The rate of First order reaction is 0.04?mol? at 10 minutes and 0.03 mol at 20 minutes after initiation. Half life of the reaction is

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Solution

Q.
The rate of First order reaction is 0.04 mol $L^{- 1}$ $s^{- 1}$ at 10 minutes and 0.03 mol $L^{- 1}$ $s^{- 1}$ at 20 minutes after initiation. Half life of the reaction is ______ minutes. (Given log2 = 0.3010, log3 = 0.4771) [2024]

Ans.
(24)

  For a first-order reaction $A \to$ products, the rate of the reaction is given by: $r = k {[A]}_{t} = k {[A]}_{0} e^{- kt}$

  Substituting value of rate at t = 10 min and t = 20 min:

at, t = 10 min :

   $0.04 mol L^{- 1} s^{- 1} = k {[A]}_{0} e^{- 10 k}$ (I)

at, t = 20 min :

$0.03 mol L^{- 1} s^{- 1} = k {[A]}_{0} e^{- 20 k}$ (II)

$I \div II$

$\frac{4}{3} = \frac{e^{- 10 k}}{e^{- 20 k}}$

   $e^{10 k} = \frac{4}{3}$

   $\ln$ $e^{10 k} = \ln \frac{4}{3}$

   $10 k$ $\ln$ $e = 2.303 \log_{10} \frac{4}{3}$

   $10 k = 2.303 \log_{10} \frac{4}{3}$    $(\ln$ $e = 1)$

$k = \frac{2.303}{10} \log_{10} \frac{4}{3}$ (III)

  Also for a first-order reaction,

   $t_{1 / 2} = \frac{2.303 \log_{10} 2}{k}$ or $k = \frac{2.303 \log_{10} 2}{t_{1 / 2}}$ (IV)

Equating III and IV:

$\frac{2.303 \log_{10} 2}{t_{1 / 2}} = \frac{2.303}{10} \log_{10} \frac{4}{3}$

   $\frac{\log_{10} 2}{t_{1 / 2}} = \frac{1}{10} (\log_{10} 4 - \log_{10} 3)$

$\frac{\log_{10} 2}{t_{1 / 2}} = \frac{1}{10} (2 \log_{10} 2 - \log_{10} 3)$

   $\frac{0.3010}{t_{1 / 2}} = \frac{1}{10} (2 \times 0.3010 - 0.4771)$

$t_{1 / 2} = \frac{0.3010}{0.01249} min = 24.1 min$

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