[DannyBoy]

Apologies in advance for the newbie question.

Say I have a basic reaction A+2B->P that is governed by the rate law R=kAB, then dA/dt=-kAB while dB/dt=-2kAB (right so far?)

My question: presumably I can write 0.5A+B->0.5P in which case the reaction rates would be half, that is, dA/dt=-0.5kAB & dB/dt=-kAB, despite being essentially the same reaction. Is it that k for the second reaction would be double that of the first reaction?

I have obviously misunderstood something fundamental here and I'd appreciate it if someone could point it out to me...

[Astrokel]

Are you sure the reaction rate would be halved?

[flightman233]

Might be mistaken here, but shouldn't your original rate law be R=kAB² ?  Need to account for the stoichiometry here.

[Astrokel]

Might be mistaken here, but shouldn't your original rate law be R=kAB² ?  Need to account for the stoichiometry here.

No, order of reactants are not determine by its stoichiometric coefficients but rather experimentally.

Kelvin