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# sn1 and sn2

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1. i dont understand what is meant by " when the reactant is in excess it is kept constant"
and therefore it is considered zero order in the rate equation...??

from my book:
----water acts as a nucleophile reagent , if the reaction is Sn2 the rate equation would be rate = k(RX)(H2O) ...... but water is in excess so h2o is effectively constant - so the rate equation would become : rate = k(prime) (RX) ,, this might lead you think the reaction was sn1-----

so WHAT if the water is in excess, molecules of water were still used in the rate determining stepand therefore water must be included in the rate equation since it is SN2!! and what is meant by kept constant if inexcess

2. It means there is so much water that the concentration change is negligible throughout the course of the reaction.
3. (Original post by alow)
It means there is so much water that the concentration change is negligible throughout the course of the reaction.
Aha okay, but water molecles are still being used in the rate determining step therefore shouldnt h2o meant to be present in the rate equation?????
4. The water concentration doesn't change* so stays constant. You already have K, the constant in the equation. Because you have two constants multiplied together, you can just say that it is a single constant, hence K x [H2O] = K'.
5. (Original post by pondsteps)
i dont understand what is meant by " when the reactant is in excess it is kept constant"
and therefore it is considered zero order in the rate equation...??

from my book:
----water acts as a nucleophile reagent , if the reaction is Sn2 the rate equation would be rate = k(RX)(H2O) ...... but water is in excess so h2o is effectively constant - so the rate equation would become : rate = k(prime) (RX) ,, this might lead you think the reaction was sn1-----

so WHAT if the water is in excess, molecules of water were still used in the rate determining stepand therefore water must be included in the rate equation since it is SN2!! and what is meant by kept constant if inexcess

When you're thinking about the rate of a reaction you're concerned with how the rate changes with changing reactant concentration.

rate = k[RX][H2O]
is the expression for a second order reaction (Sn2).
If you were to carry out this reaction in water as the solvent, then clearly water would be in vast excess. This means that the small amount of water used up in the reaction has a negligible effect on the concentration of water in the reaction vessel.
This means that the concentration of water is not a variable!
Just think, if you wanted to investigate how reaction rate changed with concentration of water, it would be impossible to change the concentration of water (in any significant way) if you were using it as the solvent. It would always be in huge excess!
This fact, that [H2O] is constant, means that it can be combined with the rate constant k. Seeing as two constants multiplied is just another constant that gives, what looks like a first order rate equation!
rate = k' [RX]

This is an example of a pseudo first order reaction. The water concentration is high, and, to all intents and purposes constant, making it impossible to tell if it is involved in the RDS by rate measurement alone!

Hope that makes sense
6. (Original post by pondsteps)
Aha okay, but water molecles are still being used in the rate determining step therefore shouldnt h2o meant to be present in the rate equation?????
The concentration of water stays effectively constant, so the reaction is approximately zero-order in H2O.

It's a (very good) approximation as the water is in excess.

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