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THE REACTION OF METHYL VIOLET WITH NaOH: A KINETIC STUDY This is a kinetic study of the reaction: C25H30N3CI + NaOH → C25H30N30H + NaCl Net ionic equation: C25H30N3+ + OH → C25H30N30H The rate law is : r=k[OH-]X[ C25H30N3+]y The experiment is performed using a large excess of NaOH. Using NaOH in excess is as if keeping the concentration of NaOH constant. The rate law is therefore reduced to r=k'[C25H30N3 The changes in the concentration of methyl violet are then observed over a period of 250 minutes using NaOH in excess with (NaOH) = 1.00 M. The collected data is shown below: Time (min) [C25H30N3+] (M) In(C25H30N3+). 1/[C25H30N3+) 0.0 1.328 0.2836 0.753 100 1.246 150 1.213 200 1.185 250 1.162 You will now use the collected data to graphically determine the value of y. Once you have “y”, you can get k' from the slope.
Determining x: k' was the rate constant when (NaOH) = 1.00 M k" was the rate constant when (NaOH) = 0.50 M r= k[C25H30N3+1. [OH-]> r=k[C25H30N3+1 [1.00]* = k' [C25H30N3+1 r=k[C25H30N3+14. [0.50]* = k” [C25H30N3+1y k'/k” = (k[C25H30N3*]\ . [1.00]/(k[C25H30N3+]Y. [0.50]*) = (1.00)/(0.50)* Use R/k”) = (1/0.5)* = (2)x . calculate x using k' and k” determined graphically Determining k: r=k[C25H30N3+LY. [1.00]* = kº [C25H30N3+ly k = k'/[1.00]* r=k[C25H30N3-L. [0.50]* = k” [C25H30N3*Lyk = k”/[0.50]* Use k = k'/1x and k = k”/(0.5)x ) the value of k is the average of the two.
Determining x: k' was the rate constant when (NaOH) = 1.00 M k" was the rate constant when (NaOH) = 0.50 M r= k[C25H30N3+1. [OH-]> r=k[C25H30N3+1 [1.00]* = k' [C25H30N3+1 r=k[C25H30N3+14. [0.50]* = k” [C25H30N3+1y k'/k” = (k[C25H30N3*]\ . [1.00]/(k[C25H30N3+]Y. [0.50]*) = (1.00)/(0.50)* Use R/k”) = (1/0.5)* = (2)x . calculate x using k' and k” determined graphically Determining k: r=k[C25H30N3+LY. [1.00]* = kº [C25H30N3+ly k = k'/[1.00]* r=k[C25H30N3-L. [0.50]* = k” [C25H30N3*Lyk = k”/[0.50]* Use k = k'/1x and k = k”/(0.5)x ) the value of k is the average of the two.