A sample of milk splits after 60 min. at 300K and after 40 min at 400K when the population of lactob — Chemical Kinetics Chemistry Question
Question
A sample of milk splits after 60 min. at 300K and after 40 min at 400K when the population of lactobacillus acidophilus in it doubles. The activation energy (in kJ/mol) for this process is closest to ---. (Given, R = 8.3 J mol K ), , e = 4.0) −1 −1 −3
💡 Solution & Explanation
**Step 1: Identify the relationship between half-life and temperature** Since the population doubles (or halves) in different times at different temperatures, we're dealing with first-order kinetics where half-life (t₁/₂) is inversely related to the rate constant k. At 300K: t₁/₂ = 60 min At 400K: t₁/₂ = 40 min **Step 2: Apply the Arrhenius equation in ratio form** For two different temperatures: $$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ Since t₁/₂ ∝ 1/k, we have k₂/k₁ = t₁/t₂ = 60/40 = 1.5 **Step 3: Calculate the natural logarithm** ln(1.5) ≈ 0.405 ≈ 0.4 (given e = 4.0, so ln(4) = ln(e) = 1, thus ln(1.5) ≈ 0.4) **Step 4: Substitute temperature values** $$0.4 = \frac{E_a}{8.3}\left(\frac{1}{300} - \frac{1}{400}\right)$$ $$0.4 = \frac{E_a}{8.3}\left(\frac{4-3}{1200}\right) = \frac{E_a}{8.3} × \frac{1}{1200}$$ **Step 5: Solve for Eₐ** $$E_a = 0.4 × 8.3 × 1200 = 3.32 × 1200 = 3984 \text{ J/mol}$$ $$E_a ≈ 3.98 \text{ kJ/mol}$$ Therefore, the answer is 3.98.