Question

Ethyl acetate hydrolyse in water into acetic acid and ethanol with $t_{1/2}$=3.00 hours. What percent of sample of ethyl acetate remains after 9 hours?

Answer 1

12.5%

Answer 2

The hydrolysis of ethyl acetate in water is a first-order reaction, as indicated by the constant half-life ($t_{1/2}$).

Given: Half-life ($t_{1/2}$) = 3.00 hours Total time elapsed (t) = 9 hours

Step 1: Calculate the number of half-lives that have passed. The number of half-lives (n) is given by the total time divided by the half-life: $n = \frac{\text{Total time}}{t_{1/2}}$ $n = \frac{9 \text{ hours}}{3.00 \text{ hours}}$ $n = 3$

Step 2: Calculate the fraction of the sample remaining. For a first-order reaction, the fraction of the sample remaining after 'n' half-lives is given by the formula: $\text{Fraction remaining} = \left(\frac{1}{2}\right)^n$ Substituting the value of n: $\text{Fraction remaining} = \left(\frac{1}{2}\right)^3$ $\text{Fraction remaining} = \frac{1}{2 \times 2 \times 2}$ $\text{Fraction remaining} = \frac{1}{8}$

Step 3: Convert the fraction remaining to a percentage. Percentage remaining $= \text{Fraction remaining} \times 100\%$ Percentage remaining $= \frac{1}{8} \times 100\%$ Percentage remaining $= 0.125 \times 100\%$ Percentage remaining $= 12.5\%$

Therefore, 12.5% of the sample of ethyl acetate remains after 9 hours.

The chemical reaction is: $\text{CH}_3\text{COOCH}_2\text{CH}_3 \text{ (Ethyl acetate)} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH (Acetic acid)} + \text{CH}_3\text{CH}_2\text{OH (Ethanol)}$