Question

Hydrolysis of sucrose follows 1st order kinetics to produce glucose and fructose. If \( t_{1/2} \) for decomposition of sucrose is 3 hr, find the % of sucrose left after 6 hr.

Answer 1

Correct Answer: 25

Step 1: Understanding the Question:
The question asks for the percentage of the reactant (sucrose) remaining after 6 hours, given that the reaction follows first-order kinetics with a half-life (\(t_{1/2}\)) of 3 hours.
Step 2: Key Formula or Approach:
For a first-order reaction, the amount remaining after \(n\) half-lives is given by:
\[ [A]_t = \frac{[A]_0}{2^n} \]
where \(n = \frac{t}{t_{1/2}}\) is the number of half-lives.
Step 3: Detailed Explanation:
Given:
Half-life \(t_{1/2} = 3\) hr.
Total time \(t = 6\) hr.
Calculate the number of half-lives (\(n\)):
\[ n = \frac{6 \text{ hr}}{3 \text{ hr}} = 2 \]
After 1 half-life (3 hr), 50% of the sucrose remains.
After 2 half-lives (6 hr), half of that amount remains:
\[ \text{Amount left} = \frac{100%}{2^2} = \frac{100%}{4} = 25%. \]
Step 4: Final Answer:
The percentage of sucrose left after 6 hours is 25%.