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14 Chemical Kinetics Solutions to Exercises Time (min) A at 608 nm [dye] In [dye] 1/[dye] 0 - 1.254 2.7 X 10⁻⁵ -10.53 3.7 10⁴ 30 0.941 2.0 X 10⁻⁵ -10.82 5.0 10⁴ 60 0.752 1.6 10⁻⁵ -11.04 6.3 10⁴ 90 0.672 1.4 X 10⁻⁵ -11.16 7.0 X 10⁴ 120 0.545 1.2 X 10⁻⁵ -11.36 8.6 X 10⁴ -10.25 9.00E+04 8.00E+04 -10.5 7.00E+04 -10.75 In[dye] -11 1/[dye] 6.00E+04 5.00E+04 4.00E+04 -11.25 3.00E+04 -11.5 2.00E+04 0 20 40 60 80 100 120 0 20 40 60 80 100 120 t, minutes t, minutes Although the graphs are not absolutely definitive, the plot of 1/[dye] time appears to be more linear. (The data point at t = 90 min us "out of line" in both plots and is sus- pect. More precision and accuracy in the experimental data would be helpful.) Assum- ing the reaction is second order with respect to the dye, the rate law is: rate = k = slope = (8.6 10⁴ - 3.7 10⁴) (120-0)min = 4.1 10² M⁻¹ min⁻¹ (The best-fit slope and k value is 3.9 10² min⁻¹) 14.109 Time (s) [C₅H₆] (M) In[C₅H₆] 1/[C₅H₆] 0.045 0.040 0 0.0400 -3.219 25.0 0.035 50 0.0300 33.3 0.030 -3.507 0.025 100 0.0240 -3.730 41.7 0.020 0.015 150 0.0200 -3.912 50.0 0 50 100 150 200 250 Time (seconds) 200 0.0174 -4.051 57.5 -3.0 60 -3.2 55 50 -3.4 [C₅H₆] -3.6 45 40 -3.8 35 30 -4.0 25 -4.2 20 0 50 100 150 200 250 0 50 100 150 200 250 Time (seconds) Time (seconds) The plot of 1/[C₅H₆] time is linear and the reaction is second order. The slope of this line is k. k = slope = (50.0 - 25.0) M⁻¹ (150-0)s = 0.167 (The best-fit slope and k value is 0.163 M⁻¹ 426