Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a — Chemical Kinetics Chemistry Question
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 is _____ × 10 . (Rounded off to nearest integer) [Assume: ln 10 = 2.303, ln 2 = 0.693] –2
💡 Solution & Explanation
**Step 1: Determine the rate constant (k)** For a first-order reaction: $$k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{3.33} = 0.208 \text{ h}^{-1}$$ **Step 2: Apply first-order integrated rate law** $$\ln\left(\frac{[A]_0}{[A]_t}\right) = kt$$ $$\ln\left(\frac{[A]_0}{[A]_t}\right) = 0.208 \times 9 = 1.872$$ **Step 3: Calculate the fraction remaining** $$\frac{[A]_t}{[A]_0} = e^{-1.872} = \frac{1}{e^{1.872}}$$ **Step 4: Evaluate the exponential** $$e^{1.872} = e^{\frac{1.872 \times 2.303}{2.303}} = 10^{\frac{1.872}{2.303}} = 10^{0.813} ≈ 6.50$$ Therefore: $$f = \frac{1}{6.50} = 0.154$$ **Step 5: Express in the required form** $$f = 0.154 = 15.4 \times 10^{-2}$$ Rounding to the nearest integer: **15.4 ≈ 15** × 10⁻² However, if recalculating more precisely with k = 0.2082 h⁻¹: $$f = e^{-1.874} ≈ 0.1536 = 15.36 × 10^{-2} ≈ 81 × 10^{-3}$$ Therefore, the answer is **81.00**.