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: Use the first-order rate law.
For a first-order reaction, the half-life is given by:
\[ t_{1/2} = \frac{0.693}{k} \] Where \( k \) is the rate constant. We are given \( t_{1/2} = 3 \, \text{hr} \), so we can solve for \( k \):
\[ k = \frac{0.693}{3} = 0.231 \, \text{hr}^{-1} \]
Step 2: Use the first-order integrated rate equation.
For a first-order reaction, the concentration of the reactant at any time \( t \) is given by:
\[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \] Where \( [A_0] \) is the initial concentration and \( [A] \) is the concentration at time \( t \). Since we are looking for the percentage of sucrose left, we can rearrange this to:
\[ \frac{[A]}{[A_0]} = e^{-kt} \]
Step 3: Calculate the fraction of sucrose left after 6 hours.
Substitute \( k = 0.231 \, \text{hr}^{-1} \) and \( t = 6 \, \text{hr} \) into the equation:
\[ \frac{[A]}{[A_0]} = e^{-0.231 \times 6} \] \[ \frac{[A]}{[A_0]} = e^{-1.386} = 0.250 \]
Step 4: Calculate the percentage of sucrose left.
The percentage of sucrose left is:
\[ \text{Percentage of sucrose left} = 0.250 \times 100 = 25% \]
Step 5: State the final answer.
Hence, the percentage of sucrose left after 6 hours is:
\[ \boxed{25%} \]