InterviewSolution
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InterviewSolution
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1.0 mL of ethyl was added to 25 mL of N/2 HCl, 2.0mL of the mixture was withdrawn from time to time during the progress of easter hydrolysis and titrated against standard NaOH solutions. The amount of NaOH required for titration at various intervals is given below: The value of at infinite time was obtained by completing the hydrolysis on boiling. Show that the reaction is of first order. Also find the average value of rate constant. |
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Answer» Solution :INITIAL concentration of ethyl acetate (a) = `V_(infty)-V_(0) = (43.95 -20.24)=23.71mL` `k=(2.303)/tlog a/(a-x) = 2.303/tlog (V_(infty)-V_(0))/(V_(infty-V_(t))` i) At t=20 min,`k=(2.303)/(20min) log (23.71)/(43.95 - 21.73) = (2.303)/(20 min)log(23.71)/(20 min) xx 0.28` `0.00322 min^(-1)` (ii) At t=75 min, k`=(2.303)/(75min) log(23.71)/(43.95-25.20)=(2.303)/(75 min)log(23.71)/(18.75)` `=(2.303)/(75min)(log 23.71 - log 18.75) = (2.303)/(75min)(1.375-1.273)=2.303/(75min) xx 0.102` `=0.00313min^(-1)` iii) At t=110 min, `k=(2.303)/(110min) log23.71/(43.95-27.60) = 2.303/(110min) log(23.71)/(43.95-27.60)= 2.303/(110min) log (23.71)/(16.35)` `=(2.303)/(110min)(log23.71-log16.35) = 2.303/(110min) (1.375-1.244)` `=2.303/(110 min) xx 0.151 = 0.00316 min^(-1)` IV) At t=183min, `k=(2.303)/(183 min)log(23.71)/(43.95-30.22) = 2.303/(183min) log(23.71)/(13.73)` `=2.303/(183 min) (log 23.71 - log 13.73) = 2.303/(183 min)(1.375-1.138)` `=(2.303).(183 min) xx 0.237 = 0.00299 min^(-1)` Since the value of k COMES out to be nearly constant, the hydrolysis of ethyl is a FIRST order reaction: Average value of RATE constant(k)`=(0.00322 + 0.0313 + 0.00316 + 0.00299)/(4) = 0.00312 min^(-1)` |
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