Question

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25 °C. After 9 h, the fraction of sucrose remaining is \( f \). The value of \( \log_{10}(1/f) \) is _______ × 10⁻². (Rounded off to the nearest integer)

Hint:

$\log_{10}(A_0/A) = \frac{kt}{2.303}$. This is the most common form for solving first-order kinetics problems.

Answer 1

Correct Answer: 81

Step 1: For first order, $A = A_0 e^{-kt}$ or $\frac{A_0}{A} = 2^{t/t_{1/2}}$.
Step 2: Fraction remaining $f = \frac{A}{A_0} = \frac{1}{2^{t/t_{1/2}}}$.
Step 3: $\frac{1}{f} = 2^{9/3.33} \approx 2^{2.702}$.
Step 4: $\log_{10} (1/f) = \frac{9}{3.33} \log_{10} 2 = 2.702 \times 0.3010 \approx 0.8135$.
Step 5: $0.8135 = 81.35 \times 10^{-2} \approx 81 \times 10^{-2}$. Thus, $x = 81$.