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Question: the hydrolysis of propyl acetate in presence of dilute HCl in aqueous solution, the following data w...

the hydrolysis of propyl acetate in presence of dilute HCl in aqueous solution, the following data were recorded:
Calculate the rate constant of decomposition and time which half of the ester was decomposed.

Explanation

Solution

There is a formula to calculate the rate constant and half-life of the ester and there are as follows.
K=2.303T2T1log(100C1100C2)K=\dfrac{2.303}{{{T}_{2}}-{{T}_{1}}}\log \left( \dfrac{100-{{C}_{1}}}{100-{{C}_{2}}} \right)
Here T2{{T}_{2}} = time in minutes after hydrolysis of ester for second time
T1{{T}_{1}} = time in minutes after hydrolysis of ester for first time
C1{{C}_{1}} = Percentage of ester at time T1{{T}_{1}}
C2{{C}_{2}} = Percentage of ester at time T2{{T}_{2}}
t1/2=0.693K{{t}_{1/2}}=\dfrac{0.693}{K} for first order chemical reaction

Complete step by step solution: - In the question it is asked to calculate the rate constant and half-life of the eater by using the given data in the question.
- First we will calculate the rate constant of the hydrolysis of the ester by using the given data and it is as follows.
K=2.303T2T1log(100C1100C2)K=\dfrac{2.303}{{{T}_{2}}-{{T}_{1}}}\log \left( \dfrac{100-{{C}_{1}}}{100-{{C}_{2}}} \right)
Here T2{{T}_{2}} = time in minutes after hydrolysis of ester for second time = 350
T1{{T}_{1}} = time in minutes after hydrolysis of ester for first time = 60
C1{{C}_{1}} = Percentage of ester at time T1{{T}_{1}} = 18.17
C2{{C}_{2}} = Percentage of ester at time T2{{T}_{2}}= 69.12
- Substitute all the known values in the above formula to get the rate constant of the hydrolysis of the ester.

& K=\dfrac{2.303}{{{T}_{2}}-{{T}_{1}}}\log \left( \dfrac{100-{{C}_{1}}}{100-{{C}_{2}}} \right) \\\ & K=\dfrac{2.303}{350-60}\log \left( \dfrac{100-18.17}{100-69.12} \right) \\\ & K=3.36\times {{10}^{-3}}/\min \\\ \end{aligned}$$ \- By using the rate constant we can calculate the half-life of the chemical reaction. $$\begin{aligned} & {{t}_{1/2}}=\dfrac{0.693}{K} \\\ & {{t}_{1/2}}=\dfrac{0.693}{3.36\times {{10}^{-3}}} \\\ & {{t}_{1/2}}=206.25\min \\\ \end{aligned}$$ **Note:** Without using the above formulas we cannot calculate the rate constant and half-life of the ester hydrolysis at different intervals of time. After hydrolysis ester is going to convert into respective alcohol and acid.