A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A

Question

A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is_____s. (Given: ln 2 = 0.693)

Prashant Kotian · Accepted Answer

Correct answer: 200. **Step 1: Set up the first-order decay equation** For radioactive decay (first-order): N(t) = N₀e^(-λt) Where λ = ln(2)/t₁/₂ **Step 2: Calculate decay constants** For A: λ_A = 0.693/100 = 0.00693 s⁻¹ For B: λ_B = 0.693/50 = 0.01386 s⁻¹ **Step 3: Write concentration expressions** Since initial moles are equal (N₀_A = N₀_B = N₀): [A] = N₀e^(-0.00693t) [B] = N₀e^(-0.01386t) **Step 4: Apply the given condition** We need [A] = 4[B] N₀e^(-0.00693t) = 4 × N₀e^(-0.01386t) **Step 5: Solve for time** e^(-0.00693t) = 4e^(-0.01386t) e^(-0.00693t + 0.01386t) = 4 e^(0.00693t) = 4 0.00693t = ln(4) = 2ln(2) = 2(0.693) = 1.386 t = 1.386/0.00693 = 200 s Therefore, the answer is 200.

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