Question

all calculations must be included! The indicator crystal violet, (V) reacts with NaOH (OH- ) according to the equation: CV(aq) + OH- (aq) → CV- (aq) (Violet) (Colorless) The rate law for the reaction is: Rate = k2 [CV] [NaOH]. In an experiment the concentration of NaOH exceeds that of the crystal violet by a factor of 1000 so the rate law becomes Rate = k1 [CV] where k1 = k2 [NaOH] . The following data were obtained for the reaction of crystal violet with the NaOH concentration at 0.00800M. The changes in [CV] were followed by measuring the absorbance (A) of the solution. To prove that the reaction is first order in CV, plot a graph of ln(Absorbance) versus time to see if it is linear. k1 is the negative slope of this graph and k2 is equal to k1 divided by the NaOH concentration used in the experiment.

Data

Time (min) Absorbance, (A) Ln(Absorbance)

0.00 0.840

8.00 0.680

12.0 0.640

17.0 0.560

21.0 0.500

25.0 0.460

30.0 0.410

Plot a graph of ln(Absorbance) versus time and determine the slope of the line. Is the reaction first order in CV? Explain your answer.

Evaluate k1 (pseudo first order rate constant) for the reaction if k1 = -(slope)

Evaluate the overall rate constant , k2, for the reaction ij k2=k1/[NaOH]

Answer 1

Since ln(absorbance) versus time gives the straight line, the reaction is first order in CV.

From the graph, the slope the line is -0.0239.

So, the k1 value = - (slope) = - (-0.0239) = 0.0239 /s

k2 = k1 / [NaOH] = (0.0239 /s)/(0.00800 M) = 2.99 M^-1 s^-1.