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# How is it possible to have 0 order?

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1. I thought that according to collision theory the greater the conc, the more collisions that overcome the activation energy per second hence the faster the reaction.
Zero order means that no matter how much you increase the conc of the reactant that is in the zero order there will absolutely no effect on the rate. How can this be?!
Does it mean that not enough of the reactants in the zero order will have the sufficient energy required to overcome the activation energy hence no affect on rate?

why the negative rating? :s
2. (Original post by A Perfect Circle)
I think it simply means that the substance which has zero order plays no part in the reaction whatsoever; it is just a bystander. This explains why collision theory is irrelevant here.

So you can add a whole load of it or you can take a load of it away, but it won't affect the reaction that this is referring to.
ahhh okay thanks, so whys it in the overall equation if its not doing anthything?
3. (Original post by A Perfect Circle)
Just because it's not doing anything, it doesn't mean it's not there.
so why would you want them there? its like inviting someone to a party and them sitting in the shed all night drinking bottles water and listening to classical music on their own... whats the point in them even being there?
4. (Original post by asaaal)
I thought that according to collision theory the greater the conc, the more collisions that overcome the activation energy per second hence the faster the reaction.
Zero order means that no matter how much you increase the conc of the reactant that is in the zero order there will absolutely no effect on the rate. How can this be?!
Does it mean that not enough of the reactants in the zero order will have the sufficient energy required to overcome the activation energy hence no affect on rate?
It is because that reaction has several steps. The step that substance is involved in is very fast compared to another step which is very slow. Hence the rate of the slow step controls the overall rate and increasing the rate of the fast step is pointless.
5. (Original post by A Perfect Circle)
I think it simply means that the substance which has zero order plays no part in the reaction whatsoever; it is just a bystander. This explains why collision theory is irrelevant here.

So you can add a whole load of it or you can take a load of it away, but it won't affect the reaction that this is referring to.
That is not true. See above.
6. (Original post by illusionz)
It is because that reaction has several steps. The step that substance is involved in is very fast compared to another step which is very slow. Hence the rate of the slow step controls the overall rate and increasing the rate of the fast step is pointless.
Is that why we never write zero order in the rate equation hence its never written in the rds as its always in the quickstep?
7. (Original post by asaaal)
Is that why we never write zero order in the rate equation hence its never written in the rds as its always in the quickstep?
On the contrary, it is quite normal to write the components with zero order in the rate equation:

rate = k[H2O2]1[I2]1[H+]0

As stated above components (reactants) with zero order are involved in mechanistic steps that are too fast to have a measurable effect on the rate of the reaction.

A good example of this is the nucleophile in tertiary halogenoalkane substitution:

R3C-X + OH- --> R3C-OH + X-

Rate = [R3C-X]1[OH-]0

Mechanism:

STEP 1: R3C-X <==> R3C+ + X-
STEP 2: R3C+ + OH- --> R3C-OH

STEP 1 is the rate determining step and STEP 2 is the fast step.

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