InterviewSolution
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InterviewSolution
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1.0 ml of ethyl acetate was added to 25 ml of N/2 HCl. 2 ml of the mixture were withdrawn from time to time during the progress of the hydrolysis of the ester and titrated against standard NaOH solution. The amount to NaOH required for titration at various intervals is given below : {:("Time (min)",,,:,,,0,,,20,,,75,,,119,,,183,,,oo),("NaOH used (ml)",,,:,,,20.24,,,21.73,,,25.20,,,27.60,,,30.22,,,43.95):} The value at oo time was obtained by completing the hydrolysis on boiling. Show that it is a reaction of the firstorder and find the average value of the velocity constant. |
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Answer» Solution :As explained above, if the REACTION is of the first order, it must obey the EQUATION `k=(2.303)/(t)log""(a)/(a-x)=(2.303)/(t)log""(V_(oo)-V_(0))/(V_(oo)-V_(t))` In the present case, `V_(0)=20.24" ml", V_(oo)=43.95" ml":.V_(oo)-V_(0)=43.95-20.24=23.71" ml"` The value of k at different instants can be calculated as FOLLOWS : `{:("t (min)",,,V_(t),,,V_(oo)-V_(t),,,k=(2.303)/(t)log""(V_(oo)-V_(0))/(V_(oo)-V_(t))),(20,,,21.73,,,43.95-21.73=22.22,,,k=(2.303)/(20min)log""(23.71)/(22.22)=0.00324min^(-1)),(75,,,25.20,,,43.95-25.20=18.75,,,k=(2.303)/(75min)log""(23.71)/(18.75)=0.00313min^(-1)),(119,,,27.60,,,43.95-27.60=16.35,,,k=(2.303)/(119min)log""(23.71)/(16.35)=0.00312min^(-1)),(183,,,30.22,,,43.95-30.22=13.73,,,k=(2.303)/(183min)log""(23.71)/(13.73)=0.00299min^(-1)):}` Since the value of k COMES out to be nearly constant, it is a reaction of the first order. The mean value of `k=0.00312min^(-1).` |
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