Kinetics - Two-Step Mechanism

[nothing123]

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Hi,

So if we have a two-step mechanism that goes something like this:

1) A + B -> C
2) C -> D
-------------
A + B -> D

and the second step is the slow step, the reaction rate would be r = k2[C] correct? Well, seeing as reactants A and B are zero order, varying their concentrations wouldn't affect the rate. However, one of the questions I encountered said if you ADD A or B or DECREASE A or B, the rate subsequently increases and decreases respectively. I understand the reasoning for this since more A or B would drive the reaction to the right and thus increasing the concentration of intermediate C, which is important in determining the rate. But really, what's the difference between varying the concentration of A or B and adding A or B; if we add A, are we not effectively increasing its concentration?

I gotta second question kind of related to this. We have a reaction A + B -> C with equilibrium constant: Kc = [C]/[A]. Now, if we add reactant A, BUT at the same concentration as the equilibrium concentration, then according to the equation, the reaction does not shift to the right, correct?

Thanks.

 

[soby10]

1st quest -are concentrations and adding reagents the same, the answer is yes as K is expressed in M, or partial pressure.
For the 2nd quest. use -le chatelier principle, addition of reagent will change concentration & system will correct itself. Add A will shift right.

 

[nothing123]

Thanks for your response but it didn't really address the questions I posed. Based on your first answer, if adding A is in fact equivalent to increasing the concentration of A, why would one affect the reaction rate and one not? My second question asks if adding A at the same concentration as its equilibrium concentration will shift the equilibrium.

Any other takers?

 

[CrimsonMirage]

For the rate law, when you 'vary the concentration', the assumption is that you are varying only that species concentration and the others are held constant.
For the equilibrium position, you must take into account concepts like Le Chatelier's, so that changing the concentration of one species would result in changes in other concentrations based on the reaction equation.

 

[TieuBachHo]

So if we have a two-step mechanism that goes something like this:

1) A + B -> C
2) C -> D
-------------
A + B -> D

and the second step is the slow step, the reaction rate would be r = k2[C] correct?
-YES.
Well, seeing as reactants A and B are zero order, varying their concentrations wouldn't affect the rate.
-True.
However, one of the questions I encountered said if you ADD A or B or DECREASE A or B, the rate subsequently increases and decreases respectively. I understand the reasoning for this since more A or B would drive the reaction to the right and thus increasing the concentration of intermediate C, which is important in determining the rate. But really, what's the difference between varying the concentration of A or B and adding A or B; if we add A, are we not effectively increasing its concentration?
-If the rate does not depend on the concentrations of the reactants as in zero-order rxn then it will not affect the rxn rate in step 1. You have more concentration of the product but you use more time to produce it.
In the enzymes and substrates analysis, there is a certain amount of substrates you can add until all the enzymes are saturated (called Vmax). After that, the rxn rate will remain constant as more substrates being added. If the relative percentage of substrates is less than its corresponding enzymes, then reaction rate varies depending upon the concentration of substrates.

****Edited: I have no idea what this question is asking? Just guessing. You might confuse the reaction rate with equilibrium constant.

I gotta second question kind of related to this. We have a reaction A + B -> C with equilibrium constant: Kc = [C]/[A]. Now, if we add reactant A, BUT at the same concentration as the equilibrium concentration, then according to the equation, the reaction does not shift to the right, correct?
-If you add A you shift the equilibrium to the right according to Le Chatelier's principle. The equilibrium constant will remain constant once the new equilibrium is establised assuming that the conditions are held constant. Now, whether you want the reaction favors more reactants or products, you can manipulate this by increasing or decreasing its temperature.