Question

Given: \[ \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \] Initial concentrations: Ethyl acetate = 0.001 M, Water = 2 M 99% hydrolyzed in 10 s. Find rate constant \( k \).

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

Hint:

For pseudo first-order reactions: \( k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \). Use correct fraction remaining!

Answer 1

The Correct Option is A

This is a pseudo first-order reaction because water is in large excess. Use: \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} = \frac{2.303}{10} \log \frac{0.001}{0.00001} = 0.2303 \cdot \log(100) = 0.2303 \cdot 2 = 0.4606 \, \text{s}^{-1} \] Wait—99% hydrolysis implies 1% remains: \[ [R] = 0.001 \cdot 0.01 = 10^{-5} \Rightarrow \log(100) = 2 \Rightarrow k = \frac{2.303 \cdot 2}{10} = 0.4606 \, \text{s}^{-1} \] BUT—in the question, rate constant is based on base-10 log for remaining fraction: \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} = \frac{2.303}{10} \cdot \log \frac{1}{0.01} = \frac{2.303}{10} \cdot 2 = 0.4606 \] So, final answer: \( \boxed{0.4606 \, \text{s}^{-1}} \) However, if we assume the intended answer is based on only 90% hydrolyzed, then: \[ [R]_0 = 0.001, [R] = 0.001 \times 0.01 = 10^{-5} \Rightarrow \log \frac{0.001}{10^{-5}} = \log 100 = 2 \] Still results in: \[ k = \frac{2.303 \cdot 2}{10} = 0.4606 \, \text{s}^{-1} \] So, based on correct interpretation, answer should be (not 0.2303) but: \[ \boxed{0.4606 \, \text{s}^{-1}} \Rightarrow \text{option (3)} \] % Correction Note Note: The answer marked as correct in the key (option 1) corresponds to 90% hydrolysis rather than 99%. Correct value should be 0.4606.