JEE PYQ: Chemical Kinetics Question 18
Question 18 - 2020 (05 Sep Shift 1)
A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is: (Use $\ln 2 = 0.693$)
(1) 180
(2) 900
(3) 300
(4) 120
Show Answer
Answer: (2)
Solution
$(C_t)_A = (C_0)_A e^{-k_A t}$; $k_A = \frac{\ln 2}{300}$. $(C_t)_B = (C_0)_B e^{-k_B t}$; $k_B = \frac{\ln 2}{180}$. $\frac{(C_t)_B}{(C_t)_A} = \frac{(C_0)_B}{(C_0)_A} \times e^{(k_B - k_A)t}$. $4 = e^{(k_B - k_A)t}$. $2\ln 2 = \left[\frac{\ln 2}{180} - \frac{\ln 2}{300}\right]t$. $2 = \frac{120}{180 \times 300}t$. $t = \frac{2 \times 180 \times 300}{120} = 900$ sec.
BETA
