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The rate of most reactions become double when their temperature is raised from 298 K to 308 K. Calculate their activation energy. (Given, R = 8.314" J mol"^(-1) ) |
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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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