At 300K, a certain reaction is `50%` complete in 20 minutes. At 350K, the same reaction is `50%` complete in 5 minutes. Calcualte the activation energy for the reaction.

At 300K, a certain reaction is `50%` complete in 20 minutes. At 350K,

1.	At 300K, a certain reaction is `50%` complete in 20 minutes. At 350K, the same reaction is `50%` complete in 5 minutes. Calcualte the activation energy for the reaction.
Answer» `k_(1) = 0.693/(20 mm) = 0.03465 mm^(-1), T_(1)=300K` `k_(2) = 0.693/(5 mm) = 0.1386 mm^(-1), T_(2)=350K` `log k_(2)/k_(1) = E_(a)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2)]` `log(0.1386 min^(-1))/(0.03465 min^(-1)) = (E_(a)xx[350K -300K])/(2.303 xx 8.314 JK^(-1)mol^(-1) xx 350K xx 300K)` `log4=(E_(a) xx 50)/(2.303 xx 8.314 xx 350 xx 300(Jmol^(-1))` `E_(a)=(0.6021 xx 2.303 xx 8.314 xx 350 xx 300)/(50) Jmol^(-1)` `24210 J mol^(-1)=24.21 kj mol^(-1)`.

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At 300K, a certain reaction is `50%` complete in 20 minutes. At 350K, the same reaction is `50%` complete in 5 minutes. Calcualte the activation energy for the reaction.

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