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The rate constant of a first order reaction increases from 2xx10^(-2) to 8xx10^(-2) when the temperature changes from 300 K to 320 K. Calculate the energy of activation. |
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Answer» SOLUTION :We shall use the Arrhenius EQUATION `"log"(k_(2))/(k_(1))=(E_(a))/(2.303xxR)[(T_(2)-T_(1))/(T_(1)T_(2))]` Given : `k_(1)=2xx10^(-2), k_(2)=8xx10^(-2), T_(1)=300K, T_(2)=320K and R=8.314JK^(-1)"MOL"^(-1)` SUBSTITUTING these values in the above equation, we get `"log"(8xx10^(-2))/(2xx10^(-2))=(E_(a))/(2.303xx8.314)[(320-300)/(320xx300)]` `or log 4=(E_(a))/(19.147)xx(20)/(96000) or 0.6021=(E_(a))/(19.147)xx(1)/(4800)` `or E_(a)=0.6021xx19.147xx4800 or E_(a)=55.34"kJ mol"^(-1)` |
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