2.2 Reaction Mechanism and the Rate-Determining Step

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Here is another quick little experiment you may want to try (and this is one you can safely do):

Find a friend, and each of you crumple up a piece of paper. Sitting facing one another, with knees almost touching, toss your pieces of paper together, trying to get them to hit while still in the air. Do this 10 times - how many successful "collisions" did you record?

Now, grab another friend with their own wad of paper. This time, a successful collision will only occur when all three pieces of paper hit together simultaneously - just two out of three won't do. Out of 10 tosses, how many are successful?

Add one more friend and repeat the exercise. Any successful collisions, where all four pieces of paper collide at the same instant? Likely not.

Chemical reactions follow the same principles. When we write a chemical equation such as:

\(\ce{2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O}\)

we see that two molecules of acetylene (C₂H₂) react with 5 molecules of oxygen. According to the collision theory, molecules must collide in order to react. But it is highly unlikely that 7 molecules would collide together all at once.

Instead, the reaction most likely occurs in a series of simple steps which only required two or three molecules colliding at any one instant. Although these steps cannot always actually be observed, chemists can often make predictions about the sequence of events.

For example, nitrogen monoxide reacts with oxygen according to the equation

\(\ce{2NO(g) + O_2 \rightarrow 2NO_2}\)

This reaction does not occur in a single step, however, but rather through these two steps:

Step 1:	\(\ce{2NO \rightarrow N_2O_2}\)
Step 2:	\(\ce{N_2O_2 + O_2 \rightarrow 2NO_2}\)

Notice that if you add these two reactions together, you end up with the overall reaction:

Step 1:

2 NO → ~~N₂O₂~~

The series of steps a reaction undergoes is called the

reaction mechanism

Step 2:

~~N₂O₂~~ + O₂ → 2 NO₂

Overall:

\(\ce{2NO(g) + O_2 \rightarrow 2NO_2}\)

Notice that in our example dinitrogen dioxide (N₂O₂) cancels out and does not appear in our overall equation. Substances such as this are called reaction intermediates and are typically short-lived.

Given an overall reaction, it will not be possible for you to predict what the reaction mechanism would be. However, if you are given the steps of a reaction mechanism you will need to be able to add together individual steps to end up with the overall reaction.

Rate Determining Step

Here is another reaction mechanism with some additional information concerning the relative rates of each of the individual steps:

Step 1:	\(\ce{HBr + O_2 \rightarrow HOOBr}\)	Slow
Step 2:	\(\ce{HOOBr + HBr \rightarrow 2HOBr}\)	Fast
Step 3:	\(\ce{HOBr + HBr \rightarrow H_2O + Br_2}\)	Fast
Step 4:	\(\ce{HOBr + HBr \rightarrow H_2O + Br_2}\)	Fast

Overall:	\(\ce{4HBr + O_2 \rightarrow 2H_2O + 2Br_2}\)

We don't have values for the actual rates of these individual steps, but here's a question for you to consider - would you consider the overall reaction to be fast or slow?

With two fast steps and only one slow step, many of you will predict that the reaction will be fast.

But let's make up some extreme numbers and ask the question again:

Step 1:	\(\ce{HBr + O_2 \rightarrow HOOBr}\)	1 year
Step 2:	\(\ce{HOOBr + HBr \rightarrow 2HOBr}\)	0.1 s
Step 3:	\(\ce{HOBr + HBr \rightarrow H_2O + Br_2}\)	0.1 min
Step 4:	\(\ce{HOBr + HBr \rightarrow H_2O + Br_2}\)	0.1 min

Overall:	\(\ce{4HBr + O_2 \rightarrow 2H_2O + 2Br_2}\)

Now would you consider the overall reaction to be fast or slow? Clearly it is a slow reaction, taking over a year to complete despite some fast steps.

The overall rate of any reaction depends on the rate of the slowest step. This slowest step is called the rate determining step. If you want to speed up a reaction, this is where you should focus your attention.

The slowest step of a reaction mechanism is the

rate determining step