1.

The rate constant of a reaction at 400 and 200 K are 0.04 and 0.02s^(-1) respectively. Calculate the value of activation energy .

Answer»

Solution :According the ARRHENIUS equation `log(k_2/k_2)=E_a/(2.303R)((T_2-T_1)/(T_1T_2))`
`T_2 =400K , k_2=0.04s^(-1)`
`T_1=200K , k_1=0.02s^(-1)`
`log(0.04s^(-1)/(0.02s^(-1)))=(E_a)/(2.303xx8.314"J K "^(-1) MOL^(-1))((400K-200K)/(200Kxx400K))`
`log(2)=E_a/(2.303xx8.314" J K "^(-1)mol^(-1))((1)/(400K))`
`E_a=log(2)xx2.303xx8.314"JK "^(-1)"mol"^(-1)xx400K`
`E_a=2305"J mol"^(-1)`


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