe I am missing something.

• I see your derivation makes sense... leaves me wondering if the factor of 1/2 is consumed in k_0. But one thing: Lets assume we have a similar problem and it is not an electrochemical one, but the reaction still has to go through the rate-determining step twice. Then we also have to add a factor of 1/2? Commented Jul 31 at 15:46
• I would think so. The 2 factor comes from the definition of reaction rate. In the case of the Tafel reaction, if you consider the formation of $\ce{H2}$, you would have $$-\frac{\ce{d|MH|}}{\ce{dt}} = \frac{1}{\nu_{\ce{H2}}} \ce{k_2 |MH|}$$ where the $\nu_{\ce{H2}}$ represent the stoichometric coefficient of MH in the reaction.
– PAEP
Commented Aug 1 at 11:46
• On the other hand, I think it is not an uncommon practice to subsume this factor into the rate coefficient (or rate constant) since you can not distinguish it experimentally .
– PAEP
Commented Aug 1 at 11:52
• MH is adsorbed hydrogen on metal surface, yes? Commented Aug 1 at 12:23
• Thanks a lot for the explanation Commented Aug 1 at 12:52