Medium
Why is the hydrolysis of an ester slow at the beginning and fast after some time?
(N/A) The hydrolysis of an ester can be represented as follows:
$Ester + H_2O \rightarrow Acid + Alcohol$
In this reaction,the acid produced acts as an autocatalyst.
Initially,the concentration of the acid is very low,so the reaction proceeds slowly.
As the reaction progresses,the concentration of the acid increases,which accelerates the rate of the reaction.
This phenomenon,where one of the products acts as a catalyst for the reaction,is known as autocatalysis.
$Ester + H_2O \rightarrow Acid + Alcohol$
In this reaction,the acid produced acts as an autocatalyst.
Initially,the concentration of the acid is very low,so the reaction proceeds slowly.
As the reaction progresses,the concentration of the acid increases,which accelerates the rate of the reaction.
This phenomenon,where one of the products acts as a catalyst for the reaction,is known as autocatalysis.
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Similar Questions
$A$ reaction is first order in $A$ and second order in $B$.
$(i)$ Write the differential rate equation.
$(ii)$ How is the rate affected on increasing the concentration of $B$ three times?
$(iii)$ How is the rate affected when the concentrations of both $A$ and $B$ are doubled?
$(i)$ Write the differential rate equation.
$(ii)$ How is the rate affected on increasing the concentration of $B$ three times?
$(iii)$ How is the rate affected when the concentrations of both $A$ and $B$ are doubled?
Difficult
View SolutionThe data for the reaction $A + B \to C$ is given below:
$Exp$ $[A]_0$ $[B]_0$ Initial Rate $1$ $0.012$ $0.035$ $0.10$ $2$ $0.024$ $0.035$ $0.80$ $3$ $0.012$ $0.070$ $0.10$ $4$ $0.024$ $0.070$ $0.80$
Determine the rate law for the reaction.
| $Exp$ | $[A]_0$ | $[B]_0$ | Initial Rate |
|---|---|---|---|
| $1$ | $0.012$ | $0.035$ | $0.10$ |
| $2$ | $0.024$ | $0.035$ | $0.80$ |
| $3$ | $0.012$ | $0.070$ | $0.10$ |
| $4$ | $0.024$ | $0.070$ | $0.80$ |
Determine the rate law for the reaction.
Difficult
View SolutionFor the following rate law,determine the unit of the rate constant: Rate $= -\frac{d[R]}{dt} = k[A]^{\frac{1}{2}}[B]^{2}$
Difficult
View Solution$A$ reaction $2NO + 2H_2 \longrightarrow N_2 + 2H_2O$ has the following mechanism:
Step-$I$: $2NO \longrightarrow N_2O_2$
Step-$II$: $N_2O_2 + H_2 \longrightarrow N_2O + H_2O$
Step-$III$: $N_2O + H_2 \longrightarrow N_2 + H_2O$
Which of the following substances is a reaction intermediate?
Step-$I$: $2NO \longrightarrow N_2O_2$
Step-$II$: $N_2O_2 + H_2 \longrightarrow N_2O + H_2O$
Step-$III$: $N_2O + H_2 \longrightarrow N_2 + H_2O$
Which of the following substances is a reaction intermediate?
Medium
View SolutionFor a certain reaction,the rate $= k[A]^2[B]$. When the initial concentration of $A$ is tripled keeping the concentration of $B$ constant,the initial rate would:
MediumNEET 2023
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