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# In a Pseudo First Order Hydrolysis of Ester in Water, the Following Results Were Obtained - Chemistry

In a pseudo first order hydrolysis of ester in water, the following results were obtained:

 t/s 0 30 60 90 [Ester]mol L−1 0.55 0.31 0.17 0.085

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

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#### Solution

(i) Average rate of reaction between the time interval, 30 to 60 seconds, = (d["Ester"])/dt

= (0.31 - 0.17)/(60 - 30)

=0.14/30

= 4.67 × 10−3 mol L−1 s−1

(ii) For a pseudo first order reaction,

k = 2.303/t log ""[R]_0/([R])

For t = 30 s, k_1 = 2.303/30 log ""(0.55)/0.31

= 1.911 × 10−2 s−1

For t = 60 s, k_2 = 2.303/60 log "" 0.55/0.17

= 1.957 × 10−2 s−1

For t = 90 s, k_3 = 2.303/90 log ""0.55/0.0855

= 2.075 × 10−2 s−1

Then, average rate constant, k = (k_1+k_2+k_3)/3

= ( (1.911xx10^(-2))+(1.957xx10^(-2))+(2.075xx10^(-2)))/3

=1.98 x 10-2 s-1

Concept: Integrated Rate Equations - First Order Reactions
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#### APPEARS IN

NCERT Class 12 Chemistry Textbook
Chapter 4 Chemical Kinetics
Q 8 | Page 118
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