Question

Calculate the rate constant, k, by using Equation 3 (k'=k[OH]^1). Keep in mind, the NaOH solution was diluted by 50% at the start of the experiment. Write a final generic rate law using Equation 1 (k[OH]^x[CV]^y) using the experimental determined values for k, and y. Find the average for CV (Crystal Violet) order (y), pseudo rate constant (k'=3.1x10^-8 Ms^-1), and the rate constant (k).
Section #: - Name: TA: Time 90 210 Table 1: Time, absorbance, and calculated absorbance data. Abs In(Abs) 1/Abs 0 1 .380 0.322 0.725 30 1.100 0.095 0.909 60 0.863 -0.147 1.159 0.677 -0.390 -0.390 1.477 120 0.516 -0.662 1.938 150 0.394 -0.931 2.538 180 0.307 -1.181 3.257 0.233 -1.457 4.292 240 0.182 -1.704 5.495 270 0.146 -1.924 6 .849 300 0.110 -2.207 9.091 330 -2.419 11.236 360 0.072 -2.631 13.889 390 0.057 -2.859 17.452 420 0.049 -3.022 20.534 450 0.042 -3.180 24.038 480 0.036 -3.324 27.778 510 0.032 -3.442 31.250 540 0.029 -3.527 34.014 570 0.026 -3.634 37.879 600 0.025 -3.685 39.841
The generic rate law is Rate = k(actual value with correct units)[CV]^(order for cv)[OH]^1.

Answer 1

CV+ + OH- -----> CVOH
The rate law for this reaction is in the form:
rate = k[CV+]m[OH-]n, We will assume that absorbance is proportional to the concentration of crystal violet (Beer’s law).
Absorbance = e [CV+]

Therefore, absorbance will be used in place of concentration in plotting the following three graphs:
• Absorbance vs. time: A linear plot indicates a zero order reaction
                         (k = –slope).

• ln(Absorbance) vs. time: A linear plot indicates a first order reaction            (k = –slope).

• 1/(Absorbance) vs. time: A linear plot indicates a second order reaction            (k = slope).

Here we see that ln(Absorbance) vs. time is A linear plot which indicates it is a first order reaction .

So, Rate Constant k = Slope (m) from the line y = mx + c

Here y = - 0.00706x + 0.12347, So rate constant k = 0.00706 sec-1

rate = k[CV+]1[OH-]1