Question

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

1. Calculate the average rate of reaction between the time interval 30 to 60 seconds.
2. Calculate the pseudo first order rate constant for the hydrolysis of ester.

Answer 1

(i) Average rate of reaction between the time interval, 30 to 60 seconds = \(\frac {d[Ester]}{dt}\)

= \(\frac {0.31-0.17}{60-30}\)

= \(0.14\)

= \(4.67×10^{-3} mol L^{-1} s^{-1}\)

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(ii) For a pseudo first order reaction,

\(k =\frac { 2.303}{t }log \frac {[R]_0}{[R]}\)

For t = 30 s

\(k_1 = \frac {2.303}{30} log\  \frac {0.55}{0.31}\)

= \(1.911\times 10^{-2} s^{-1}\)

For t = 60 s,

\(k_2 = \frac {2.303}{60} log \ \frac {0.55}{0.17}\)

= \(1.957\times 10^{-2}s^{-1}\)

For t = 90 s,

\(k_3 = \frac {2.303}{90} log \ \frac {0.55}{0.085}\)

= \(2.075\times 10^{-2} s^{-1}\)

Then, average rate constant,

\(k =\frac { k_1+k_2+k_3}{3}\)

\(k = \frac {(1.911\times10^{-2}) + (1.957\times10^{-2}) + (2.075\times 10^{-2})}{3}\)

\(k = 1.98 \times  10^{-2} s^{-1}\)