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The question: In the lab described below, why did the decolorization reaction described, which should be first order in each of the two reactants, go most slowly when the reactants were most concentrated?

As part of the introduction to our kinetics unit, I asked my students for factors that they thought would influence the amount of time it takes a reaction to occur. Not surprisingly, they came up with a variety of things, including temperature and the concentration of reactants. I set up a simple lab to compare temperature and concentration conditions based on the reaction of crystal violet with sodium hydroxide. Later in the unit we will come back to the reaction and analyze its kinetics more carefully, but for the moment, I just wanted my students to do a more basic analysis.

The students did trials for 4 conditions, and each involved the same quantity of reactants: 25 ml of 0.10 mM crystal violet and 25 ml of 1.0M sodium hydroxide. The reaction was performed by adding the crystal violet to a flask, adding water to dilute, adding the sodium hydroxide to the flask, stirring, and timing the reaction until it no longer showed visible purple. (Yes, that's a sloppy endpoint, but it was sufficient for this activity's purposes.)

Trial 1: 25 ml crystal violet, 150 ml room temp distilled water, 25 ml sodium hydroxide (a 1:8 dilution of stock reactants)

Trial 2: 25 ml crystal violet, 150 ml hot distilled water, 25 ml sodium hydroxide

Trial 3: 25 ml crystal violet, 50 ml room temp distilled water, 25 ml sodium hydroxide (a 1:4 dilution of stock reactants)

Trial 4: 25 ml crystal violet, 0 ml distilled water, 25 ml sodium hydroxide (a 1:2 dilution of stock reactants)

The results:

  1. The hot reaction went much faster than the cold.
  2. The 1:4 dilution went faster than the 1:8 dilution. (~6 minutes vs 9 minutes, respectively)
  3. The 1:2 dilution went much, much more slowly than any other. (~16 minutes)

Result #3 is the anomaly. It involved the most concentrated solutions and so should have been the fastest of the 3 room temperature reactions, but it was far and away the slowest.

Things I've considered:

  • The reaction isn't significantly endo- or exothermic, so the solution didn't have a different temperature due to its minimal mass. (All of the reactions except for the one with hot water added were within a couple of degrees of each other.)

  • The students didn't make a mistake in execution - this was observed by multiple groups over multiple trials.

  • The difference was not due to the sloppy endpoint of the reaction. Though the reduced volume meant that a given amount of remaining crystal violet would be most purple in this trial, the difference in color was enormous and obvious.

  • The reverse reaction never becomes significant with such a massive excess of sodium hydroxide.

  • New idea: The unusual result is the only trial without distilled water - maybe the water is contaminated or something like that, but in a way that is promoting the reaction? I was excited when I thought of this (I actually popped up out of a dead sleep with it in my head,) but it seems unlikely, I used the exact same container of water an hour earlier to make each of the stock solutions...

I've wracked my brain, but I can't figure out a reasonable explanation for what we observed. Any suggestions would be welcome.

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  • $\begingroup$ Have you tried with still more concentrated solution ? For example with no dilution at all ? Or with 2.0 M or 4.0 M solution ? $\endgroup$ – Maurice Jan 4 at 10:16
  • $\begingroup$ @Maurice The unusual result is essentially no dilution - the only dilution that occurs is the mixing of the two stock solutions at in equal volume, so that the reaction mixture is half of the stock's concentration for each. I didn't run the trial with higher concentrations. $\endgroup$ – Jason Patterson Jan 4 at 11:55
  • $\begingroup$ Don't you think there is a different reaction or a new equilibrium at high concentrations of NaOH ? $\endgroup$ – Maurice Jan 4 at 12:34
  • $\begingroup$ @Maurice It's entirely possible. I will test on Monday. $\endgroup$ – Jason Patterson Jan 5 at 0:19
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Your analysis of the expectation is correct, and we can look at it more quantitatively in this way: Since the hydroxide concentration is much higher than the CV concentration in all of the reactions, let's approximate it as unchanging. Assuming the reactions are first-order in each reactant, we have a rate law for the concentrated reaction of:

$$-\frac{d[\ce{CV}]}{dt}=k[\ce{HO-}][\ce{CV}]\approx k[\ce{HO-}]_{initial}[\ce{CV}]$$

For simplicity, we'll define a $k'$ as $k'=k[\ce{HO-}]_{initial}$ and treat the reaction as pseudo-first-order.

The half-life of a first-order reaction is given by $t_{1/2}\approx \frac{0.7}{k}$, so for the concentrated reaction we have $t_{1/2}=\frac{0.7}{k'}$.

For the 1:4 reaction, the initial hydroxide concentration is halved, so we have a pseudo-first-order rate constant of $0.5k'$ and $t_{1/2}=\frac{0.7}{0.5k'}$ or 2 times the half-life of the concentrated reaction.

Likewise, the half-life of the 1:8 reaction is twice that of the 1:4 reaction and four times that of the 1:2 reaction, so it's $t_{1/2}=\frac{0.7}{0.25k'}$.

Let's say that the concentrated reaction needs 6 half-lives for the concentration to get to the point where the color disappears. Since the 1:4 reaction starts at half the initial CV concentration, it only needs 5 half-lives. But the half-lives are each twice as long, so it should take a time equivalent to 10 half-lives of the concentrated reaction, so 10/6=1.7 times as long.

For the 1:8 reaction, the same analysis says that it will take 4 half-lives, but each is 4 times as long as that of the 1:2 reaction, so it takes 16 equivalent half lives, meaning 16/6=2.7 times as long.

The only instance in which the more dilute reaction will be faster is if the concentrated reaction requires less than two half-lives to be observed to be complete. That can't be the case here or there would be no color at the start of the 1:8 reaction.

So we must look for an alternative explanation for your observations. 0.5 M NaOH is a very concentrated solution, and it is possible that other reactions are occurring besides the intended one. To test that, I would follow Maurice's suggestion and run the reaction at a wider range of base concentrations. It would also be preferable to have the same concentration of CV in every reaction, so only one variable is involved. To do this, used a fixed volume of (NaOH + water) in each reaction.

UPDATE: This paper1 reports doing this reaction at much lower hydroxide ion concentrations and also notes that the reaction rate decreases with increasing ionic strength, which could be at least a partial explanation for why your observed rate at high concentrations is lower than expected.

1 Felix, L.D. Kinetic Study of the Discoloration of Crystal Violet Dye in Sodium Hydroxide Medium. J Chem Appl Chem Eng Vol: 2 Issue: 1

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  • $\begingroup$ Yes, typically the reaction is done in a spectrometer cell over the course of an hour or so. At the end the solution is still purple. In order to get decolorization in a reasonable time I raised the [NaOH] substantially. It is usually treated as pseudo-first order in order to find the order of CV and the concentration of NaOH is varied to find its order. We'll be doing that later in the unit. (This is the most common kinetics lab that high schoolers do these days - I was trying to expand its use as an intro to the unit.) The ionic strength idea sounds promising. I'll test on Monday. $\endgroup$ – Jason Patterson Jan 5 at 0:16
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All the reaction have different concerntration at start . As you have taken so much excess NaOH it would remain constant. We could look it as you have a solution of 50 ml with 0.05 mM violet red and 0.5M NaOH . You dilute the N times and then to have to compare rates. Threotically we could find that but when we say discolouration in the solution does it mean the reaction has been 100% complete? Actually there is a particular concerntration at which we will see discolouration . Lets say that concerntration is 0.001mM of violet red.

Time taken in completion of reaction (assuming the reaction is 1st order with respect to violet red and also note that pseudo rate constant depends on intial concertration of NaOH)= $\frac{ln(\frac{intial\,contretaion}{final\,concentration})}{k(concentration\,of\,NaOH)}$ = $\frac{ln(\frac{0.05*1000}{n})}{(k*0.5/n)}$=$\frac{2n*ln(\frac{50}{n})}{k}$=F(n)

Here k is the kinetic rate of the 2nd order reaction.

F'(n) = 2ln($\frac{50}{n}$) -2

As you can the derivative of F is greater than zero till 50/e and after that less than zero (You could assume the complete reaction concerntration be anything but assume it be constant)

So whatever concentration you choose the function comes out that time of completion of reaction increases(F'(n) may come out be negative at the very start though) till a particular value of n and after that decrease as n increase(n is directly proportional to dilution)

If we assume the kinetic 2nd order law to be completely valid we can find the concerntration of violet red at which our eyes think the reaction has been completed.Yeah for that you will need to conduct many experiments.

The result is not unusual and its perfectly fine. If you find the rate of the reaction at intial points I m sure you will find it to be greatest in most concerntrated solution.

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  • $\begingroup$ Please use MathJax to format your mathematical formulae. See the help guide for more info. $\endgroup$ – Aniruddha Deb Jan 4 at 10:26
  • $\begingroup$ Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown. Refrain from using colloquialisms and slang such as "u" in place of "you" in you posts. $\endgroup$ – andselisk Jan 4 at 10:32
  • $\begingroup$ Is it ok or needs more modifications? $\endgroup$ – aryan bansal Jan 4 at 10:50
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    $\begingroup$ I'm struggling to see how this explains the event. In each case we used the same number of moles of reactant. The order of the reaction is first in NaOH and first in CV, but even if I'm wrong about that and it's one or the other, the most concentrated solution should react fastest at any particular number of remaining CV moles. If it's always fastest and the reactions all consume equal quantities of CV, how could it do anything other than finish first? $\endgroup$ – Jason Patterson Jan 4 at 12:01
  • $\begingroup$ Exactly what i m saying is that rate depends on concerntration , and from a concerntration(which depends on how much we diluted the intial mixture) we have to reach a particular concerntration. If u add less water more moles need to react because u have to reach one particular concerntration which means if less water implies less moles left at the end(intial moles were the same but final moles are not same so we cant say anything about time). $\endgroup$ – aryan bansal Jan 4 at 12:50

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