Question

A student performs a trial using 50mL of 35 µM Yellow Dye and 5 mL of concentrated bleach (1.124 M). What is the concentration of the dye after 5 min? Rate = 0.1031 M-1min-1 [Yellow Dye] [Bleach]

Answer 1

The reaction follows 2 order kinetics since rate = k[A][B] is given as rate law.

The initial concentration of dye = 50ml x35M =1750 mmol

Initial concentration of bleach = 1.124ml x 5M =5.62 mmol

And k = 0.1031min^-1 (if the reaction were following 2 order K should have units conc^-1time-1)

The integrated rate law for 2 order reaction is

K= 2.303/t(a-b) log[b(a-x)/a(b-x)

However as the concentration of dye (a ) is very large compared to concentration of bleach , we can have (a-b) = a and (a-x) =a

Then the rate expression becomes k = 2.303/ta log (b/b-x) that is pseudo first order reaction, first order with respect to bleach and zero order with respect to dye, where the concentration of dye remains almost constant.

Thus the concentration of yellow dye remains as it is after 5 min.

NEW

Now also the same argument is validwith the bleach in excess which makes the reaction first order with respect to dye and zero order with respect to bleach.

Taking the first order rate equation and substituting the values we get

log (a/a-x )= 0.2238 thus (50x35x10^-6)/(50x35x10^-6-x) = 1.66 and calculating for x and the remaining dye concentration is 21microM.