Question

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1/2} = 3$ hour. The percentage of sucrose remaining after 6 hours is ______. (Nearest integer)
(Given : $\log 2 = 0.3010$ and $\log 3 = 0.4771$)

Hint:

Calculate the number of half-lives that have passed in 6 hours. Since one half-life is 3 hours, two half-lives have passed. Each half-life reduces the concentration by half.

Answer 1

Correct Answer: 25

The hydrolysis of sucrose follows first-order kinetics. In first-order reactions, equal percentages of the reactant decompose in equal intervals of time.

Step 1: Understand the meaning of half-life ($t_{1/2}$).
The half-life is given as 3 hours. This means that every 3 hours, the amount of sucrose currently present will be reduced by half (50%).

Step 2: Track the concentration over time.
Let the initial percentage of sucrose be 100%.

After the first half-life (at $t = 3$ hours):
Amount remaining = $100\% / 2 = 50\%$

After the second half-life (at $t = 3 + 3 = 6$ hours):
Amount remaining = $50\% / 2 = 25\%$

Step 3: Conclusion.
Since the question asks for the percentage remaining after exactly 6 hours, which corresponds to exactly two half-lives, the remaining amount is 25%.

Calculated result: 25%.