What is the reaction order of SN1

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The discussion centers on the rate-determining step of SN1 mechanisms, with a debate on whether it is first order or zeroth order. One participant argues that since SN1 is unimolecular substitution, the rate expression should be first order, as only the substrate is involved in the reaction, while the solvent merely modifies the equilibrium constant. In contrast, the teacher claims it is zeroth order because the leaving group can leave spontaneously in the right solvent, independent of its concentration. However, this view is challenged, as spontaneity does not accurately describe reaction order. The consensus leans towards first order for SN1, emphasizing that while solvent effects can influence reaction rates, they do not change the fundamental order of the reaction. The importance of the rate-determining step is highlighted, noting that if the formation of the carbocation were spontaneous, it would alter the overall reaction dynamics, potentially leading to a second-order reaction, contradicting the zeroth-order claim.
shredder666
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Sorry the title is bit misleading, I'm actually talking about the rate determining step of SN1 mechanisms. I thought it was first order, but my teacher disagrees.

he argues that it is zeroth order because in the appropriate solvent, the leaving group leaves spontaneously regardless of its concentration, thus the rate expression

rate=k[substrate]^0

my argument is that first, it is unimolecular substitution, thus the name SN1 (I'm only talking about the rate determining step, not the whole reaction).
Second only the substrate was involved in the reaction itself. The solvent just modifies the K value. Thus the rate expression
rate=k[substrate]^1

I do realize that a lot of texts (that I know of) tends towards my argument, but please ignore it for now, for most mechanisms are only proposed.
thank u
 
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As a your teacher says, that the leaving group will leave spontaneously in the appropriate solvent. So that isn't good enough to generalisation that all SN1 reactions are of zeroth order.
But let's assume that the Formation of the carbocation,i.e, the rate-determining step, is spontaneous. Then it will no longer remain the slowest step. It will, in fact, proceed faster than the second step of the reaction. And automatically the second step will become the slowest step and determine the rate. So, you end up with a second order reaction, anyway. A total contradiction.
Further, The rate of a reaction depends on all the steps in the reaction. If you have a step which is prominently slower than all others, it becomes the rate determining step, because here, the rates of the other steps do not greatly affect the rate of the overall reaction. If the rates of the steps involved are comparable, you will have a complicated rate law equation. So even by this argument you may get a complicated rate law, but zero order isn't probable.
 
shredder666 said:
...he argues that it is zeroth order because in the appropriate solvent, the leaving group leaves spontaneously regardless of its concentration, thus the rate expression...

If that is what your teacher said, then teacher is wrong. Spontaneity is not the correct term to use in a reaction order or rate of reaction discussion. How the concentration of the species that is ionizing affects the overall rate of the reaction is the only pertinent point to discuss.

It is well known that solvent effects affect the rate of reaction but not the reaction order. A high dielectric solvent might lower the energy barrier leading to the reactive intermediate and thus increase the rate, but how the concentration of that species affects the overall rate remains unchanged... first order for SN1.
 
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