Question

Sucrose hydrolyses in acidic medium to form glucose and fructose which follows first order kinetics. If the half-life of sucrose is 3 hrs, find the % of sucrose left after 6 hrs.

Answer 1

Correct Answer: 25

Step 1: Use the concept of half-life in first order kinetics.
For a first order reaction, the amount of reactant becomes half after every one half-life.
Here, the half-life of sucrose is given as:
\[ t_{1/2} = 3 \text{ hrs} \] Step 2: Calculate how many half-lives have passed in 6 hours.
The total time given is 6 hours. Therefore, the number of half-lives elapsed is:
\[ n = \frac{6}{3} = 2 \] Step 3: Determine the fraction of sucrose left after 2 half-lives.
After one half-life, the amount left becomes:
\[ \frac{1}{2} \] After two half-lives, the amount left becomes:
\[ \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] Step 4: Convert the fraction into percentage.
The fraction of sucrose left after 6 hours is:
\[ \frac{1}{4} \times 100 = 25% \] Step 5: State the final result.
Hence, the percentage of sucrose left after 6 hours is:
\[ \boxed{25%} \]