Crystal Violet Formal Lab Report

Name: Gordon WangTeacher: Mr. Falat

Class: Chemistry 2APDate: 13 February 2014Partners: Jeeyong Shin

Ezra BrooksNoah Ritz

On my honor, I have neither given nor received any unauthorized aid on this lab report,

Objective:

The objective of this experiment is to use photocolorimetric techniques to experimentally determine the rate law of a bimolecular system. Another objective was to determine the overall order of the reaction and order with respect to each reactant through integrated rate laws, A final objective was to determine the value of the true reaction rate constant as well as the pseudo rate constants for both Samples 1 and 2.

Theory:

This lab dealt with chemical kinetics. This is the area of chemistry that concerns reaction rates. It is the study of the rate of reaction or how quickly the reaction takes place. The reaction rate of a chemical reaction is defined as the change in concentration of a reactant or product per unit time where concentration is measured in molarities.

One thing that can be seen through experiments is that the rate of a reaction is not constant, but changes with time because the concentrations change as the reactants are consumed as the reaction proceeds. There are many factors that affect rate so describing rates must be done very specifically.

Chemical reactions are reversible. There is a forward reaction where two reactants combine to form a product but also a reverse reaction where the product decomposes back into the two reactants. Both have rates of reaction. Studying reaction rates is done by choosing conditions that make the rate of reaction of the reverse reaction negligible, and the reaction rate only depend upon the concentrations of the reactants. For example, the Rate = k[A]x. An expression that shows how the rate depends on the concentrations of reactants is called a rate law. “k”, a proportionality constant called the rate constant, and x, the order of the reactant, must both be determined by experiment.

The order of a reaction can be any integer including zero or a fraction. Only the concentrations of the reactants appear in the rate law, not the concentrations of the products, because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate.

There are 2 different types of rate laws. The differential rate law expresses how a change in concentration changes the rate. The integrated rate law expresses how the concentration changes over time. The integrated rate law is a major focus of this lab, and it will be covered more in detail later.

This lab studies the reaction between crystal violet and sodium hydroxide. The net ionic equation is:

(1) CV+ + OH- CV-OH

The purple CV+ cation slowly combines with hydroxide ions to form a neutral product, CVOH, which is colorless.

The reaction rate for the Eq. (1) depends on the concentrations of [CV+] and [OH-]. A differential rate law can be written for this reaction in the general form of:

(2) r = -∆[CV+]/∆t = k2[CV+]x[OH-]y

In this generalized rate law, k2 is the true rate law constant for this reaction, x is the order with respect to CV+ and y is the order with respect to OH-. The values of x and y are to be determined experimentally through two trials of the reaction.

A useful trick for this experiment is to keep the initial [OH-] many times larger than the initial [CV+]. When this is done, the [OH-] change over the course of the reaction as CV+ is consumed, is almost negligible compared to the high initial [OH-]. So it can be assumed that the [OH-] is constant during the experiment, which allows the rate law expression to be further simplified for easier analysis.

Eq. (2) can be rewritten as:

(3) -∆[CV+]/∆t = k1[CV+]x

“k1” is equal to k2 [OH-]y. “k1” is called a pseudo rate constant since it is a constant that relates the reaction rate to [CV+]x only if the [OH-] remains constant. So if the [OH-] were to change, the k1 value would also change, meaning it isn’t really a constant.

In order for the reaction order, x, of CV+ and the pseudo rate constant to be found, the differential rate law expressed in Eq.(3) must be integrated. Integrated rate laws, mentioned earlier, express the [CV+] as a function of time, given the differential rate law for the reaction.

For zero, first and second order reactions, there are different concentration-time relationships derived using calculi. Integrated rate laws can be arranged in line equation form where the result is a function of concentration on the y axis and time on the x axis and the absolute value of the slope will reveal the value of the pseudo rate constant. The zero order integrated rate law in line equation form has the [CV+] on the y axis, time on the x axis, and a slope of –k1. The first order integrated rate law in line equation form has ln[CV+] on the y axis, time on the x axis, and a slope of –k1. Finally, the second order integrated rate law in line equation form has the 1/[CV+] on the y axis, time on the x axis, and a slope of +k1. To determine whether [CV+] is 0, 1st, or 2nd order, the [CV+] vs. time data taken during the experiment must be graphed on 3 separate plots, [CV+] vs. time, ln[CV+] vs. time, and 1/[CV+] vs. time and which ever one of these graphs gives the data in a straight line determines the order according to the line equation forms of each different order integrated law. The absolute value of the slope of the best fit line for the proper graph should also give the value of the pseudo rate constant.

Two trials varying the concentration of [OH-] should be done. Since k2, the true rate constant remains constant between trials, y, the order of reaction with respect to OH– can be found from taking the ratio of the pseudo rate constants and the ratio of the dilution adjusted concentrations of hydroxide ion used in trials 1 and 2 and then applying the law of logarithms. After, y, is determined, it is quite easy to determine the value of k2, the true rate constant for each trial and take the average and write an experimentally determined rate law for the reaction in Eq.(1)

A final note concerns that of Beer’s Law, which is of much importance to this lab. Beer’s Law states that the absorbance of a sample is directly proportional to the concentration, according to A = abc where A is absorbance, a is the molar absorptivity specific to the measured substance, b is the path length of the light through the sample, and c is the varying concentration of the compound that absorbs. A plot of absorbance vs. concentration will therefore be linear with a slope of ab since a and b are both constants. So if the absorbance of crystal violet solutions with specific known concentrations is measured, a Beer’s Law calibration curve can be prepared with absorbance on the y axis and concentration of crystal violet on the x axis and the equation of the line for that Beer’s Law calibration curve can be used to determine the concentration of CV from the measured absorbance of the solution during the reaction.

This experiment utilizes photocolorimetric techniques, in which a piece of equipment called a Spec. 20 that measures the %T, or the fraction of light passes through a sample. For each time period of 30 seconds during the experiment, the %T for a sample in which the reaction indicated by Eq. (1) is occurring is measured. The %T can easily be converted to Absorbance through the formula Absorbance = 2 – log(%T). Absorbance can then be easily converted to [CV+] by using the slope of the Beer’s Law calibration curve, and all graphing and calculations can be done from here.

Sources:

Zumdahl TextbookKinetic Study of Crystal Violet and OH1- lab handouthttp://www.flinnsci.com/media/850290/cf7644.pdfhttp://faculty.icc.edu/bcook/c132xp1.pdfhttp://www.uccs.edu/Documents/chemistry/nsf/106%20Expt2V-KineticsII.pdfhttp://course1.winona.edu/wng/C213S13-Labs/Lab%202%20Kinetics%20CV/Student%20-%20Kinetic_CV1_S13.pdfhttp://faculty.sites.uci.edu/chem1l/files/2011/03/E07MANCVRateLaw.pdf

Materials:

2 – 250 mL beakers2 – 100 mL grad. Cylinder2 – 10 mL grad. Cylinders1 – Spectronic 201 – kim wipes250 mL of 0.100 M NaOH2 – 50 mL vol. Flasks1 – 10 mL pipet6 – Spec 20 curvettes1 – clock with second hand50 mL of 0.000050 M crystal violet

Procedure:

Part I: Beer’s Law Curve

1. Turn on the Spec. 20 and allow a warm up period.2. Use a 50 mL vol. Flask to prepare the following crystal solutions. Fill each

curvette up to the white dot. Discard the excess. Label each curvette.0.000005 M CV0.000004 M CV0.000003 M CV0.000002 M CV0.000001 M CV

3. Set the Spec 20 to 570 nm4. Close the empty sample department and set to 0% T (left dial).5. Place the blank and set to 100% T (right dial).6. Record the % T for each [CV], convert to absorbance and plot the 5 data points.7. Determine the slope for the curve.

Part II. Integrated Rate Law

1. Use a 10-mL graduated cylinder to add 10.0 mL of 0.000050 M crystal violet solution to two 100-mL graduated cylinders. (0.000050 M CV = 0.0204 g/L)

2. Add 85 mL of DH2O to the first cylinder and 80 mL of DH2O to the second cylinder.

3. Set up Spec – 20 and make sure it is calibrated. (0% and 100%)4. As nearly simultaneously as possible use a 10-mL cylinder to add 5.0 mL of 0.100

M NaOH to the first cylinder and 10.0 mL of 0.100M NaOH to the second.5. Immediately pour each solution back and forth into a 250-mL beaker to

thoroughly mix each reaction mixture.

6. Transfer enough of each mixture from step 4 to two different curvettes up to the white dot.

7. Place one curvette of CV reactants into the Spec – 20 and record the %T in the box for “0” time. Note the time to the nearest five seconds on the clock. Immediately remove this curvette and place the second curvette of the CV reactants into the curvettes and place them in a dry 150-mL beaker. Take a %T reading every two minutes for a total of twenty minutes.

Note 1: About thirty seconds before each reading use the blank and check for 100%.

Note 2: Do not leave the curvettes in the Spec – 20. The heat generated by the machine will change the reaction rate.

8. Clean all equipment.9. For each set of data convert %T to A, ln(A) A-1.10. Plot (A), ln(A) and A-1 vs time for each set of data.11. Calculate the slope of each line (curve fit).12. Solve for k2 using the two different concentrations of [OH]1- .13. Compare your k2 values with your classmates. What order of reaction is it?

Data:

Table 1: %T and Absorbance for 5 Crystal Violet solutions of Different Concentrations

Concentration (M) %T Absorbance

0.000005 51 0.29243

0.000004 56 0.251812

0.000003 69 0.161151

0.000002 78.8 0.103474

0.000001 86.8 0.06148

0 100 0

Table 2: %T, A, [CV] , ln[CV], and [CV]-1 for Sample 1

Sample 1 Containing 85 mL DH2O, 10 mL 0.00005 M CV, 5 mL 0.100 M NaOH

Time (min) % T A [CV] Ln[CV] [CV]^-10 52.88 0.276709 4.72449E-06 -12.2628 211663.13532 59 0.229148 3.91244E-06 -12.4513 255594.65054 62 0.207608 3.54468E-06 -12.5501 282112.98416 63.6 0.196543 3.35575E-06 -12.6048 297996.03388 68.2 0.166216 2.83795E-06 -12.7724 352367.5941

10 73.8 0.131944 2.25279E-06 -13.0033 443894.080912 75 0.124939 2.13319E-06 -13.0579 468781.753314 78 0.107905 1.84236E-06 -13.2045 542781.00516 79.6 0.099087 1.6918E-06 -13.2897 591087.024918 83.2 0.079877 1.3638E-06 -13.5052 733242.851520 83 0.080922 1.38165E-06 -13.4922 723771.865

Table 3: %T, A, [CV] , ln[CV], and [CV]-1 for Sample 2

Sample 2 Containing 80 mL DH2O, 10 mL 0.00005 M CV, 10 mL 0.100 M NaOH

Time (min) % T A [CV] Ln[CV] [CV]-1

0 53 0.275724 4.70768E-06 -12.2663 212418.8402

2 59.4 0.226214 3.86234E-06 -12.4642 258910.2143

4 65.6 0.183096 3.12616E-06 -12.6757 319881.0931

6 70.6 0.151195 2.58149E-06 -12.8671 387373.1552

8 76 0.119186 2.03497E-06 -13.105 491406.7059

10 82.2 0.085128 1.45347E-06 -13.4416 688009.5206

12 86 0.065502 1.11837E-06 -13.7036 894162.0635

14 87.6 0.057496 9.81678E-07 -13.834 1018664.049

16 89.8 0.046724 7.97754E-07 -14.0415 1253519.006

18 93.2 0.030584 5.22189E-07 -14.4652 1915015.438

20 93.6 0.028724 4.90433E-07 -14.528 2039015.86

Graphs:Part 1:

Graph 1: Beer's Law Calibration Curve for Crystal Violet Solution

y = 58569xR2 = 0.9887

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 0.000001 0.000002 0.000003 0.000004 0.000005 0.000006

Concentration (M)

Abs

orba

nce

(A)

Slope = ab

Part 2:Sample 1 Graphs:

Graph 2A: [CV] vs. Time for Sample 1

R2 = 0.9539

0.000001

0.0000015

0.000002

0.0000025

0.000003

0.0000035

0.000004

0.0000045

0.000005

0 2 4 6 8 10 12 14 16 18 20

Time (min)

[CV] -k1

Graph 2B: ln[CV] vs. Time for Sample 1

y = -0.0639x - 12.287R2 = 0.987

-13.6

-13.4

-13.2

-13

-12.8

-12.6

-12.4

-12.20 2 4 6 8 10 12 14 16 18 20

Time (min)

ln[C

V]

-k1

Graph 2C: [CV]-1 vs. Time for Sample 1

R2 = 0.9658

200000

300000

400000

500000

600000

700000

800000

0 2 4 6 8 10 12 14 16 18 20

Time (min)

[CV]

-1

-k1

Graph 3B: ln[CV] vs. Time for Sample 2

y = -0.1179x - 12.22R2 = 0.9936

-15

-14.5

-14

-13.5

-13

-12.5

-120 2 4 6 8 10 12 14 16 18 20

Time (min)

ln[C

V]

-k1

Graph 3A: [CV] vs. Time for Sample 2

R2 = 0.9237

0

0.000001

0.000002

0.000003

0.000004

0.000005

0 2 4 6 8 10 12 14 16 18 20

Time (min)

[CV] -k

Graph 3C: [CV]-1 vs. Time for Sample 2

R2 = 0.8969

0

500000

1000000

1500000

2000000

2500000

0 2 4 6 8 10 12 14 16 18 20

Time (min)

[CV]

-1

-k1

Graph Analysis:

Graph 1 is the Beer’s Law Calibration plot for Crystal violet solution. It shows the relationship between concentration of crystal violet solution and the absorbance is linear. This confirms Beer’s Law which states that A = abc, where A is absorbance, a is molar absorptivity, b is path length, and c is concentration. For this graph, c was known and A was known from calculations, so the slope of the trendline of the graph is equal to ab. The slope of the trendline for the Beer’s Law plot was 58569. “ab” is constant. The importance of this slope is that throughout the rest of the experiment, this slope of 58569 can be used to convert from absorbance to [CV].

Graphs 2A, 2B, and 2C and Graphs 3A, 3B, and 3C are integrated rate laws that express how the [CV] changes over time or how time changes the rate of the reaction. From the graphed relationship between [CV] and time, the reaction order with respect to [CV] can be determined along with the value of the pseudo rate constant. Determining the reaction order involves plotting 3 different graphs: 1 with [CV] vs. time to see if order is 0, 1 with ln[CV] vs. time to see if order is 1, and 1 with 1/[CV] vs. time to see if order is 2. The one graph with the data expressed most linearly will identify the reaction order with respect to [CV] and the slope of the trendline of that graph with give the value of the pseudo rate constant.

Since there were two samples, the order has to be determined twice with a total of 6 graphs, and 3 graphs each and also 2 different pseudo rate constants have to be determined because the concentration of [OH-] varied between samples and pseudo rate constant depends on [OH-]. To determine the graph for which the data was most linear,

the R-squared values for each linear trendline had to be compared. An high R-squared value close to 1 shows how well the linear trendline fits the data so the graph with the highest R-squared value for its trendline was the most linear graph. For Sample 1, Graph 2B has the most linear data since its R-squared value of 0.987 was larger that Graph 2A’s R-squared value of 0.9539 and Graph 2C’s R-squared value of 0.9658. Since Graph 2B graphed ln[CV] against time, x, the reaction order with respect to [CV] for Sample 1 was 1 and the pseudo rate constant for Sample 1, according to the slope of the trendline was 0.0639 min-1. For Sample 2, Graph 3B’s R-squared value of 0.9936 was larger than Graph 3A’s R-squared value of 0.9237 and Graph 3C’s R-squared value of 0.8969. Since Graph 3B graphed ln[CV] against time, x, the reaction order with respect to [CV] for Sample 2 was 1 and the pseudo rate constant for Sample 2, according to the slope of the trendline was 0.1179 min-1.

In summary:ab= 58569x = reaction order with respect to [CV] for Sample 1 and Sample 2 = 1pseudo rate constant for Sample 1 = 0.0639 min -1 pseudo rate constant for Sample 2 = 0.1179 min -1

Data Analysis:

1. Sample Calculation of Absorbance from %T

A = log (100/%T) = 2 – log(%T)

%T = 59.4A = ?

A = 2 – log (59.4)A = 0.226214

2. Sample Calculation of k1

Slope = -k1

Slope = -0.0639

-k1 = -0.0639

k1 = 0.0639 min -1

3. Sample Calculation of [CV] from Absorbance

A = ab[CV]

ab = 58569 (from slope of Beer’s Law plot)A = 0.226214[CV] = ?

0.226214 = 58569[CV][CV] = 3.86234E-06 M

4. Sample Calculation of ln[CV]

Ln[CV] = ln[CV]

[CV] = 3.86234E-06 M

Ln[3.86234E-06] = -12.4642

5. Sample Calculation of 1/[CV]

1/[CV] = ?

[CV] = 3.86234E-06 M

1/[3.86234E-06] = 258910.2143

6. Calculation of Reaction Order of [OH-]

A) Calculation of [OH-] in Sample 1

i) Calculation of Total L of Solution

80 mL Water + 10 mL CV + 5 mL NaOH = 100 mL total solution

100 mL x 1 L / 1000 mL = 0.1 L total

ii) Calculation of moles of OH-

5 mL of 0.100 M NaOH

5 mL x 1 L/1000 mL x 0.100 mol NaOH/L = 5E-4 mol NaOH = 5E-4 mol OH -

iii) Calculation of Molarity

M = mol OH- / L solution

M = 5E-4 mol / 0.1 LM = 0.005 M OH -

B) Calculation of [OH-] in Sample 2

i) Calculation of Total L of Solution

80 mL Water + 10 mL CV + 10 mL NaOH = 100 mL total solution

100 mL x 1 L / 1000 mL = 0.1 L total

ii) Calculation of moles of OH-

10 mL of 0.100 M NaOH

10 mL x 1 L/1000 mL x 0.100 mol NaOH/L = 0.001 mol NaOH = 0.001 mol OH -

iii) Calculation of Molarity

M = mol OH- / L solution

M = 0.001 mol / 0.1 LM = 0.01 M OH -

C) Calculation of Reaction Order with Respect to [OH-]

In Sample 1 when [OH-] = 0.005 M:

k1 = k2[OH-]y

k1 = pseudo rate constant for Sample 1 = 0.0639 min-1

k2 = true rate constant for reaction = ?[OH-] = 0.005 My = reaction order with respect to [OH-]

0.0639 = k2[0.005]y

In Sample 2 when [OH-] = 0.01 M

k1 = k2[OH-]y

k1 = pseudo rate constant for Sample 2 = 0.1179 min-1

k2 = true rate constant for reaction = ?[OH-] = 0.01 My = reaction order with respect to [OH-]

0.1179 = k2[0.01]y

Taking the ratio of the pseudo rate constants Sample 1: Sample 2:

0.5419847328 = 0.5y

log(0.5419847328) = y log(0.5)y = (log(0.5419847328))/(log(0.5))

y = 0.884y ≈ approx. 1

The order with respect to [OH - ] is 1 .

6. Calculation of, k2, Value of True Rate Constant

A) Calculation of k2 for Sample 1

k1 = k2[OH-]y

k1 = 0.0639 min-1

k2 = ?[OH-] = 0.005 My = 1

0.0639 = k2[0.005]1

k2 = 12.78 M -1 min -1

B) Calculation of k2 for Sample 2

k1 = k2[OH-]y

k1 = 0.1179 min-1

k2 = ?[OH-] = 0.01 My = 1

0.1179 = k2[0.01]1

k2 = 11.79 M -1 min -1

C) Calculation of Average True Rate Constant

Avg. True Rate Constant = (k2 of Sample 1 + k2 of Sample 2) /2

Avg. True Rate Constant = (12.78+11.79)/2

Avg. True Rate Constant = 12.285 M -1 min -1 7) Summary of Experimental Rate Law for the Reaction between CV+ and OH-

r = k2[CV+]x[OH-]y

k2 = Experimental Avg. True Rate Lawx = Order with respect to CV+, according to analysis of experimental datay = Order with respect to OH-, according to analysis of experimental data

r = k2[CV+]1[OH-]1

Where k2 = 12.285 M-1min-1

Overall Order of Reaction is the sum of the orders with respect to each reactant’s concentration, so the overall order of this reaction is equal to two.

Conclusion:

The objectives of this experiment were:

(1) to use photocolorimetric techniques to experimentally determine the rate law of a bimolecular system.

(2) to determine the overall order of the reaction and order with respect to each reactant through integrated rate laws

(3) to determine the value of the true reaction rate constant as well as the pseudo rate constants for both Samples 1 and 2.

It was expected that the above objectives be achieved and this lab experiment was successful in achieving all those objectives. The experimentally determined rate law of the bimolecular system is r = k2[CV+]1[OH-]1. The order with respect to both reactants was experimentally determined through integrated rate laws and pseudo rate constants to be 1st order. The overall order, being the sum of the orders with respect to reach reactant’s concentration, is therefore 2nd order. The experimentally determined pseudo rate constant for Sample 1 when [OH-] = 0.005 M was 0.0639 min-1. The experimentally determined pseudo rate constant for Sample 2 when [OH-] = 0.01 M was 0.1179 min-1. The pseudo rate constants were calculated from the slope of the linear equation form of the correct integrated rate law. The calculated averaged value of the true reaction rate constant for the reaction investigated in this experiment was 12.285 M-1min-1 from taking the average of the two true rate constants for each sample that were found by plugging in known concentrations and pseudo rate constants into the modified form of the differential rate law.

There may have been some errors in this lab. One piece of evidence for error was that the value of y, the order with respect to the hydroxide ion was calculated to be around 0.884 when it should have have been approximately equal to 1, since it should be 1st order. With 1 as the accepted value and 0.884 as the observed value of y, the percent error calculation below for the determination of y gives a percent error of 11.6%.

% Error = = = 11.6%

Another piece of evidence for error was that the two true rate constants calculated using the [OH-] and pseudo rate constants for each sample should have been the same or very close together, but they weren’t. Sample 1’s calculated true rate constant was 12.78 M-1min-1, while Sample 2’s calculated true rate constant was 11.79 M -1min-1. The true rate constant for the reaction should remain constant for all trials but it didn’t. A percent difference calculation below where the difference between the two true rate constants is divided by the average of the true rate constants (12.285) and is multiplied by 100 gives a percent difference of 8.059%.

% Diff. = = = 8.059 %

There were several possible sources of error in this lab. One source of error was that when measuring the %Transmittance, the curvettes may not have been properly handled. If a person touches the sides of the curvettes with their fingers, he might leave fingerprints. Smudged fingerprints on the curvettes would lower the %Transmittance, because the light passing through the sample would also be blocked by the fingerprints. Not only would the % Transmittance be affected, the absorbance values in the experiment, and the concentration values determined from the concentration would be altered. The slope of the graphs would be altered, and the values for the pseudo rate constants would change, ending up that that the orders may not be easily determined and that the true rate constant value would be obviously different between the two trials. Another possible source of error was that the %Transmittance was supposed to be measured exactly every 2 minutes for each Sample during the course of the experiment. However, the lab group wasn’t able to measure the %Transmittance at the right time intervals due to factors such as distraction, and the time it takes to identify the proper curvette when alternating taking measurements for the two samples and place the curvette into the Spec.20. By the time the %Trans. is actually measured, a couple of seconds may have passed. This affected the slope of the trend line of the graph for the determination of the pseudo rate constant and order of the reaction with respect to the crystal violet ion and the rest of the results. Another source of error was that the reaction mixture may have not been well mixed, which slows down the rate of reaction by reducing chances of an effective collision, thus interfering with the results. In the future, these errors may be corrected by measuring the %Transmittance for Sample 1 every two minutes for twenty minutes and then measuring the %Transmittance for Sample 2 every 2 minutes for twenty minutes after the first trial is done instead of measuring the %Transmittance for Sample 1 and Sample 2 alternating every other minute. This

would simplify things and allow more careful measurements to be taken, leading to more accurate results.

In conclusion, this lab was a success in achieving all its objectives despite some sources of error. The rate law for a reaction was successfully determined through experimentation, and in the process I have learned a lot about Chemical kinetics and how experimentation can be used to determine the values of rate constants and the order with respect to each reactant. I learned about differential rate laws and integrated rate laws and how to identify the order of a reactant from a series of integrated rate law graphs. I learned how scientists can simplify a complicated rate law by using pseudo rate constants and assuming one reactant’s concentration is constant through the course of a reaction by making the concentration of this reactant much larger than that of the other reactant. Finally I learned how absorbance and concentration are directly proportional through Beer’s Law and how photocolorimetric techniques can be applied to investigate chemical kinetics.