2) (4 pts total) The following shows results from experiments similar to those o

Question

2) (4 pts total) The following shows results from experiments similar to those of the online tutorial, but with different dyes and bleach The reaction has 1:1 stoichiometry dye+ bleach colorless products The stock solutions have the following concentrations Blue dye: 4.50 x 10M Red dye 1.30x 10M Bleach: 0.200 M The absorbance was measured at various dye concentrations and the results are summanzed below. (Note: This data also available in the file: dyeData.xls on the Blackboard site.) Blue Dye Red Dye Conc (M) Red Dye abs Blue Dye Conc (M) abs OOCE +00 6.00E-06 1.20E-05 1.80E-05 2.40E-05 3.00E-05 3.60E-05 4.20E-05 4.80E-05 5.40E-05 6.00E-05 0.000 0.085 0.170 0.255 0.340 0,425 0.510 0.595 0.680 0.765 0.850 0.00E+00 1.50E-06 3.00E-06 4.50E-06 6.00E-06 7.50E-06 9.00E-06 1.05E-05 1.20E-05 1.35E-05 1.50E-05 0.000 0.080 0.160 0240 0.320 0.400 0.480 0.560 0.640 0.720 0.800 Experiments were then performed by mixing the following volumes of dye and bleach solutions and monitoring absorbance versus time. experiment 2 experiment Experiment : blue dye: 9.50 mL bleach: 0.500 mL Experiment 2: blue dye: 9.00 mL bleach: 1.00 ml Experiment 3 red dye: 9.50 mL bleach: 0.500 mL Experiment4 ed dye: 9.00 mL bleach 1.00 ml The results of the experiments are available in the dyeData.xls excel file accompanying this homework, and are plotted to the right. Page 2 of5

Explanation / Answer

Abs = M (constant) * C (concentration)

Blue dye:

0.85 = M1 * 6 x10-5

M1 = 1.42 x 104

Red dye:

0.8 = M2 * 1.5 x 10-5

M2 = 5.33 x 104

Let B denote blue dye, R denote R dye, L denote bleach

a)

In Exp 1 absorbance or [B] at 255s ~ Exp 2 absorbance or [B] at 60 s:

rate = K [B]x [L]y

At this point, rate1/rate2 = [L1/L2]y

Since L is in much excess,

[L1] = (0.2 M * 0.5 mL) / 10

= 0.01 M

[L2] = (0.2 M * 1 mL) / 10

= 0.02 M

Rate1/rate2 = Abs at 255s in Exp 1 / Abs at 60s in Exp 2 = [L1/L2]y

= (0.189348 – 0.176108) / (0.23906 – 0.178618) = [L1/L2]y

0.22 = 0.5y

y ~ 2

Initial rate1/Initial rate2 = Abs at 0 s in Exp 1 / Abs at 0s in Exp 2

= (0.605625 – 0.563046) / (0.57375 – 0.42843) = [B1/B2]x [L1/L2]y

0.293 = [(4.5 x 10-5 M * 9.5 mL/ 10 mL)/(4.5 x 10-5 M * 9 mL/ 10 mL)]x [0.5]y

0.293 = 1.056x * 0.25

x ~ 3

A = M * C

Rate = -dC/dt = 1 / M1 * -dA/dt

Initial rate1 = 1/ 1.42 x 104 M *(0.605625 – 0.563046)/ 15 s

= 2 x 10-7 M s-1

= k [B]x [L]y

= k [9.5 x 4.5 x 10-5 / 10]3 [0.2 * 0.5 / 10]2

2 x 10-7 M s-1 = k * (4.28 x 10-5 )3 * 0.012

k = 2.56 X 1010 M-4 s-1

b)

In Exp 3 absorbance or [r] at 210 s ~ Exp 4 absorbance or [R] at 105 s:

rate = K [R]x [L]y

At this point, rate1/rate2 = [L1/L2]y

Since L is in much excess,

[L1] = (0.2 M * 0.5 mL) / 10

= 0.01 M

[L2] = (0.2 M * 1 mL) / 10

= 0.02 M

Rate1/rate2 = Abs at 210s in Exp 3 / Abs at 105s in Exp 4 = [L1/L2]y

= (0.204653 – 0.195058) / (0.200959 – 0.183236) = [L1/L2]y

0.54 = 0.5y

y ~ 1

Initial rate1/Initial rate2 = Abs at 0 s in Exp 3 / Abs at 0s in Exp 4

= (0.658667 – 0.568449) / (0.624 – 0.479521) = [R1/R2]x [L1/L2]y

0.624 = [(1.3 x 10-5 M * 9.5 mL/ 10 mL)/(1.3 x 10-5 M * 9 mL/ 10 mL)]x [0.5]y

0.624 = 1.056x * 0.5

x ~ 4

A = M * C

Rate = -dC/dt = 1 / M2 * -dA/dt

Initial rate1 = 1/ 5.33 x 104 M * (0.658667 – 0.568449)/ 15 s

= 1.13 x 10-7 M s-1

= k [R]x [L]y

= k [9.5 x 1.3 x 10-5 / 10]4 [0.2 * 0.5 / 10]1

1.13 x 10-7 M s-1 = k * ( 1.235 x 10-5)4 * 0.011

k = 4.85 x 1014 M-4 s-1

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