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A first order reaction is 50% completed in 45 minute at $87^oC$ and in 9minutes at $127^oC$.The energy of activation of the reaction is

$\begin{array}{1 1}(a)\;2.82RKJ/mol&(b)\;5.81RKJ/mol\\(c)\;6.65RKJ/mol&(d)\;11.97RKJ/mol\end{array}$

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$ln\large\frac{k_2}{k_1}=\frac{E_a}{R}\big(\large\frac{1}{T_1}-\frac{1}{T_2}\big)$
$T_1=360K$
$T_2=400K$
$ln\large\frac{(t_{1/2})_1}{(t_{1/2})_2}=\frac{E_a}{R}\big(\frac{1}{360}-\frac{1}{400}\big)$
$ln \large\frac{45}{9}=\frac{E_a}{R}\big(\frac{1}{3600}\big)$
$E_a=5.81RKJ/mol$
Hence (b) is the correct answer.
answered Dec 17, 2013 by sreemathi.v
 

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