TBR: Rate Law Question

[justadream]

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TBR GC Book II page 241 #71



Step I: H2O + CO => H2CO2



Step II H2CO2 => H2 + CO2



“If Step II is the rate-determining step, than all of the following predictions are valid except”

A. Labeled Oxygen in CO would not be found in CO2 [CORRECT – I agree]

B. Adding H2O would increase the rate, because [H2CO2] would increase

C. Removing CO would decrease the rate, because [H2CO2] would decrease



Choices B and C are incorrect (that means, they are TRUE)



The answer makes sense intuitively but how can you reconcile it with the rate-law?



The rate-law for the overall reaction would be: rate = k[H2CO2]



It does not have any value for H2O or CO.

 

[Czarcasm]

TBR GC Book II page 241 #71



Step I: H2O + CO => H2CO2



Step II H2CO2 => H2 + CO2



“If Step II is the rate-determining step, than all of the following predictions are valid except”

A. Labeled Oxygen in CO would not be found in CO2 [CORRECT – I agree]

B. Adding H2O would increase the rate, because [H2CO2] would increase

C. Removing CO would decrease the rate, because [H2CO2] would decrease



Choices B and C are incorrect (that means, they are TRUE)



The answer makes sense intuitively but how can you reconcile it with the rate-law?



The rate-law for the overall reaction would be: rate = k[H2CO2]



It does not have any value for H2O or CO.

This is one of those odd-ball scenarios where the slow-step is actually not the first step (as typically presented) but rather, the second step of the mechanism. Usually, when we over simplify that the overall rate of the reaction equals the rate of the slow step, it's generally when the slow step is also the first step. However, in this scenario, that's not the case. The first step, even though it's not the slow-step is essential and still relevant, because for the slow step to even occur, the products of step 1 are the reactants of the slow-step. Ultimately, what we do in this scenario is assume that the first step occurs so quickly relative to the slow step (we assume it reaches equilibrium). We then include this in the rate law by basically taking the equilibrium constant expression (which equals rate forward/rate reverse for reaction 1) and then use that expression in substitution for one of the variables in the slow-step. It would look something like this:

Step I: H2O + CO => H2CO2 -- Rate: Keq (ratef/rater) = [H2CO2]/[CO][H2O]
Step II H2CO2 => H2 + CO2 -- Rate: kf[H2CO2]

The rate law for step Ii is: kf[H2CO2] but then using the expression above to substitute for [H2CO2], we get: Keq[CO][H2O] = [H2CO2]
So the overall expression would be: kf x Keq[CO][H2O]. ( Some professors go further and sum kf x Keq into one constant variable).

 

[JMMTB]

Another way to approach this question is applying Le Chatlier's Principle.
Because the first step is fast Q is near Keq and shifting the concentrations of the reactants will change the concentrations of the product (H2CO2).
If we add H2O or CO we will have more H2CO2. H2CO2 is in the rate law. So the rate goes up as its concentration goes up.
And the reverse is true too, remove H2O or CO and [H2CO2] falls.

 

[justadream]

@Czarcasm

Why do you use Keq for the first step? Shouldn't you be using rate-laws (which only look at reactants)?

Here's another approach used by TBR which I have generalized:

A + B ==> Intermediate + C [SLOW STEP]
Intermediate ==> Product [FAST STEP]

Let k1 be the rate constant for the top step (and k-1 be the rate constant for the reverse)
Let k2 be the rate constant for the second step

We can write:
k1[A] = k-1[Intermediate][C] + k2[Intermediate]