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Q.
A first order reaction is $50 \%$ completed in $20\, \min$ at $27^{\circ} C$ and in $5\, \min$. at $47^{\circ} C$ the energy of activation of the reaction is :-
Solution:
$t _{1 / 2} \propto \frac{1}{ k }$
$\log \frac{ k _{2}}{ k _{1}}=\frac{ E _{ a }}{2.303 R }\left(\frac{ T _{2}- T _{1}}{ T _{1} T _{2}}\right)$
$\log \frac{20}{5}=\frac{ E _{ a }}{2.303 \times\left(\frac{8.314}{1000}\right)}\left(\frac{320-300}{300 \times 320}\right)$