Date Sheet 1 Solution concentration: (crystal violet)-15X10 5 [OH-)= 0.020 Elaps

Question

Date Sheet 1 Solution concentration: (crystal violet)-15X10 5 [OH-)= 0.020 Elapsed time (min)A 1/A 0.661 Ln(0.661)-0.414 1/0.661-1.512 0.446 0.479 0.509 0.544 0.583 0.612 0.644 0.681 0.715 0.753 0.787 0.823 0.858 0.896 0.931 0.973 1.008 1.047 1.085 1.124 0.640 0.619 1.562 0.580 0.558 0.542 0.525 0.506 0.489 0.471 0.455 0.439 0.424 0.408 0.394 0.378 0.365 0.351 0.338 0.325 1.616 1.664 1.724 1.792 1.845 1.905 1.976 2.045 2.123 2.198 2.278 2.358 2.451 2.538 2.646 2.739 2.849 2.958 3.077 10 12 13 14 15 16 17 18 19 20 KINE0504: Determining the Rate Law for the Crystal Violet-Hydroxide lon Reaction plot of absorbance function that yields the best straight line for the data: order of reaction with respect to crystal violet: slope of straight line: versus

Explanation / Answer

since concentration is proportional to absorbance and concentration is related to rate of species A (-dA/dt) as

-dA/dt= K[A]n, n is the order of reaction. K is the rate constant

for n=1, the reaction is 1st order and for 1st order, -dA/dt= K[A], when integrated noting at t=0 , A=[A]0 and t= t, [A]=[A]

lnA= ln[A]0-Kt, so a plot of lnA vs t gives straight line whose slope will be-K for1st order.

for second order, n=2 and the intergrated equation is 1/A= 1/[A]0+Kt, so a plot of 1/A vs t has to be straight line.

Since here concentration of A is measured in terms of Absorbance, 1st order and 2nd order are verfied by drawin either a plot of lnA vs t or 1/A vs t. The generated plots are shown below.

from the plots it can be seen that the plot of lnA( A refers to absorbance which is proportional to concentration) vs t is having high R2 value and gives the best fit. Hence the reaction 1st order. The slope value is -K and gives the rate constant value K=0.035/min

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