A flask contains a mixture of compounds A and B . Both compounds decompose by first-order kinetics. The half-lives are 54.0 min for A and 18.0 min for B . If the concentrations of A and B are equal in

Question

A flask contains a mixture of compounds A and B . Both compounds decompose by first-order kinetics. The half-lives are 54.0 min for A and 18.0 min for B . If the concentrations of A and B are equal initially, how long (in minute) will it take for the concentration of A to be four times that of B ?

TestHub · Accepted Answer

Containing mixture of compound A and B both decomposed with first order kinetics

$t_{1/2}$ for $A = 54 min t_{1/2}$ for $B = 18 min$

$A_{t} = A_{oe} e^{- k_{1} t}$ .....(i)

$B_{t} = B_{oe} e^{- k_{1} t}$ .....(ii)

$A_{o} = B_{o}$

$A_{t} = 4 B_{t}$

by (i)/(ii)

$\frac{4 B _{t}}{B _{t}} = (\frac{B _{0} e ^{- k_{1} t}}{B _{0} e ^{- k_{2} t}})$

$4 = e^{(k_{2} - k_{1}) t}$

$ln 4 = (k_{2} - k_{1}) t$

$t = \frac{l n 4}{( k _{2} - k _{1} )} = \frac{l n 4}{( k _{2} - k _{1} )} = \frac{l n 4}{\frac{0.6932}{18} - \frac{0.6932}{54}} = 54 min$ .

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