Question

For the reaction \[ CH_3CO_2C_2H_5+H_2O \rightarrow CH_3CO_2H+C_2H_5OH \] Rate is given by \[ \text{Rate}=k[H_2O][CH_3CO_2C_2H_5] \] The initial concentrations of ethyl acetate and water are \(0.001\,M\) and \(2\,M\) respectively. The value of rate constant, \(k\), of the reaction, if \(99\%\) of ethyl acetate is hydrolyzed in \(10\) seconds is

• A. \(0.2303\,s^{-1}\)
• B. \(2.303\,s^{-1}\)
• C. \(4.606\,s^{-1}\)
• D. \(3.303\,s^{-1}\)

Hint:

When one reactant is present in large excess, its concentration remains almost constant and the reaction becomes pseudo-first-order. Use \[ k'=\frac{2.303}{t}\log\frac{a}{a-x} \] and then relate it with the actual rate constant using \[ k'=k[\text{excess reactant}] \]

Answer 1

The Correct Option is A

Step 1: Identify the nature of the reaction.
The rate law is \[ \text{Rate}=k[H_2O][CH_3CO_2C_2H_5] \] Here, water is present in very large excess: \[ [H_2O]=2\,M \] while ethyl acetate concentration is very small: \[ [CH_3CO_2C_2H_5]=0.001\,M \] So, the concentration of water remains nearly constant during the reaction.
Therefore, the reaction behaves as a pseudo-first-order reaction with respect to ethyl acetate.

Step 2: Use the pseudo-first-order rate equation.
For a first-order reaction, \[ k'=\frac{2.303}{t}\log\frac{a}{a-x} \] where \(a\) is the initial concentration and \(a-x\) is the remaining concentration after time \(t\).
Given that \(99\%\) of ethyl acetate is hydrolyzed, so only \(1\%\) remains.
Therefore, \[ \frac{a}{a-x}=\frac{100}{1}=100 \] Also, \[ t=10\,s \]

Step 3: Calculate the pseudo-first-order rate constant.
\[ k'=\frac{2.303}{10}\log(100) \] Since \[ \log(100)=2 \] we get \[ k'=\frac{2.303}{10}\times 2 \] \[ k'=0.4606\,s^{-1} \]

Step 4: Relate \(k'\) with the given rate constant \(k\).
Since \[ \text{Rate}=k[H_2O][CH_3CO_2C_2H_5] \] and water concentration is constant, we write \[ k'=k[H_2O] \] Thus, \[ k=\frac{k'}{[H_2O]} \] Substituting values, \[ k=\frac{0.4606}{2} \] \[ k=0.2303\,s^{-1} \]

Step 5: Final conclusion.
Therefore, the value of rate constant is \[ \boxed{0.2303\,s^{-1}} \]