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The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_(a)) of the reaction aassuming that it does not change with temperature. |
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Answer» Solution :Given if rate at 293 K is R thus at 313 K rate BECOMES = 4R `LOG.(k_(2))/(k_(1)) = (E_(a))/(2.303 R)[(T_(2)-T_(1))/(T_(1) xx T_(2))]` `log.(4R)/(R ) = (E_(a))/(2.303 xx 8.314)[(313 - 298)/(293xx 313)]` `log4 = (E_(a))/(19.1471)[(20)/(91709)]` `0.6021 = (E_(a))/(19.1471)[(20)/(91709)]` `(0.6021 xx 19.1471 xx 91709)/(20) = E_(a)` `E_(a) = 52863.2177 J mol^(-1)` or `= 52.863 kJ mol^(-1)` |
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