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The activation energy for the reaction, 2HI_((g))rarr H_(2(g))+I_(2(g)) is 209.5 KJ mol^(-1) at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy. |
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Answer» Solution :In the given case `E_(a)=209.5 "KJ mol"^(-1)=209500J mol^(-1)` T = 581 K `R = 8.314 JK^(-1) mol^(-1)` Now the FRACTION of molecules of reactants having energy equal to or greater than activation energy is given as. `x = e^(-E_(a)//RT)` `rArr log x = - E_(a//RT)` `rArr log x =-(E_(a))/(2.303 RT)` `rArr log x = (209500 "J mol"^(-1))/(2.303xx8.314 JK^(-1)mol^(-1)xx581)` `= 18.8323` x = Antilogs (18.8323) = Antilogs `BAR(19).1677` `= 1.471xx10^(-19)`. |
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