2
$\begingroup$

Recently I heard my teacher say that when increasing the total pressure of the system, the reaction which is "more dependent on collisions" is favoured.

This seems to make sense, although I've looked at many resources online and on SE, and people have either given a mathematical explanation or simply talked about Le Chatelier's principle (increasing pressure favours reaction to produce less moles)

My question is: is it correct to say that increasing the pressure favours the reaction which involves more molecules colliding?

Take for example the forward reaction here which is favoured upon pressure increase. (4 molecules need to collide for forward and 2 molecules need to collide for reverse reaction)

$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$

I'd appreciate an intuitive explanation about collisions and interactions between molecules rather than something mathematical. Thanks in advance!

$\endgroup$
4
  • 4
    $\begingroup$ Be aware molecularity of a particular reaction step ( number of colliding molecules ) is not the same as the sum of the stoichiometric coefficients of reactants. Reactions, where 4 molecules have to collide, are very rare and their rate must be minimal unless under high pressure. But what you teacher says is rather right, even if said by somewhat obfuscated way. The side where is more molecules, has steeper dependence of the reaction rate with the increasing pressure. Also, if youwrite the equilivrium equation, you can see which way the reaction shifts with pressure. $\endgroup$
    – Poutnik
    Jul 8 '20 at 11:54
  • 2
    $\begingroup$ The concentration of a species in an ideal gas is directly proportional to the pressure, which means increasing the pressure (while keeping mole fractions constant) will increase concentrations. $\endgroup$
    – Buck Thorn
    Jul 8 '20 at 12:00
  • $\begingroup$ reaction shift in the direction,where no. of gaseous molecules are more $\endgroup$
    – Jack Rod
    Jul 8 '20 at 12:53
  • 1
    $\begingroup$ Your teacher might have a weird way of expressing himself, or not know what he´s talking about, in any case that statement is imo not worth giving it another thought. Collisions are important for kinetics, not for thermodynamics, at least not on a high school level. $\endgroup$
    – Karl
    Jul 8 '20 at 18:49
2
$\begingroup$

Yes, I agree with your teacher's comment that upon increasing the total pressure of a system, there are select reactions "more dependent on collisions", and such reactions have been a matter of study.

For example, per this 2008 work reported in 'The Journal of Physical Chemistry', The Temperature and Pressure Dependence of the Reactions H + O2 (+M) → HO2 (+M) and H + OH (+M) → H2O (+M), where '+M' here denotes the catalytic presence of a surface, as can be provided by another gas molecule.

To quote from the abstract:

The reactions $\ce{H + O2 (+M) → HO2 (+M)}$ and $\ce{H + OH (+M) → H2O (+M)}$ have been studied using high-level quantum chemistry methods. On the basis of potential energy hypersurfaces obtained at the CASPT2/aug-cc-pVTZ level of theory, high-pressure limiting rate coefficients have been calculated using variable reaction coordinate transition state theory. Over the temperature range 300−3000 K, the following expressions were obtained...Available experimental data on the pressure dependence of the reactions were analyzed using a two-dimensional master equation.

Note, reactions are being studied via surface collisions involving the addition of Ar and N2 gases. The selection of the gas relates to chemical inertness and the total collisional cross-section of the particular gas particles, all of which are components of more complex theories surrounding collision frequency and collision theory (see, for example, discussion here).

$\endgroup$
1
$\begingroup$

I know you wanted something non-mathematical, but I think the best way to understand this intutively requires some simple math.

Suppose you have the following elementary reaction:

$$\ce{A(g) + B(g) <=> C(g)}$$

By "elementary reaction" I mean that this shows the actual reaction mechanism, such that the rate equation can be obtained directly from it.

Let the rate constants for the forward and reverse reactions be $k_f$ and $k_r$, respectively.

Since A and B are produced from C, the rate at which A and B are produced is proportional to [C]. And since A and B are consumed by a biomolecular reaction involving both A and B, the rate at which each is consumed is proportional to $\ce{[A]\times[B]}$. We can thus express the rate at which [A] and [B] change as follows:

$$\ce{\frac{d[A]}{dt}=\frac{d[B]}{dt}}=k_r[C]-k_f[A][B]$$

And, using the same reasoning for C, we have: $$\ce{\frac{d[C]}{dt}=k_f[A][B]-k_r[C]}$$

At equilibrium, the rates of the forward and reverse reactions are equal, i.e.:

$$\ce{k_f[A][B]=k_r[C]}$$

Let's suppose you are at equilibrium, and suddenly double the pressure (before the reaction has a chance to respond). In that case, the rate of the forward reaction ($\ce{k_f[A][B]}$) will initially increase four-fold (because both [A] and [B] double, and $2 \times 2 = 4$), while the rate of the reverse reaction ($\ce{k_r[C]}$) will only double. Consequently the reaction will shift to the right (towards the product side). Thus you can see what your teacher (probably) meant when he or she said that the reaction shifts away from the side more dependent on collisions—because the rate at which the species on that side are consumed will be greater.

Eventually, the reaction will requilibrate. The equilibrium constant will be unchanged, but the relative concentrations of species will have changed to favor the product.

Note this would not apply to liquid-phase reactions, since an increase in pressure will have only a negligible effect on concentration. In this case, you need to consider the relative volumes of the reactants and products.

$\endgroup$

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.