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Offline numbersixman

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Kinetics Lab Question
« on: August 26, 2012, 04:01:04 PM »
The following data was collected at 20 degrees Celsius.



Determination 1,      [HBPB^2-] = 7.22x10^-6 mol/L,        [OH^-] = 1.00 mol/L,                   time = 75 s
determination 2,      [HBPB^2-] = 7.22x10^-6 mol/L,        [OH^-] = 0.25 mol/L,                   time = 290 s
Determination 3,      [HBPB^2-] = 3.63x10^-6 mol/L,        [OH^-] = 1.00 mol/L,                   time =152 s



Assume that the small, constant amount of HBPB^2- consumed in the experiment described above corresponds to an HBPB^2- concentration change of 7.22x10^-7 mol/L. Find the rate constant for this reaction.

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I'm pretty confused by this question. I'm getting conflicting information that isn't helping, so some clarification would be greatly appreciated.


With respect to HBPB, I get a -1 order reaction:

7.22x10^-6/3.63x10^-6 = 1.99^m         =  75/152 = .493

Solving for m gives -1. The numbers are slightly off, but a -1 order is the closest I can possibly get.


With respect to OH:

1.00/.25 = 4^m                   75/290 = .258

Again, solving for m gives -1.

This gives a rate law of: rate = k[HBPB]^-1[OH]^-1


Assuming I've done everything right up to this point, this is where I am confused by the question. I substituted the concentration change of 7.22x10^-7 M and dived that by the given times for each experiment, which gives me a reaction rate for each experiment. But when I plug the numbers in, using -1 orders for both HPBP and OH, I get different rate constants for each one, and I know this shouldn't be the case.


I realize this question is incredibly involved, but I have no clue what I'm doing wrong. Any help would be much appreciated.

Offline ramboacid

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Re: Kinetics Lab Question
« Reply #1 on: August 26, 2012, 09:30:45 PM »
The concentrations aren't raised to -1, but rather just 1. Remember:
(reaction rate) = (amount of reactant consumed) / (time)

The method you are using in your calculations is for an equation relating
(reaction rate) = k([HBPB]x)([OH]y)

but since it's time that you're given and not reaction rates, you have to substitute (amount of reactant consumed) / (time) for (reaction rate):

(amount of reactant consumed) / (time) = k([HBPB]x)([OH]y)

Now look at your data and form a ratio using data for two reactions in which one of the reactants has the same concentration. For our purposes let's use determinations 1 and 2, where HBPB is constant.

(amount of HBPB consumed) / (time1)       k([HBPB]x)([OH]1y)
-------------------------------------- = -----------------------
(amount of HBPB consumed) / (time2)       k([HBPB]x)([OH]2y)

Notice that since [HBPB] is constant in both the numerator and denominator, the amount of HBPB consumed is also constant for both the numerator and denominator since the reactions go to completion. Now we can cancel out like terms in the numerator and denominator and simplify:

(time2)      [OH]1y
------- =  ----------
(time1)      [OH]2y

time2 over time1 is approximately 4. [OH]1 over [OH]2 is also 4. Now we write out the equation again:

4 = 4y

and we see y=1. Therefore the order of the reaction with respect to hydroxide is 1. You can repeat the same thing with your other determinations to find out the reaction order for HBPB, though this method will become more intuitive and you won't have to write out all the steps.

Hope this helps!
"Opportunity is missed by most people because it is dressed in overalls and looks like work." - Thomas Edison

Offline numbersixman

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Re: Kinetics Lab Question
« Reply #2 on: August 26, 2012, 10:07:25 PM »
Okay, I see what you did.

I managed to get as far as doing (reactant consumed)/(time) to get the reaction rates, but since I did the initial step of finding the reaction order wrong, it threw me off.

For some reason we were assigned the harder of two similar kinetics labs first, and this is infinitely more complicated than the one we're doing later.

Thanks again for the help.
« Last Edit: August 26, 2012, 10:28:45 PM by numbersixman »

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