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The activation energy for the reaction `:` ltbr. `2Hl(g) rarr H_(2)(g)+I_(2)(g)` is `209.5 kJ mol^(-1)` at `581K`. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ? |
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Answer» Correct Answer - `1.47xx10^(19)` Fraction of molecules having energy equal to or greater than activation energy is given by `:` `(n)/(N)=(x)=e^(-Ea//RT)` `:.ln (x)=(-Ea)/(RT)` `or log (x)=(-E_(a))/(2.303RT)` `=(209.5xxJ mol^(-1))/(2.303xx8.314J K^(-1) mol ^(-1)xx581K)` `=-18.8323` `:. x=Antilog(-18.8323)` `=Antilog(-18.8323+1-1)` `=Antilogbar(19).1677=1.471xx10^(-19)` |
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