Crystal violet hydroxide ion reaction rate law determination

Chemical kinetics is the study of the rates of chemical reactions. The rate of a chemical reaction describes how fast a reaction takes place, or in other words, how fast the reactant and product concentrations change. Factors such as reactant concentration, temperature, and presence of a catalyst effect reaction rate.

            In this experiment, we examine the effects of changes in initial reactant concentration on reaction rate. Usually, an increase in reactant concentration causes an increase in reaction rate. This is because the more reactant molecules that are in a given volume of reaction mixture, the greater the likelihood that the molecules will collide and react. The purpose of the experiment is to determine the complete (including k with units) rate law for the reaction.

            The specific reaction rate we study is this experiments that of crystal violet dye, [(H3C)2NC6H5]3C+ and hydroxide ion, OH-: rxn.gif

Crystal violet is purple, whilst both OH- and the product are colorless. Therefore,  if we measure the rate at which the purple color disappears, we can determine how fast the crystal violet reacts with OH-.

            We use a rate law to describe the dependence of a reaction rate on the corresponding reactant concentration. For the general reaction of two reactants, R and Q to form product P (R + Q --> P) we define its rate law as

rate = k [R]x[Q]y

The proportionality constant, k, is called the rate constant, [R], and [Q] are the molar concentrations of R and Q, x and y are the orders in respect to the two reactants.

            The order with respect to a reactant quantifies the effect of charges in the concentration of just that reactant on the reaction rate. They are not necessarily the coefficients in the balanced chemical equation.

Determining Reaction Order: The Graphic Method

            To use the graphic method for determining reaction order with respect to a specific reactant, we must plot changes in the concentration of that reactant against time. We begin by measuring the concentration of that reactant, [R], during the course of a controlled reaction in which we keep all factors as constant as possible. One technique for achieving this consistency is to use a large excess of the other reactant, Q. During the course of the reaction, only a relatively tiny portion of the excess reactant Q will be consumed, so its concentration will remain relatively constant. Thus, any changes in the reaction rate will be due only to the decreasing concentration of R.

            Once we have obtained the concentration data for reactant R under such controlled conditions, we plot the data, using the following  graphics treatments, no more than one of which can yield a straight line graph: [R] versus time (true if the reactions is zeroth order in R), ln[R] vs time (straight of 1st order in R), or 1/[R] vs. time(straight of 2nd order in R). Remember, Beer's law coorelates Abs and concentration (A =ebc) Knowing e = 5.0 x 104 L/(cm mol) and b = 1 cm for this experiment, you can convert (or have Excel do it) each Abs measurement into concentration. We can then simply plot []  vs. time, ln [] vs. time, and 1/[] vs. time. Hopefully one of them will be linear and we will have the order of reaction for crystal violet determined.

Determining Reaction Order: The Method of Initial Rates

            So this graphical method procedure works swimmingly for determining the order of reaction for the colored reactant, crystal violet. But what about the order of reaction for the OH-, the other reactant? For that determination, we must go the old fashioned way, the method of initial rates. To determine the rate with respect of a certain reaction, we do two runs of a reactions by changing the concentration of one of the reactants by a prescribed amount, usually twice or half that of the original concentration. We then look at how the rate changed to see what the order of reaction was. 
   We will need to know the rate of reaction of the 0.100 M run and the rate of reaction of the 0.05 M reaction. The slope of the [] vs. time line is in fact the reaction rate. So we just conmpare those slopes, and holefully they will be something convient in comparison to one another (the same, double, 4x, whatever)

Our Reaction

            In this experiment, we investigate the rate of reaction between crystal violet and OH-, as shown earlier. Crystal violet is purple, while OH- and the reaction product are clear and colorless. The color of crystal violet is due to the absorbance of visible light at a wavelength of 595 nm. The absorbance of this solution is proportional to the concentration of the solution.

            In the first trial, we will mix a crystal violet solution with an OH- solution that is about 1000 times that of the crystal violet solution. The [OH-] will not change appreciably during the course of the reaction, so we will assume it is constant. We will measure the absorbance of the solution (the crystal violet) as time passes, during which the absorbance (concentration) will fall. When we plot A, ln A and 1/A vs. time, one of which will yield a straight line giving us the order of the reaction.

            We will then do a second trial, changing only the initial OH- concentration. We again monitor the absorbance of the reaction mixture over time, but in this case, we only want the 'initial' rate, we allready know the curve that will give a straight line, but rate of reaction is equal to the slope of the time vs. [ ] graph, so that is what we will produce for the second run. The slope of that line is directly proportional to the rate. With the second reaction, you need only generate one graph, that of time vs. [ ]. Here is a screencast showing you how to do the graphs.

Screencast description of the whole experiment  (11 min)

Determining k is a bit of a trick. We have never (up until this point) calculated the actual rate of reaction because we did not calculate change in concentration vs. time, we calculated change in Abs vs. time. To determine k, we need one 'run' knowing the actual change in concentration vs. time. Remember, Beer's law coorelates Abs and concentration (A =ebc) Knowing e = 5.0 x 104 L/(cm mol) and b= 1 cm for this experiment, you can 'convert' the slope of abs vs time into change in conc vs time by simply diving by eb. Now that you have change in concentration vs time (that's rate baby!) plug it into your partial rate law (after you know the orders of each reactant) and calculate k with units just like you did a few times in lecture. Note: [CV]ini is NOT 1.5 x 10-5 M. It is roughly half that, why?

Procedure:
 

1. Pick out your spectrometer and get it set up. Calibrate, set for 1 reading/second and collect data for 240 seconds.

2. Using a volumetric pipet, put 5.00 mL of 0.050 M NaOH solution in a clean dry test tibe. Obtain 5.00 mL of 1.5 x 10-5 M crystal violet into a second dry test tube.

4. To begin the reaction, pour the samples of crystal violet and NaOH together and pour back and forth 3 times (to mix). Start you timer immediately upon first mixing.

5. After the solution is mixed, rinse a cuvet twice with a small amount of the reaction mixture. Then fill the cuvet and dry the outside.

6. Place the cuvet in the spectrometer and start the Abs-time readings at exactly 90 seconds past the time the solutions were mixed.

7. After the run, process the saved data as explained on the data sheet.

8. Repeat steps 2 through 7 with 5.0 mL of 0.100 M NaOH and 5.0 mL of 1.5 x 10-5 M crystal violet.

10. Open a data file in MS Excel, create 1 graph containing three sets of data, A vs. time, Ln A vs. time, 1/A vs. time for the lower concentration run (0.050 M NaOH) . Make sure you add a trend line and R2 value to each line. They should all be well labeled. A second graph of just A vs. time is needed for the second (0.100M NaOH) run.

 Report: You will fill out an answer sheet that will be available soon... 

 Prelab questions (on recycled piece of paper, properly labeled)

1. Write a very brief summary of the entire procedure. It should be less than 5 sentences.
2. The following data were obtained for the reaction: 2 NO2(g)-->2 NO(g) + 1 O2(g)

time (s)

[NO2] M

ln [NO2]

1/[NO2]

0

0.8333

 

 

20

0.4167

 

 

40

0.2778

 

 

60

0.2083

 

 

a. Determine the ln [NO2] and 1/[NO2] for each data point.
b. Make a computer generated graph of [NO2] vs time, ln [NO2] vs time, and 1/[NO2] vs time. (one graph, 3 lines) Determine the slope and R2 value of each line. Here is a screencast showing you how to do the graphs.
c. Which graph yields the best straight line (R2 value closest to 1) ?
d. What is the reaction order with respect to NO2?
e. Write the rate law for this reaction including the numerical value for k.
3. Come up with a data sheet for the appropriate needed data. Show your instructor on the way into lab.
4. Bring your portable memory drive (flash drive thingie) to extract data from lab.

~MEO 2.22.06