Integrated Rate Laws Crystal Violet Reaction

CRYSTAL VIOLET REACTION Integrated Rate Laws: Crystal Violet Reaction CRYSTAL VIOLET REACTION PURPOSE: To find the rate constant, k, and the half-life for a crystal violet reaction as well as find the order of the reaction. PROCEDURE: Obtain 4.0 mL of 0.10 M NaOH solution in a 10 mL graduated cylinder. Use another 10 mL graduated cylinder with 4.0 mL of 2.25 X 10^-5 M crystal violet solution. Fill a cuvette ¾ full with distilled water. Record the temperature. Connect the Spectrometer to the computer and open a new file on the LoggerPro software. Calibrate the Spectrometer. Empty the cuvette and fill it about ¾ full with the crystal violet solution and place it in the Spectrometer. Click “collect” and a graph of the solution will appear. The peak absorbance should be near 590 nm. Click “stop.” Click on the Configure Spectrometer Data Collection icon, then click Abs vs. Time. Click “ok.” Remove the cuvette from the Spectrometer and return the crystal violet solution to your graduated solution. Click on the button that looks like a graph with a clock over it and change the time of collection from 200 to 300 seconds. Pour the 4 mL of crystal violet and sodium hydroxide solutions into the 100 mL beak and swirl it. Fill the cuvette ¾ full with this reaction mixture and place it in the Spectrometer. Click “collect.” After 5 minutes, click “stop” and pour the contents into the waste beaker. Before repeating the experiment, chill the reagents in an ice bath. Store the latest run when you click “collect.” Export the data as an Excel CSV and open it in Excel. Save it as a Workbook file. Add trendlines with the equation and R squared values to find the order of the reaction. CRYSTAL VIOLET REACTION CHEMICALS USED: • C25H30N+ • H2O • OH- GRAPHS: CONCLUSIONS: The order of this reaction is first order because ln/[A] produces a linear plot which indicates a first order reaction. The rate law is k[CV+]. The value for the rate constant of the first trial is 2.19 X 10^-5. The value for the rate constant for the second trial is 5.31 X 10^-5. The half-life is 31643.84 seconds. The activation energy of this reaction is 8.28 X 10^+4 J/mol. CRYSTAL VIOLET REACTION POST LAB QUESTIONS: 1. Report the reaction order with respect to crystal violet, write the rate law, and give the values of the rate constant (with proper units) for both temperatures you studied today. The order of this reaction is first order because ln/[A] produces a linear plot which indicates a first order reaction with respect to crystal violet. The rate law is k[CV+]. The value for the rate constant of the first trial is 2.19 X 10^-5. The value for the rate constant for the second trial is 5.31 X 10^-5. 1st Trial: Ln[0.456] = -k(298.3 K) + ln[0.459] -0.785 = -k(298.3 K) - 0.778 + 0.778 + 0.778 -6.55 X 10-3 = -k(298.3 K) (-6.55 X 10-3)/298.3 K = -k(298.3 K)/298.3 K -1(-2.19 X 10^-5) = -1(-k) 2.19 X 10^-5 = k 2nd Trial: Ln[0.578] = -k(290.6 K) + ln[0.587] -0.548 = -k(290.6 K) - 0.532 + 0.532 + 0.532 -0.154 = -k(290.6 K) (-0.154)/290.6 K = -k(290.6 K)/290.6 K -1(-5.31 X 10^-5) = -1(-k) 5.31 X 10^-5 = k 2. Using your data table or graph of Abs vs Time, estimate the half-life of the reaction: select two points, one with an absorbance value that is about half of the other absorbance value, and compare the time between these points. The time it takes the absorbance (or concentration) to be halved is known the half-life for the reaction. Do this multiple times and calculate the average. T ½ = 105 T ½ = 104 T=0 Abs = 0.459 T = 123 Abs = 0.2 T = 105 Abs = 0.227 T = 227 Abs = 0.1 105 + 104 + 104 = 313/3 = 104.33 average T ½ T = 195 T = 299 T ½ = 104 Abs = 0.124 Abs = 0.062 CRYSTAL VIOLET REACTION 3. Now calculate the half-life from the rate constant, k, using the appropriate formula. Does your value agree with the one determined above? T ½ = 0.693/(2.19 X 10^-5) T ½ = 31643.84 seconds This value does not agree with the one determined above. 4. Since you have a value for the rate constant at two or three temperatures, you can make an Arrhenius Graph: ln(k) vs. 1/T. It will only have 2 points, but the slope of this line can be used to determine the Activation Energy for the reaction: Ea (in J/mol) = |m*R| where R = 8.314 J/(K·mol) Calculate the Activation Energy of this reaction and report it in units of J/mol or kJ/mol. 1/T 1/298.3 = 0.00335233 1/290.6 = 0.003441156 Ea = 9971 X 8.314 J/(Kmol) Ea = 8.28 X 10^+4 J/mol Ln k ln(2.19*10^-5) = -10.72902392 ln(5.31*10^-5) = -9.84333363