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 : In 2 = 0.693)

Answer 1

Correct Answer: 200

To determine the time required for the concentration of A to be four times that of B, follow these steps:

1. **Understanding the decay process:**
For radioactive decay, the concentration of substance at time \(t\) is given by:
\[ [X]_t = [X]_0 \times e^{-kt} \] where \([X]_0\) is the initial concentration, \(k\) is the decay constant, and \(t\) is time.

2. **Decay constants for A and B:**
Given the half-life formula \(t_{1/2} = \frac{0.693}{k}\), we can calculate them:
For A: \(t_{1/2} = 100\) s, so \(k_A = \frac{0.693}{100} = 0.00693 \text{ s}^{-1}\).
For B: \(t_{1/2} = 50\) s, so \(k_B = \frac{0.693}{50} = 0.01386 \text{ s}^{-1}\).

3. **Initial conditions:**
\([[A]_0 = [B]_0\) because the flask is filled with equal moles of A and B initially.

4. **Finding the time when \([A] = 4[B]\):**
\[[A]_t = [A]_0 \times e^{-k_A t} = 4[B]_t = 4 \times [B]_0 \times e^{-k_B t}\]
\[e^{-0.00693 t} = 4 \times e^{-0.01386 t}\]
Taking natural logarithm on both sides:
\(-0.00693 t = \ln(4) - 0.01386 t\)
\(\ln(4) = 2 \times \ln(2) = 2 \times 0.693 = 1.386\)
\(-0.00693 t + 0.01386 t = 1.386\)
\(0.00693 t = 1.386\)
\(t = \frac{1.386}{0.00693}\)
\(t \approx 200 \text{ s}\)

5. **Verification:**
The calculated time, \(t \approx 200\) s, is within the expected range (200,200) as specified.