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At page 556 I found this question being analyzed and the rate law is provided. However, when I apply steady state approximation, I can't find the result without the factor 2. Is it me being wrong?

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    $\begingroup$ Edition number? $\endgroup$ – orthocresol Mar 23 at 19:19
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    $\begingroup$ I guess I'm a dolt, but I don't understand "I can't find the result without the coefficient 2." $\endgroup$ – MaxW Mar 23 at 21:03
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    $\begingroup$ It's edition 11. $\endgroup$ – David Li Mar 24 at 5:14
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The factor of 2 originates from the definition of $k_b$ (I thought it was an error, but it turns out IUPAC guidelines lead to its presence).

If $k_\mathrm{a}' \gg k_\mathrm{b}[\ce{O2}]$ and defining

$$K_\mathrm{a} = \frac{k_\mathrm{a}}{k_\mathrm{a}'}$$

it follows from the equation posted in the OP that

$$\frac{\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = 2k_\mathrm{b}K_\mathrm{a}[\ce{NO}]^2[\ce{O2}]$$

However, starting from scratch we could write that

$$\frac{\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = k_\mathrm{b}[\ce{N2O2}][\ce{O2}]$$

(this is not the standard IUPAC convention however). Then, if

$$K_\mathrm{a} = \frac{k_\mathrm{a}}{k_\mathrm{a}'} = \frac{[\ce{N2O2}]}{[\ce{NO}]^2}$$

it follows that

$$\frac{\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = k_\mathrm{b}K_\mathrm{a}[\ce{NO}]^2[\ce{O2}]$$

and the factor of $2$ is not there. In the fourth version of Atkins' book (which I parallel in this derivation from "scratch") the factor of 2 is missing, but in that book the analysis does not proceed by dissecting the full rate law before imposing the pre-equilibrium approximation. The OP presumably sources a later version of the book. As noted in a comment, IUPAC guidelines call for incorporation of that factor of 2. One should write

$$\frac{\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = 2k_\mathrm{b}[\ce{N2O2}][\ce{O2}]$$ because that step of the reaction reads $$\ce{N2O2 + O2->2NO2}$$ The stoichiometric factor of 2 means two moles of $\ce{NO2}$ are produced for every mole of reaction.

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    $\begingroup$ Um. It sort of depends on how you define the rate equation for the third reaction in the series; if you define it instead as $$\frac{\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = 2k'_\mathrm{b}[\ce{N2O2}][\ce{O2}]$$ then you end up with the correct factor of 2. That simply necessitates that my rate constant $k'_\mathrm b = k_\mathrm b / 2$. In fact, this is more in line with IUPAC: goldbook.iupac.org/html/R/R05156.html There's a question about this on SE, but I don't have the time to find it now. $\endgroup$ – orthocresol Mar 24 at 15:15
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    $\begingroup$ @NightWriter I think it would be helpful if you edited your answer to indicate more clearly that including the factor of 2 is in fact the standard convention and not an error, rather than just acknowledging that at the end. $\endgroup$ – Andrew Mar 26 at 12:46
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    $\begingroup$ @orthocresol This one? $\endgroup$ – hBy2Py Mar 26 at 15:34
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    $\begingroup$ @hBy2Py I was thinking of this one chemistry.stackexchange.com/q/38167/16683, but yours fits as well and I wouldn't be surprised if there are even more. $\endgroup$ – orthocresol Mar 26 at 15:55
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    $\begingroup$ @orthocresol Ah, yeah, yours looks rather more canonical and detailed than mine. $\endgroup$ – hBy2Py Mar 26 at 17:13

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