Rate Kinetics

[ThiaminePyroPhosphate]

Random example:
X+Y -> A+G (fast)
A -> C + D (SLOW)
C+E -> F (fast)

So the rate law equation would be rate = k [A] - first order kinetics. Now what if a question asked what changes would affect the rate of the overall reaction? According to the textbooks, the reactants of the rate determining step (RDS) largely dictate the overall rate. What about changes made to the products of the RDS?

For example, lets say I add more C and D as the reaction occurs, wouldn't that slow the reaction down as it pushes the reaction to the left? Le Chatlier principle is applied during equilibrium, but can it also be applied in this case?

Furthermore, what about the amount reactants and products in the other steps? Let's say I increase the concentration of X and Y, therefore the reaction produces more A, thereby affecting the rate of the overall reaction.


Thank you.

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[NextStepTutor_1]

Yes, I think you're thinking about this well.
Le Chatelier's principal is for when the reaction is at equilibrium. This principal tells us the consequential result of adding or subtracting some variable in order for the reaction to re-attain the lost equilibrium.
In your example (and as you explain) the rate of the reaction would be directly dependent on the concentration of A. So while altering the concentration of A will absolutely change the rate of the reaction, it is wise of you to keep in mind how the concentration of A could be changed (i.e. altering X + Y -> A + G rxn.).


Hope this helps some.

 

[justadream]

@NextStepTutor_1

So to clarify, would you agree that:

If you add more C and D, the reaction would slow down

If you added more X and Y, the reaction speeds up.

 

[NextStepTutor_1]

If you add more X and Y, the rxn (presuming it was in equilibrium) will shift to the right increasing the amount of A. Increasing A will increase the rate of the reaction by the reaction rate equation.

However, increasing the amount of C and D is murkier to me. If this rxn was in equilibrium we'd see a shift to the left and an increase in the amount A. However, if this is not an equilibrium reaction (meaning the concentrations of the variables aren't of importance for the reaction, just their presence) then I'd say you'd still have the normal reaction occur just a larger end amount of C and D because you simply added more.

I'd say if this type of question comes up, pay special attention to their wording.

 

[ThiaminePyroPhosphate]

@NextStepTutor_1
Thanks for the explanation!

Just to make sure, the determined rate law equation of a reaction, e.g. r=k[A] will be true from the start of the reaction, and at equilibrium?

 

[NextStepTutor_1]

Yes, as long as that reaction maintains being the slowest step.