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A flask contains a mix of compounds a and b. both compounds decomposed by 1st order kinetics. the half lives are 54 min for a and 18 min for b. if the

conc. of a and bare equal initially, how long will it take for the conc. of a to be 4 times of b.



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For first order reaction,
ln C = ln C₀ - Kt
⇒ ln (C/C₀) = - Kt
where C is concentration of the compounds.

half life of A = 54 min
ln(1/2) = -Ka×54
⇒ln 2 = Ka×54
⇒Ka = (ln 2)/54

half life of B = 18 min
ln(1/2) = -Kb×18
⇒ln 2 = Kb×18
⇒Kb = (ln 2)/18

Initial concentration was same
A₀ = B₀
Let at time t, concentration of A becomes 4 times of B
At = 4 Bt

A_t =A_0\  e^{-k_at} =A_0\  e^{- \frac{ln2}{54} t}

B_t =B_0\  e^{-k_bt} =B_0\  e^{- \frac{ln2}{18} t}

Dividing both, we get

 \frac{A_t}{B_t} = \frac{A_0\ e^{- \frac{ln2}{54} t} }{B_0\ e^{- \frac{ln2}{18} t} }= e^{-ln2*t( \frac{1}{54} - \frac{1}{18} )}  \\  \\ 4=e^{0.693t( \frac{1}{18} - \frac{1}{54} )}=e^{0.693t( \frac{2}{27} )}= e^{0.0513t}  \\  \\ ln\ 4=0.0513t \\  \\ t =  \frac{1.386}{0.0513} =27\ minutes

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