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Chemistry
Chemical Kinetics
in a pseudo first order hydrolysis of ester in wat...

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

(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.

Answer:

(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 - 3mol L^{- 1}s ^{- 1}

(ii) For a pseudo first order reaction,

k = 2.303/ t log [R]_º / [R]

For t= 30 s, k₁ = 2.303/ 30 log 0.55 / 0.31

= 1.911 × 10 ^{- 2}s^{- 1}

For t = 60 s, k₂ = 2.303/ 60 log 0.55 / 0.17

= 1.957 × 10^{- 2}s ^{- 1}

For t= 90 s, k₃ = 2.303/ 90 log 0.55 / 0.085

= 2.075 × 10 ^{- 2}s ^{- 1}

Then, average rate constant, k = k₁ + k₂+ k₃/ 3

= 1.911 × 10 ^{- 2}+ 1.957 × 10^{- 2} + 2.075 × 10 ^{- 2} / 3

= 1.981 x 10^-2s ^{- 1}

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