Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with hours. What fraction of sample of sucrose remains after 8 hours?
The fraction of sucrose remaining after 8 hours is approximately 0.157 (or 15.7%).
Explanation
This is a first-order reaction problem. The context confirms that sucrose decomposition into glucose and fructose follows a first-order rate law. For first-order reactions, the rate constant k can be calculated from half-life using k = 0.693/t½. Then the integrated rate law ln([A]₀/[A]ₜ) = kt is used to find the fraction remaining.
Solution Steps
Step 1: Calculate the rate constant (k) from the given half-life. For a first-order reaction: k = 0.693/t₁/₂ k = 0.693/3.00 hours = 0.231 h⁻¹
Step 2: Apply the first-order integrated rate law. For first-order reactions: [A]ₜ/[A]₀ = e⁻ᵏᵗ Substituting values: [A]ₜ/[A]₀ = e⁻⁽⁰·²³¹⁾⁽⁸⁾
Step 3: Calculate the final value. [A]ₜ/[A]₀ = e⁻¹·⁸⁴⁸ = 0.157
Therefore, the fraction of sucrose remaining after 8 hours is 0.157 or approximately 15.7%.