[Rutherford]

To apply the rate-determining-step approximation take the reaction rate r as equal to the rate of the rate-determining step (divided by the stoichiometric number s_rds of the rate-determining step, if s_rds≠1).
What does the bold text mean? I need some explanation.

[Rutherford]

For example:
The rate of the reaction: I₂+2N₃^-+CS₂ 3N₂+2I^-+CS₂ is half the rate of the rate-determining step: CS₂+N₃^- CS₂N₃^-, so r=1/2k[N₃^-][CS₂]

In the rate-determining step one azide molecule participates, while in the overall reaction two molecules participate. Shouldn't then the rate be multiplied by two: r=2k[N₃^-][CS₂]?

[mjc123]

No. Think about it like this:
For each mole of reaction, 1 mole of I₂ is consumed, and 2 moles of N^3- are consumed. So the rate of consumption of N^3- is twice that of I₂:
d[N^3-]/dt = 2 d[I₂]/dt
So if we define the rate of reaction as the rate of consumption of I₂, then
rate = -d[I₂]/dt = -1/2 d[N^3-]/dt - i.e. you divide by the stoichiometric coefficient, not multiply.
So if the rate of the RDS is r' = -d[N^3-]/dt
then rate of reaction r = 1/2 r'.

I strongly dislike this "reaction rate r" language, though I suppose if your textbook or prof uses it, you have to go with it. I much prefer to talk about the rate of change of a specific reagent (e.g. -d[I₂]/dt), and keep this in view throughout the calculation. That's how we did kinetics when I learnt it, many moons ago. I think it is easier this way to avoid confusion.

[Rutherford]

I read that in a textbook.
Thanks for the explanation.
I like it most when I don't need to think about these things, i.e. they are included in the rate constant.

[mjc123]

Isn't thinking about things the reason for studying a science?
Generally, what I would want to say to some of the people who post problems on these forums is: think about what's going on physically, don't just blindly try to apply a formula. If they aim at understanding the science, not just passing an exam, isn't that what they would want to do anyway?

( By the way, I meant N₃^- of course, not N^3-!)