Question

Two substances A (t1/2 = 5 min) and B (t1/2 = 15 min) are taken in such a way that initially [A] = 4[B]. The time after which both the concentration will be equal is : (Assume that reaction is first order)

• A. 5 min
• B. 15 min
• C. 20 min
• D. Concentration can never be equal

Answer 1

Answer: (B)

15 min

Answer 2

$C_{t}\text{ = }\text{C}_{0}\text{ }\text{e}^{\text{-Kt}}$

According to question

$C_{A't} = C_{B't}$

$C_{A}e^{\text{-}\text{K}_{A}t}\text{= }\text{C}_{B}e^{\text{-}\text{K}_{B}t}$

$4 = e^{\left\lbrack \frac{In2}{5} - \frac{In2}{15} \right\rbrack \times t}$

$In4 = \left\lbrack \frac{In2}{5} - \frac{In2}{15} \right\rbrack t$

$In(2)^{2} = \left\lbrack \frac{In2}{5} - \frac{In2}{15} \right\rbrack t$

$2In2 = \left\lbrack \frac{In2}{5} - \frac{In2}{15} \right\rbrack t$

$2 = \left\lbrack \frac{1}{5} - \frac{1}{15} \right\rbrack t$

$2 = \frac{2}{15} \times t$

$t = 15$ minute