3
$\begingroup$

If we have a zero order complex reaction (although uncommon in chemical practices), say the decomposition of ammonia:

$$\ce{2NH3 -> N2 + 3H2}$$

This is a zero order complex reaction. In my textbook, it says:

For complex reaction, the order is given by the slowest step and molecularity of the slowest step is the same as the order of the overall reaction. (sic)

At the same time, molecularity of an elementary reaction cannot be zero.

How is this possible and what am I missing?

$\endgroup$
4
$\begingroup$

Terminology There are two term which you must understand and use carefully.

  • Molecularity - The [integer] number of reactant molecular entities that are involved in the 'microscopic chemical event' constituting an elementary reaction. (From IUPAC Gold Book)

  • Order of reaction - 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 $\ce{k = [A]^\alpha[B]^\beta\text{ ...}}$ (expressing in full the dependence of the rate of reaction on the concentrations A, B, ...) where $\alpha$, $\beta$ are constant exponents (independent of concentration and time) and k is independent of A and B etc. (rate constant, rate coefficient), then the reaction is said to be of order $\alpha$ with respect to A, of order $\beta$ with respect to B, ... , and of (total or overall) order $n = \alpha +\beta + ...$ The exponents $\alpha$, $\beta$, ... can be positive or negative integral or rational nonintegral numbers.

Equilibrium reaction

For a reaction like the decomposition of ammonia you could write a bimolecular reaction like:

$$\ce{2NH3 -> N2 + 3H2}\tag{1}$$

Often chemists consider such a reaction reaching equilibrium such as:

$$\ce{2NH3 <=> N2 + 3H2}\tag{2}$$

where there is a forward reaction which (naively) could be consider as the forward reaction

$$\ce{2NH3 ->[$k_\mathrm{f}$] N2 + 3H2}\tag{2.a}$$

with the forward rate given by $r_\mathrm{f} = k_\mathrm{f}[\ce{NH3}]^2$ and the reverse reaction

$$\ce{N2 + 3H2 ->[$k_\mathrm{r}$] 2NH3}\tag{2.b}$$

with the reverse rate given by $r_\mathrm{r} = k_\mathrm{r}\ce{[N2][H2]}^3$.

However, if you consider the reverse reaction, you can see that there are four molecules involved. It is highly unlikely that four molecules would ever collide in just the right way to cause a reaction. So the reverse reaction occurs in multiple steps with intermediate species involved. The exponents $x$ and $y$ would be determined experimentally and would almost certainly not be integer values. Thus more realistically:

$$r_\mathrm{r} = k_\mathrm{r}[\ce{N2}]^x[\ce{H2}]^y\tag{3}$$

Catalytic reaction

Now, back to the more salient point of the question. The decomposition of ammonia is typically not done just by heating ammonia to some high temperature to form an equilibrium, but rather it is done selectively with a catalyst and a lower temperature.

$$\ce{2NH3 ->[catalyst] N2 + 3H2}\tag{4}$$

The catalytic surface absorbs ammonia preferentially (and very quickly compared to the reaction time of the ammonia decomposition reaction) and can only absorb a small amount of ammonia. Hence the rate of decomposition doesn't depend on the concentration of ammonia so the rate of decomposition is essentially constant until almost all of the ammonia is consumed. So with a catalyst there is a zero order kinetic reaction with the equation:

$$r_\mathrm{f} = k_\mathrm{f}\tag{5}$$

$\endgroup$
  • $\begingroup$ So, to recap: since the decomposition of Ammonia is conducted with a catalyst that is inherently preferential to the amount of Ammonia that decomposes on its surface, and which is why the concentration of Ammonia is irrelevant, we cannot talk of its molecularity. Am I correct? $\endgroup$ – Aditya Shukla Jun 26 '18 at 9:09
  • $\begingroup$ Not quite. The catalyst reaction has a molecularity of zero. $\endgroup$ – MaxW Jun 26 '18 at 12:16
  • $\begingroup$ How is that possible? Molecularity cannot be zero or a non-integer value. $\endgroup$ – Aditya Shukla Jun 26 '18 at 13:45
  • $\begingroup$ There is molecularity and there is kinetics. So if A + B -> C, the reaction is bimolecular. But rate for c, $r_c = k_f[A]^x[B]^y$ where x and y are not necessarily integers. $\endgroup$ – MaxW Jun 26 '18 at 15:20
  • 3
    $\begingroup$ @MaxW Process, has zeroth order kinetics, because it's not reaction, but diffusion controlled. Actual reaction is multistep and definitely doesn't have molecularity of zero. $\endgroup$ – Mithoron Jun 26 '18 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.