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Q.

For acid catalysed hydrolysis of ester, rate law obtained is rate= k [ester] H+

where k=0.01 M−1hr−1. What is the half-life of this reaction, if the initial concentrations are 0.02 M for the ester and 0,05 M for the catalyzing acid?

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a

5000 hours

b

1386 hours

c

2 hours

d

1429 hours

answer is C.

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Detailed Solution

The value of rate constant,

k′=0.01×0.05

 The half-life of this reaction

t1/2=ln⁡2k′ t1/2=0.6930.01×0.05 t1/2=1386 hrs

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