1.

The rate constant of a reaction at 500 K and 700 K are 0.02 s^(-1) and 0.07 s^(-1) respectively .Calculate the values of E_(a) and A.

Answer»

SOLUTION :Calculation of `E_(a)`:
`log(K_(2))/(k_(1))=(E_(a))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))`
`log (0.07 s^(-1))/(0.02 s^(-1))=(E_(a))/(2.303xx8.314 JK^(-1)mol^(-1))xx((700-500)/(700xx500))`
`therefore log 3.5=(E_(a)xx200)/(2.303xx8.314xx700xx500)J mol^(-1)`
`therefore E_(a)(0.5441xx2.303xx8.314xx700xx500)/(200) J mol^(-1)` Calculation of A:
`K=Ae^(-(E_(a))/(RT))`
Where,`E_(a)=18231 J mol^(-1)`
500 K temp.`K_(1)=0.02 s^(-1)`


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