For the first order reaction A → 2B, 1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half-life of the reaction is ...... min. (Round off to the nearest integer). [Use: ln 2 = 0.69, ln

Question

For the first order reaction A → 2B, 1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half-life of the reaction is ...... min. (Round off to the nearest integer). [Use: ln 2 = 0.69, ln 10 = 2.3, ln 3 = 1.1]

Prashant Kotian · Accepted Answer

Correct answer: 690.00. **Step 1: Set up the first-order integrated rate law** For a first-order reaction: ln([A]₀/[A]ₜ) = kt where [A]₀ = initial moles of A = 1 mole [A]ₜ = moles of A at time t **Step 2: Find moles of A remaining after 100 minutes** From stoichiometry A → 2B: - If 0.2 moles of B formed, then 0.1 moles of A reacted - Moles of A remaining = 1 - 0.1 = 0.9 moles **Step 3: Calculate the rate constant k** ln(1/0.9) = k × 100 ln(10/9) = k × 100 ln(10) - ln(9) = k × 100 2.3 - 2(1.1) = k × 100 2.3 - 2.2 = k × 100 0.1 = k × 100 k = 0.001 min⁻¹ **Step 4: Apply the half-life formula for first-order reactions** t₁/₂ = 0.693/k = ln(2)/k t₁/₂ = 0.69/0.001 = 690 minutes Therefore, the answer is 690.00.

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