9
$\begingroup$

I am doing a chemistry review about rates, and I get asked "Which of these statements is false?"

The answer is

In reactions that are second order in one reactant and first order in another, the slow step generally involves a three-body collision of the reactants.

I imagine it looks like

$$ rate = k[A]^2[B]^1 $$

So at the slowest elementary step there should be $$ 2A + B = something $$

Which is a three way collision between two A molecules and a B.

What am I doing wrong?

EDIT:

Question number 2 (http://www.mesacc.edu/~paudy84101/CHM152F2005/Exam1%20Key.pdf)

This is not my review, in case you are wondering. I just have the same question.

$\endgroup$
  • $\begingroup$ Can you present the complete question? $\endgroup$ – Koba Jan 5 '16 at 0:52
14
$\begingroup$

Let's disprove that statement by counterexample. Here, the slow step is a collision of some intermediate and another thing, but the maths work out that the overall reaction order is third.

Let's have a two step reaction with the stoichiometry:

$$\ce{2A + B -> C}$$

The mechanism looks like this. The first step is fast and reversible and rapidly establishes equilibrium between $\ce{A + B}$ and an intermediate $\ce{I}$.

$$\ce{A + B <=>[k_1][k_{-1}] I}$$

The second step is slow, not so reversible and involves the intermediate $\ce{I}$ reacting with the other equivalent of $\ce{A}$.

$$\ce{I + A ->[k_2] C}$$

Now the rate of the reaction shall be written as the rate of appearance of $\ce{C}$, because it makes our lives a little easier.

$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=k_2 [\ce{A}][\ce{I}]$$

However, it's messy to have the intermediate in the rate law. Since the first step was fast and in equilibrium, we can get an expression for $\ce{[I]}$ in terms of $[\ce{A}]$ and $[\ce{B}]$.

$$K_{eq}=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{I}]}{[\ce{A}][\ce{B}]}\\ [\ce{I}]=K_{eq}[\ce{A}][\ce{B}]$$

A little substituting and we get:

$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=\dfrac{k_2 k_1}{k_{-1}} [\ce{A}]^2 [\ce{B}]=k_{obs}[\ce{A}]^2 [\ce{B}]$$

$\endgroup$
  • 1
    $\begingroup$ $\ce{NO}^\bullet$ autoxidation? :-) $\endgroup$ – hBy2Py Jan 5 '16 at 2:28
  • 2
    $\begingroup$ This illustrates nicely why time is important here. The chance of 3 reactants interacting at the same instance is virtually zero, but if two of them can collide to form an intermediate (I), that lasts for some small but finite period, then this has a chance to collide with another reactant (A) before it falls apart. If I is very short lived, i.e. $k_{-1} $ is huge then the rate tends to zero. $\endgroup$ – porphyrin Dec 3 '16 at 14:43

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.