Determine the rate constant K from only the reaction data below. The reaction is

Question

Determine the rate constant K from only the reaction data below. The reaction is an irriversible first order reaction, A(t)->B(t) acquired from a UV-Vis instrument.

Time 420nm 700nm 300nm 580nm 0 0.002732 0.042346 5.21E-09 7.19E-29 50 0.002279 0.035317 0.036041 0.001361 100 0.001901 0.029455 0.066101 0.002495 50 0.00 1585 (0.024566 0.091171 0.003442 R200 0.001322 0.020489 0.11208 0.004231 250 0.001103 0.017088 0.129518 0.004889 300 0.00092 0.014252 0.144062 0.005438 350 0.000767 0.011886 0.156192 0.005896 400 0.00064 0.009913 0.166308 0.006278 450 0.000533 (0.008268 0.174745 0.006597 500 0.000445 0.006896 0.181782 0.006862 550 0.000371 (0.005751 0.187651 0.007084 600 0.000309 0.004796 0.192546 0.00 7269 650 0.000258 0.004 0.196628 0.007423 700 0.000215 (0.003336 0.200033 0.007551 750 0.000 18 0.002783 0.202873 0.007658 800 0.000 15 0.002321 0.205241 0.007748 850 0.000125 0.001936 0.207216 0.007822 900 0.000104 0.001614 0.208864 0.007885 950 8.69 E-05 0.001346 0.210238 0.007937 1000 7.25E-05 0.001123 0.211383 0.00 798

Explanation / Answer

The absorbance data with time is given

This is a first order reaction

For a first order reaction,

rate constant k = (ln[Ao] - ln[A])/dt

So,

at 300 nm

rate constant k = (ln(0.002732) - ln(0.001901)/100 = 5.84 / time units

at 420 nm

rate constant k = [(ln(0.042346) - ln(0.029455)]/100 = 3.63 x 106-3 /time units

at 580 nm

rate constant = [ln(5.21 x 10^-9) - ln(0.066101)]/100 = 0.163 /time units

at 700 nm

rate constant k = [ln(7.19*10^-29) - ln(0.002495)]/100 = 0.59 /time units

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