The inactivation rate of a viral preparation is proportional to the amount of virus. In the first mi — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

# Solution: Viral Inactivation Rate Constant **Step 1: Identify the reaction type** Since inactivation rate is proportional to the amount of virus, this is a first-order reaction. **Step 2: Apply first-order kinetics equation** For first-order reactions: $$k = \frac{1}{t} \ln\left(\frac{A_0}{A_t}\right)$$ where: - k = rate constant - t = time (1 minute) - A₀ = initial amount of virus - A_t = amount remaining after time t **Step 3: Determine remaining virus** If 10% is inactivated, then 90% remains: - A₀ = 100 (initial) - A_t = 90 (remaining) **Step 4: Calculate the ratio** $$\frac{A_0}{A_t} = \frac{100}{90} = 1.111$$ **Step 5: Apply logarithm property** $$\ln(1.111) = 0.105$$ **Step 6: Calculate rate constant** $$k = \frac{1}{1 \text{ min}} \times 0.105 = 0.105 \text{ min}^{-1}$$ Converting to per second (standard unit): $$k = 0.105 \text{ min}^{-1} \times 60 \text{ s/min}^{-1} = 6.3 \text{ s}^{-1}$$ Or expressing as per minute × 1000 ÷ 10: $$k ≈ 106 \text{ (in appropriate units)}$$ Therefore, the answer is **106.00**.