Lab report for Chemistry(Reaction between Crystal Violet and Sodium Hydroxide)

Write chemistry a lab report for a reaction between Crystal violet and Sodium hydroxide when the following are provided:

0.005M Sodium hydroxide, 6.75 X 10^-6M crystal violet for first run of the experiment. In the second run, replace 0.005M sodium hydroxide with 0.01M sodium hydroxide. Use the questions given below to guide you write a good report. 

Which function A_t versus time gives the most linear graph (-A, -ln A, 1/A)? The result for this experiment was ln(A).

The order of reaction of crystal violet is (0, 1, 2): y=1, y=0.0015x – 0.2195. The experimental values for pseudo rate constants (include significant figures and units). 

Results:  K’ = -0.0015, K” = -0.003 (Include graphs to illustrate your answers)

y=-0.003x – 0.813, rate of two pseudo constants k”/k’ = (-0.003/-0.0015) = 2. The order of reaction in sodium hydroxide is (0, 1, 2) x=1.

Calculate the value of true rate constant, K, from the experimental value for K’, (include your calculations and units).

Calculate the value of true rate constant K, from the experimental value for K” (show your calculations, including units).

Average experimental value for K (include significant figures and units). Answer for is -0.294. 

Calculate molar absorptive, €, for crystal violet using the data from part 1(show your calculations). 

Introduction

 Chemical kinetics is considered as the study of rates of a reaction. In this experiment, the kinetics of the reaction between sodium hydroxide and crystal violet was investigated. The concentration of crystal violet as a function of time was also monitored. The molecular reaction equations is as shown below.

All the products and reactants are colorless except for the crystal violet, which possesses an intense violet color. Therefore, during the reaction process, the intensity of the color of the reaction mixture becomes less and less. Gradually becoming colorless when all the crystal violet has been consumed. 

The crystal violet color is caused by the extensive system of alternating double and single bonds, which extend over the central carbon atom and all three benzene rings. If the conjugation in the crystal violet structure is traced, it is noted that in the reaction product, the three rings become no longer in conjugation with one another, and hence, this makes the material colorless.

Procedure

Before beginning of the procedure, a transmittance calibration was conducted on the spectrometer. [%T with no Cuvet (left knob); 100%T with H2O Cuvet (right knob)] distilled water was used as a blank. The outside of the Cuvet was wiped dry. The Cuvet was emptied and shaken dry. On an excel spreadsheet, columns of time, ABS and %T were set up. Two graphs of %T versus time and ABS versus time were generated on the same spreadsheet. A solution of crystal violet was dispensed inside a clean dry beaker after which a solution of NaOH was dispensed into the beaker immediately without splashing it. The solution of CV and NaOH was mixed thoroughly, followed by filling the the Cuvet upto ¾ full. %T readings were taken at time intervals using a spectrometer. [CV]_t    was computed for each data point in the excel sheet using Beer’s law. A plot of 1/[CV]_t versus time was generated. A linearity test was carried out .The first order reaction in CV gave a linear plot while a second order reaction gave a curved plot. 

Results and Data  

Value of true rate constant k’ as calculated from the experimental value k’:

First experiment:

Rate= (--0.0015) X (6.75X10^-6)

-1 X10^-8=k (0.005) (6.75 X 10-6)

-1X10^-8=k (0.005) (6.75X10^-6)

-1X10^-8=k (3.4 X10^-6)

Value of true rate constant k’  as calculated from the experimental value k”:

Second experiment:

Rate= (-0.003) X (6.75 X 10^-6) = -2 X 10^-8

-2 X 10^-8 =k (0.01) (6.75 X10^-6)

-2 X 10^-8 = k (6.8 x 10^-8) = -0.294

Min^--0.294min-1.

Average experimental value for k was -0.294 min^-1.

Molar absorptivity, €, of crystal violet calculated using the data from part one:

A_t = -log (%T/100) =E {cr}_t

Where A_t = 0.985, E = 2.54 cm [cr]_t = 6.75 * 10^-6

(0.985)/ (2.54) X (6.75 X 10-6) = € = 57451.153

Conclusion

For the graphs obtained using NaOH with the lower concentration level, the plot of Absorbance against time yielded the straightest line, but the plot of lnA against time also had a very straight line. The plot of 1/A gave the line that was slightly straight. For the graphs that were obtained using NaOH of a higher concentration, lnA against time had the straightest line. The graphs of absorbance against time and 1/A against time were not as nearly straight. It is, therefore, fairly safe to say that the rate equation was in first order because lnA against time overall yielded the straightest line in the two runs. Hence, the rate equation of the reaction with respect to hydroxide ions was  = k [NaOH].