Percentage of Sucrose Remaining After Hydrolysis Reaction

To find the percentage of sucrose remaining after $6\text{ hours}$, we use the principles of first-order reaction kinetics.

Step 1: Determine the number of half-lives ($n$) The half-life of the reaction is given as $t_{1/2} = 3\text{ hours}$. The total time elapsed is $t = 6\text{ hours}$.

The number of half-lives that have passed is: $n = \frac{t}{t_{1/2}} = \frac{6\text{ hours}}{3\text{ hours}} = 2$

Step 2: Calculate the fraction of sucrose remaining For a first-order reaction, the concentration of the reactant remaining after $n$ half-lives, relative to its initial concentration $[A]_0$, is given by: $\frac{[A]}{[A]_0} = \left(\frac{1}{2}\right)^n$

Substituting $n = 2$: $\frac{[A]}{[A]_0} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

Step 3: Calculate the percentage remaining The percentage of sucrose remaining is: $\text{Percentage remaining} = \frac{[A]}{[A]_0} \times 100 = \frac{1}{4} \times 100 = 25\%$

Thus, the percentage of sucrose remaining after $6\text{ hours}$ is 25.