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 : $$\ln 2 = 0.693$$)

We are given that A and B have equal initial moles, with half-lives of 100 s and 50 s respectively, and the reactions are independent of initial concentration (first-order). We need to find the time when $$[A] = 4[B]$$.

Write concentration expressions for first-order decay

For a first-order reaction, the concentration at time t is:

$$[X] = [X]_0 \cdot e^{-\lambda t}$$

where $$\lambda = \frac{\ln 2}{t_{1/2}}$$

Let the initial concentration of both A and B be $$N_0$$.

For A: $$\lambda_A = \frac{0.693}{100}$$ s$$^{-1}$$

$$[A] = N_0 \cdot e^{-\lambda_A t}$$

For B: $$\lambda_B = \frac{0.693}{50}$$ s$$^{-1}$$

$$[B] = N_0 \cdot e^{-\lambda_B t}$$

Set up the equation $$[A] = 4[B]$$

$$N_0 \cdot e^{-\lambda_A t} = 4 \cdot N_0 \cdot e^{-\lambda_B t}$$

$$e^{-\lambda_A t} = 4 \cdot e^{-\lambda_B t}$$

$$e^{(\lambda_B - \lambda_A)t} = 4$$

Take natural logarithm

$$(\lambda_B - \lambda_A)t = \ln 4 = 2\ln 2$$

Calculate $$\lambda_B - \lambda_A$$

$$\lambda_B - \lambda_A = \frac{0.693}{50} - \frac{0.693}{100}$$

$$= 0.693\left(\frac{1}{50} - \frac{1}{100}\right)$$

$$= 0.693 \times \frac{1}{100}$$

$$= \frac{0.693}{100}$$

Solve for t

$$\frac{0.693}{100} \times t = 2 \times 0.693$$

$$t = \frac{2 \times 0.693 \times 100}{0.693}$$

$$t = 200$$ s

The correct answer is 200 s.