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.
Answer
599.1k+ views
Hint: There is a formula to calculate the rate constant and half-life of the ester and there are as follows.
\[K=\dfrac{2.303}{{{T}_{2}}-{{T}_{1}}}\log \left( \dfrac{100-{{C}_{1}}}{100-{{C}_{2}}} \right)\]
Here ${{T}_{2}}$ = time in minutes after hydrolysis of ester for second time
${{T}_{1}}$ = time in minutes after hydrolysis of ester for first time
${{C}_{1}}$ = Percentage of ester at time ${{T}_{1}}$
${{C}_{2}}$ = Percentage of ester at time ${{T}_{2}}$
\[{{t}_{1/2}}=\dfrac{0.693}{K}\] 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.
\[K=\dfrac{2.303}{{{T}_{2}}-{{T}_{1}}}\log \left( \dfrac{100-{{C}_{1}}}{100-{{C}_{2}}} \right)\]
Here ${{T}_{2}}$ = time in minutes after hydrolysis of ester for second time = 350
${{T}_{1}}$ = time in minutes after hydrolysis of ester for first time = 60
${{C}_{1}}$ = Percentage of ester at time ${{T}_{1}}$ = 18.17
${{C}_{2}}$ = Percentage of ester at time ${{T}_{2}}$= 69.12
- Substitute all the known values in the above formula to get the rate constant of the hydrolysis of the ester.
\[\begin{align}
& K=\dfrac{2.303}{{{T}_{2}}-{{T}_{1}}}\log \left( \dfrac{100-{{C}_{1}}}{100-{{C}_{2}}} \right) \\
& K=\dfrac{2.303}{350-60}\log \left( \dfrac{100-18.17}{100-69.12} \right) \\
& K=3.36\times {{10}^{-3}}/\min \\
\end{align}\]
- By using the rate constant we can calculate the half-life of the chemical reaction.
\[\begin{align}
& {{t}_{1/2}}=\dfrac{0.693}{K} \\
& {{t}_{1/2}}=\dfrac{0.693}{3.36\times {{10}^{-3}}} \\
& {{t}_{1/2}}=206.25\min \\
\end{align}\]
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.
\[K=\dfrac{2.303}{{{T}_{2}}-{{T}_{1}}}\log \left( \dfrac{100-{{C}_{1}}}{100-{{C}_{2}}} \right)\]
Here ${{T}_{2}}$ = time in minutes after hydrolysis of ester for second time
${{T}_{1}}$ = time in minutes after hydrolysis of ester for first time
${{C}_{1}}$ = Percentage of ester at time ${{T}_{1}}$
${{C}_{2}}$ = Percentage of ester at time ${{T}_{2}}$
\[{{t}_{1/2}}=\dfrac{0.693}{K}\] 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.
\[K=\dfrac{2.303}{{{T}_{2}}-{{T}_{1}}}\log \left( \dfrac{100-{{C}_{1}}}{100-{{C}_{2}}} \right)\]
Here ${{T}_{2}}$ = time in minutes after hydrolysis of ester for second time = 350
${{T}_{1}}$ = time in minutes after hydrolysis of ester for first time = 60
${{C}_{1}}$ = Percentage of ester at time ${{T}_{1}}$ = 18.17
${{C}_{2}}$ = Percentage of ester at time ${{T}_{2}}$= 69.12
- Substitute all the known values in the above formula to get the rate constant of the hydrolysis of the ester.
\[\begin{align}
& K=\dfrac{2.303}{{{T}_{2}}-{{T}_{1}}}\log \left( \dfrac{100-{{C}_{1}}}{100-{{C}_{2}}} \right) \\
& K=\dfrac{2.303}{350-60}\log \left( \dfrac{100-18.17}{100-69.12} \right) \\
& K=3.36\times {{10}^{-3}}/\min \\
\end{align}\]
- By using the rate constant we can calculate the half-life of the chemical reaction.
\[\begin{align}
& {{t}_{1/2}}=\dfrac{0.693}{K} \\
& {{t}_{1/2}}=\dfrac{0.693}{3.36\times {{10}^{-3}}} \\
& {{t}_{1/2}}=206.25\min \\
\end{align}\]
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.
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