1.

A first order reaction is 50% complete in 50 minutres at 300K and the same reaction is again 50% complete in 25 minutes at 350K . Calculate activation energy of the reaction .

Answer»

Solution : The RELATION between HALF- life period and rate CONSTANT for a first order reaction is GIVEN by
`t_(1//2)= (0.693)/(K)`
Substitutiong the value , we have
`K_(2)(350K)= (0.693)/(25)` and `K_(1)(300K)= (0.693)/(50)`
`therefore (K_(2))/(K_(1)= 2)`
Applying Arrhenius equation , we have
log `(K_(2))/(K_(1)) = (e_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
Or log `2= (E_(a))/(2.303xx8.314)[(350-300)/(350xx300)]`
Or `e_(a)=12.104Kj//mol`


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