1.

The rate constants of a reaction at 500K and 700K are 0.02s^(-1) and 0.07s^(-1), respectively. Calculatevalue of activation energy for the reaction.[ Given R=8.314JK^(-1)MI^(-1)].

Answer»

SOLUTION :Use the Arrhenius EQUATION
`"LOG"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
Given : `T_(1)=500K, T_(2)=700K, k_(1)=0.02s^(-1) and k_(2)=0.07s^(-1)`
Substituting the values in the equation above, we have
`"log"(0.07)/(0.02)=(E_(a))/(2.307xx8.314)[(700-500)/(500xx700)] or 0.5441 = (E_(a))/(19.1804) xx(200)/(350000)`
or `E_(a)=(0.5441xx19.1804xx3500)/(2)o=18263J=18.263kJ`.


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