Question

When does stoichiometric coefficient become the order of the reaction?

Answer 1

This confusion pops up often around here.

The problem usually comes down to insufficient consideration of units.

When you say "rate", what do you mean? It has units of mol per volume per second, e.g. molL⋅s. But that isn't enough information. You need to know moles of what.

First example: if the rate is defined as loss of A

For example, if you say that the "rate" means the disappearance of A, then the units of the rate are really mol of AL⋅s. If so, then as long as:

The reaction is 2A⟶B; and

The rate as you have defined it can be expressed by =1 [A]2 ...

Then the rate of appearance of B will be =121 [A]2

Second example: if the "rate" is defined as appearance of B

If conversely you suppose that the "rate" refers to the appearance of B, then the units of rate are really mol of BL⋅s, and if so, then as long as

The reaction is 2A⟶B; and

The rate as you have defined it can be expressed by =2 [A]2 ...

Then the rate of disappearance of A will be =22 [A]2

Third example: a totally separate definition of rate

You say its an elementary reaction. Let's say that we didn't know that and wrote the reaction as 200A⟶100B. The rate, according to this new example definition, is the rate of the reaction. The units of the rate are mol of "reaction"L⋅s. If the reaction is still 2nd-order then the rate of the reaction will be

=3 [A]2

According to our reaction, 200 moles of A disappear for each mole of reaction, so

=2003 [A]2

and 100 moles of B appear, so

=1003 [A]2

Final thoughts

You will notice that I used different subscripts to distinguish between the values for the different assumptions. That's because they aren't equal to each other. The value of the "rate constant" depends on how you define the rate.

No matter how we define "the rate" and what we define as the stoichiometry (e.g. 2A⟶B or A⟶12B or 200A⟶100B, the rates must be equal. Therefore:

=2003 [A]2=22 [A]2=1 [A]2

That is,

2003=22=1

But no matter which way we decide to write it, it is still a second-order reaction because "the rate", no matter how we define it, is proportional to [A]2.

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