A flask contains a mixture of A and B. Both the compounds decompo… - Vidyadip

Question

A flask contains a mixture of $A$ and $B.$ Both the compounds decompose by first order kinetics. The half$-$lives are $60 \ min$ for $A$ and $15 \ min$ for $B.$ If the initial concentrations of $A$ and $B$ are equal, how long will it take for the concentration of $A$ to be three times that of $B$?

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Answer

Given:
For $A : tm = 60 \ min$ For $B: t_{1/2} = \ 15 \ min$
Let initial concentrations of
$[A] = [B]_o = M \ mol \ dm^{-3}$
After time t, let the concentrations be, $[B]_t = x_t$ then $[A]_t = 3x$
$k_{ A }=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{60}=0.01155 \ min^{-1}$
$k_{ B }=\frac{0.693}{15}=0.0462 \ min^{-1}$
$k_{ B }=\frac{2.303}{t} \log _{10} \frac{[ B ]_0}{[B]_t}$
$=\frac{2.303}{t} \log _{10} \frac{ M }{x}$
$\therefore \log _{10} \frac{ M }{x}=\frac{k_{ B } \times t}{2.303}=\frac{0.0462 \times t}{2.303}=0.02 t$
Now,
$k_{ A }=\frac{2.303}{t} \log _{10} \frac{[ A ]_0}{[A]_t}$
$=\frac{2.303}{t} \log _{10} \frac{ M }{3 x}$
$=\frac{2.303}{t}\left[\log _{10} \frac{M}{x}+\log _{10} \frac{1}{3}\right]$
$=\frac{2.303}{t}[0.02 t+(\overline{1} .5224)]$
$=\frac{2.303}{t}[0.02 t -0.4776]$
$=2.303 \times 0.02-\frac{2.303 \times 0.4776}{t}$
$0.01155=0.04606-\frac{1.1}{t}$
$\therefore \frac{1.1}{t}=0.04606-0.01155=0.03451$
$\therefore t=\frac{1.1}{0.03451}=31.8 \ min$
After $31.8 \ min,$ concentration of $A$ will be three time that of $B.$

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