1.

A first order reaction is 50% completed in 20 minutes at 27^(@)C. The energy of activation of the reaction is-

Answer»

43.58 KJ
55.14 KJ
11.97 KJ
6.65 KJ

Solution :`"Log"((K_(2))/(K_(1)))=(Ea)/(2.303R)((1)/(T_(1))-(1)/(T_(2)))`
`Log4=(Ea)/(2.303xx8)((1)/(300)-(1)/(320))`
`{K_(1)=(1n2)/(20),K_(2)=(n2)/(5)}`


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