Quick rate law question

[September24]

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Lets say the slow step is:
aA+bB—>cC

Is the rate law simply: Rate=k1[A]ab

I thought that this "technique" could only be used for the elementary step (first step) and all other rate laws had to be determined experimentally.

 

[September24]

Sorry formatting was messed up

I meant rate law is : Rate=k[A]^a^b

 

[Willy38]

That only works for reaction quotients and equilibrium constants (or solubility products). Rate is always experimentally determined. BUT it can be used in this case because you know what the "slow" step (Rate Determining Step) is. It has nothing to do with what number step it is in the reaction, it's totally dependent upon which step determines rate.

Sorry formatting was messed up

I meant rate law is : Rate=k[A]^a^b