1.

Rates of reaction double with every 10^@ rise in temperature. If this generalization holds for a reaction in the temperature ranges 298 K to308 K, what would be the value of activation energy for their reaction ?R = 8.314 J K^(-1) "mol"^(-1).

Answer»

Solution :`logK_2/K_1=E_a/(2.303R)[1/T_1-1/T_2]`
Here , `T_1`=298 K , `T_2`=308 K, `R=K^(-1) "mol"^(-1)`
`K_2/K_1=2`
`log2=E_a/(2.303xx8.314)[1/298-1/308]`
`0.3010=E_a/(2.303xx8.314)[10/(298xx308)]`
`E_a=(0.3010xx2.303xx8.314xx298xx308)/10`
`=52898 "J mol"^(-1)`
`=52.898 "KJ mol"^(-1)`


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