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Rate of reaction increases by factor of 10 when temperature increased from $300K$ to 400K.Then find activation energy of reaction

$\begin{array}{1 1}(a)\;20.23KJ\\(b)\;21.9KJ\\(c)\;22.97KJ\\(d)\;23.97KJ\end{array}$

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$K=Ae^{-E/RT}$
Applying Arrehenius equation
At two point we will get
$\log\large\frac{K_2}{K_1}=\frac{E_a}{2.303R}\bigg[\large\frac{1}{T_1}-\frac{1}{T_2}\bigg]$
Here $\large\frac{K_2}{K_1}$$=10$
$T_1=300K$
$T_2=400K$
So $\log 10=\large\frac{E_a}{2.303\times 8.314}$$\bigg[\large\frac{1}{300}-\frac{1}{400}\bigg]$
$E_a=\large\frac{1\times 2.303\times 8.314\times 300\times 400}{100}$
$E_a=22.97KJ$
Hence (c) is the correct answer.
answered Dec 18, 2013 by sreemathi.v
 

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