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प्रश्न
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 |
- Calculate the average rate of reaction between the time interval 30 to 60 seconds.
- Calculate the pseudo first order rate constant for the hydrolysis of ester.
संख्यात्मक
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उत्तर
i. Average rate during the interval 30 to 60 seconds
= `"Change (decrease) in concentration"/"Time interval involved in the change"`
= `- (0.17 - 0.31)/(60 - 30)`
= `- (-14)/30`
= `14/30`
= 4.67 × 10−3 mol L−1 s−1
ii. For a first-order reaction,
k = `2.303/t log_10 [A]_0/([A])`
In the present case, [A]0 = 0.55 M
When t = 30 s, k = `2.303/30 log_10 0.55/0.31` = 1.91 × 10−2 s−1
When t = 60 s, k = `2.303/60 log_10 0.55/0.17` = 1.96 × 10−2 s−1
When t = 90 s, k = `2.303/30 log_10 0.55/0.085` = 2.07 × 10−2 s−1
∴ Average value of k = `(1.91 xx 10^-2 + 1.96 xx 10^-2 + 0.07 xx 10^-2)/3`
= 1.98 × 10−2 s−1
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