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The rates of most the reactions double when the temperature is raised from 298 K to 308 K Calculate their activation energy. |
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Answer» Solution :From ARRHENIUS EQUATION ,we know `"LOG"K_1/K_2=E_a/(2.303R)[1/T_1-1/T_2]` Here `K_2/K_1=2,T_1=298K,T_2=308K`.SUBSTITUTING the values,we get `"log"2=E_a/(2.303xx8.314JK^-1mol_^-1)[(T_2-T_1)/(T_1T_2)]` `E_a=`52897.7mol^-1=52.89kj` MOL. |
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