1.

The rate constant of a particular reaction doubles when the temperature changes from 300K to 310 K, calculate the energy of activation.

Answer»

Solution :DATA: constant at 300 K is `K_(1)`
Rate constant at 310 K is `K_(2), "" K_(2) = 2K, "" R = 8.314 J K^(-1)MOL^(-1)`
Formula: `(k_(2))/(k_(1)) = (Ea)/(2.303R) {(T_(2) - T_(1))/(T_(1)T_(2))}`
Substitution: `log ""(2k_(1))/(K_(1)) = (E_(a))/(8.314 xx 2.303) [(310 - 300)/(300 xx 310)]`
`log 2 = (E_(a))/(8.314xx2.303)xx(10)/(300xx310)`
`E_(a) = (0.3010xx8.314xx2.303xx300xx310)/(10)`
`E_(a) = 53.598 kJ mol^(-1)`


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