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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 ?A. `1.82xx10^(-18)`B. `1.47xx10^(19)`C. `2.67xx10^(16)`D. `3.89xx10^(19)` |
Answer» Correct Answer - B Fraction of molecules (x) having energy equal to or greater than activation energy may be calculated as follows: `x=n//N=e^(-Ea)"//RT"` In `x=-E_(a)/(RT)or logx=-E_(a)/(2.303RT)` or `logx=-(209xx10^(3)J" mol"^(-1))/(2.303xx(8.314Jk^(-1)" mol"^(-1))xx581k)` `=-18.8323` `x=Antilog(-18.8323)` `=1.471xx10^(-19)` Fraction of molecules `=1.47xx10^(-19` |
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