2
$\begingroup$

I read the answer to Rate Law vs Law of Mass Action and it's written in the answer:

In the rate law, the rate equation consists of actual factors that changes with change in experimental conditions. These may or may not be equal to stoichiometric factors. These are generally responsible for determining the order of reaction.

Both the laws are correct because both have a wide area of application. Rate law is although quite experimental in nature and is generally considered to be more appropriate.

Why do reactions not obey the law of mass action? And if that's the case, then how does it even have any applications? Shouldn't all problems be solved using the rate law only?

$\endgroup$
  • $\begingroup$ There is no special "rate law". A rate law is just an empiric mathematical description of a process. $\endgroup$ – Karl Sep 27 '17 at 8:16
13
$\begingroup$

The rate of mass action is true: the rate of reactions is proportional to the activity of its reactants.

On the other hand, from a macroscopic point of view, it is also true that reaction rates may not correspond to apparent stoichiometric proportions.

To understand this disparity, we need to refer to the concept of mechanisms.

The rate of mass action applies to elementary reactions, that is, to reactions that happen the way we write them at the microscopic level.

If the molecular, physical reaction consists on two molecules colliding and reacting, and we write it as

$$\ce{A + B -> C}$$

The rate of that reaction will, per the law of mass action, be

$$r = k · a_A · a_B$$

However, most macroscopic chemical processes aren't that simple, and in fact are the product of multiple steps; for instance, in the previous example, perhaps A undergoes a transformation into an intermediate which then goes on to collide with B to form C:

$$\ce{A -> D}$$

$$\ce{D + B -> C}$$

Macroscopically, we still have $\ce{A + B -> C}$, but the kinetics of the reaction will be very different.

For instance, imagine that the second step ($\ce{D + B -> C}$) is much faster than the first, so that any amount of D created reacts with B almost immediately to produce C. Then the first reaction will act as a bottleneck, and the overall reaction rate will be roughly the rate of the first step:

$$r = k_1 · a_A$$

If we do not know the underlying mechanism, this will be an apparent contradiction between the law of mass action (from which we could naively conclude that $r = k · a_A · a_B$) and the actual rate law ($r = k_1 · a_A$).

Much of the work on chemical kinetics consists in proposing mechanisms, testing these proposals (for instance, looking for experimental evidence that D exists as an intermediate) and deriving rate laws that are consistent with the law of mass action and experimental observation (which applies to elementary steps).

$\endgroup$

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.