I have the reaction:

$$\ce{Hb + O2 <=>[$k$][$k'$] HbO2}$$

where $\ce{Hb}$ is hemoglobin, $k$ is the rate constant of the forward reaction, $k'$ is the rate constant of the inverse reaction. On my book there is the rate law:

$$W = kP(\ce{O2})[\ce{Hb}] - k'[\ce{HbO2}]$$

where $W$ is in $\pu{mol m-3 s-1}$, $P(\ce{O2})$ is a partial pressure in Pascal ($\pu{kg m m-2 s-2 = kg m-1 s-2}$).

The inverse reaction is a first order reaction because:

$$\pu{mol m-3 s-1} = [k'] (\pu{mol m-3})$$

thus $[k'] = \pu{s-1}$.

I am in trouble with the forward reaction because I got a strange unit of measurement:

$$\frac{\pu{mol}}{\pu{m3 s}} = [k] \frac{\pu{kg}}{\pu{m s2}} \frac{\pu{mol}}{\pu{m3}}$$

$$[k] = \frac{\pu{m s}}{\pu{kg}}$$

Which is the order of the forward reaction?

  • 1
    $\begingroup$ This looks like homework, and not exactly good one. Haemoglobin is tetrameric and therefore binding of oxygen is 4 step process with 4 (significantly different) constants. $\endgroup$
    – Mithoron
    Dec 27 '17 at 23:03
  • $\begingroup$ Hello @Mithoron, my reaction represents the first step. $\endgroup$ Dec 28 '17 at 7:11

You seem to be overthinking this a great deal. The order of a reaction is not defined by the units of the rate constant, but rather by the sum of the exponents in the empirical rate equation. From the IUPAC Gold Book:

If the macroscopic (observed, empirical or phenomenological) rate of reaction ($v$) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $k[\ce{A}]^\alpha[\ce{B}]^\beta\cdots$ [...] then the reaction is said to be of [...] (total or overall) order $n = \alpha + \beta + \cdots$

Considering that the partial pressure of $\ce{O2}$ is directly proportional to its concentration, the forward reaction has a rate equation


where the proportionality factor between $P(\ce{O2})$ and $[\ce{O2}]$ is absorbed into the new rate constant $k'$. The order is simply 2.

  • $\begingroup$ Excellent reply @orthocresol! Thank you very very much! $\endgroup$ Dec 28 '17 at 15:57

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