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The rate of a reaction increases to four times when the prevailing temperature is raised from 300 K to 320 K. Calculate the energy of activation of this reaction assuming that it does not change with temperature.""[R=8.314"J mol"^(-1)K^(-1)] |
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Answer» Solution :Arrhenius EQUATION relating the RATE CONSTANT and temperature is given below: `"log"(k_(2))/(k_(1))=(E_(a))/(2.303R) xx ((1)/(T_(1)) - (1)/(T_(2)))` SUBSTITUTING the values, we get `log 4= (E_(a))/(2.303xx8.314)xx ((1)/(300)-(1)/(320)) or log 4= (E_(a))/(19.147) ((20)/(300xx320))` `or E_(a)=(19.147 xx 32xx3xx0.6021)/(20)=55.336"kJ mol"^(-1)`. |
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