kinetics/rate question

[km1865]

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I just wanted to clarify the following:

If we have an overall reaction,

A + B ---> C + D

with a 3 step mechanism,
(Fast) A + B ---> X + H20
(slow) X---> Y + D
(fast) Y + H20 ---> C

why would adding water to the reaction change the rate of the overall reaction (and if it does change it how will it affect the reaction rate)? I dont see why water should have any effect on the reaction rate at all since only the concentration of reactants in the rate determining step should determine the rate of a reaction. since rds is rate= k [ x] since it is the rate limiting step, shouldn't only [x] affect reaction rate and not water since water doesnt go into the slow step? In one of their questions TBR says that water does affect the reaction rate in this reaction which is why im a little confused...

Also, if B in the reaction was OH- then why would increasing the pH affect the rate of the reaction, b/c similarly, B or (OH-) isn't involved in the rate determining step?

Im getting chatlier's confused with kinetics
Any help would be appreciated, thanks!

 

[plzNOCarribbean]

The reaction can only go AS FAST AS THE RATE determining step. Thats not to say that changing the concentration of the other species in the reaction, and in other steps does NOTHING to affect the overall rate of the reaction. You can only tell how much the rate will be affected by experimental data and specifically by changing the concentration of one of the reactants, while keeping the concentration of the other reactants constant. This way you can measure, quantitatively, the effect that a change in that species will have on the rate of rxn.

A chart is always provided that looks like this. When you look at this, notice that when the concentration of NO is kept constant between trial 1 and 2, while the H2 concentration doubles, the rate doubles. This shows that H2 is first order.

http://02.edu-cdn.com/files/static/mcgrawhillprof/9780071624770/AP_CHEMISTRY_PRACTICE_EXAM_2_07.GIF

 

[neurodoc]

I just wanted to clarify the following:

If we have an overall reaction,

A + B ---> C + D

with a 3 step mechanism,
(Fast) A + B ---> X + H20
(slow) X---> Y + D
(fast) Y + H20 ---> C

why would adding water to the reaction change the rate of the overall reaction (and if it does change it how will it affect the reaction rate)? I dont see why water should have any effect on the reaction rate at all since only the concentration of reactants in the rate determining step should determine the rate of a reaction. since rds is rate= k [ x] since it is the rate limiting step, shouldn't only [x] affect reaction rate and not water since water doesnt go into the slow step? In one of their questions TBR says that water does affect the reaction rate in this reaction which is why im a little confused...

Also, if B in the reaction was OH- then why would increasing the pH affect the rate of the reaction, b/c similarly, B or (OH-) isn't involved in the rate determining step?

Im getting chatlier's confused with kinetics
Any help would be appreciated, thanks!

Sounds like adding H20 should push the equilibrium to the left...note that this is an equilibrium product and not a rate issue.🙂

 

[km1865]

thanks. Also, I came across a question asking in which graph the intermediate would be in the highest concentration.. I chose the right answer, which was an energy diagram (free energy vs. reaction coordinate) of a non-concerted reaction in which the intermediate was the lowest energy... I just wanted to confirm that the other graph, in which the intermediate was of higher energy compared to the correct graph, would be the wrong answer BECAUSE the intermediate is higher in energy, and therefore would NOT be present in a higher concentration correct? Just making sure Im thinking of it in the right way..

 

[whiteshadodw]

thanks. Also, I came across a question asking in which graph the intermediate would be in the highest concentration.. I chose the right answer, which was an energy diagram (free energy vs. reaction coordinate) of a non-concerted reaction in which the intermediate was the lowest energy... I just wanted to confirm that the other graph, in which the intermediate was of higher energy compared to the correct graph, would be the wrong answer BECAUSE the intermediate is higher in energy, and therefore would NOT be present in a higher concentration correct? Just making sure Im thinking of it in the right way..

don't worry about it too much man, just move on. remember it and move on. there are some questions on there, while valid, they just focus too much much on details.

if you add water, you already decrease the concentration of x. that's the simplest way to look at it, but adding water also shifts the first fast reaction to the left, decreasing the concentration of x. its just small detail. if there is a question this detailed on the MCAT, there will be like one or two at most.

 

[MShopes]

I just wanted to clarify the following:

If we have an overall reaction,

A + B ---> C + D

with a 3 step mechanism,
(Fast) A + B ---> X + H20
(slow) X---> Y + D
(fast) Y + H20 ---> C

why would adding water to the reaction change the rate of the overall reaction (and if it does change it how will it affect the reaction rate)? I dont see why water should have any effect on the reaction rate at all since only the concentration of reactants in the rate determining step should determine the rate of a reaction. since rds is rate= k [ x] since it is the rate limiting step, shouldn't only [x] affect reaction rate and not water since water doesnt go into the slow step? In one of their questions TBR says that water does affect the reaction rate in this reaction which is why im a little confused...

Also, if B in the reaction was OH- then why would increasing the pH affect the rate of the reaction, b/c similarly, B or (OH-) isn't involved in the rate determining step?

Im getting chatlier's confused with kinetics
Any help would be appreciated, thanks!

You seem to be mastering most of the concepts here but you have a small gap between concepts. The rate determing step (slow step) does indeed determine the rate of the overall reaction but what is important to mention as well is that if the slow step is the first step in the reaction, then you are right that any other increase in concentrations of reactants or products in the other fast steps are irrelevant and has nothing to do with the rate of reaction. But if the slow step is not the first step (here is second step), then the increase in concentration of reactants or products of the fast step will affect the rate by increasing the concentration of reactants in the slow step and thus the rate of the slow step will increase which increases the overall reaction and vice versa. Think about it this way, if there is an increase in A or B concentration, then the rate of reaction will increase and the products concentration of the fast step will increase. Those products of the fast step are the reactants of the slow step and therefore the reactants concentrations of the slow step will increase, increasing the rate!

Sorry if I sounded confusing but yea that is the concept behind this. If you have any other questions, let me know!

 

[km1865]

cool, that makes it clear, thanks!