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The activation energy for the reaction,2HI(g)toH_2(g)+I_2(g)is 209.5kj mol^(-1)at 581 K.Calculate the fraction of molecules of reactant having energy equal to or greater than activation energy. |
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Answer» SOLUTION :Let the fraction be X `E^(E//RT)` i.e.,x=`e^(E//RT)` In `x= -E_a/(RT)log x=-E_a/(2.303RT)=(-209.5xx10^3J)/(2.303xx8.314xx581)=-18.8323` `thereforex=Antilog` of -18.8323=Antilog of `bar19 .1677=1.471xx10^(-19)` |
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