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The rate of a particular reaction quadruples when the temperature changes from 293 K to 313 K. Calculate activation energy in KJ/mol. |
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Answer» Solution :`K_2//K_1`=4 `T_1`=293 K , `T_2`=313 K `logK_2/K_1=-(E_a)/(2.303R)[1/T_1-1/T_2]` Thus, on calculating and substituting values, we GET : `E_a=52.86 KJ "MOL"^(-1)` |
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