1.

The activation energy for the reaction 2HI_((g))toH_(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 freater than activation energy.

Answer»

SOLUTION :FRACTION of molecules =`e^(-E_(a))/(RT)`
`X=e^((209.5xx1000)/(8.314xx581))`
`X=e^(-43.37)`
`=1.461xx10^(-19)`
OR
`therefore K=A e^(-(E_(a))/(RT))`
`therefore (k)/(A)=e^(-(E_(a))/(RT))`
[But ,`(k)/(A)`=Fraction of molecule=X]
`therefore X=-e^(-(E_(a))/(2.303RT)`
`therefore log X=-(209.5xx10^(3)J mol)/(2.303xx8.134 J K^(-1)mol^(-1)xx581K)` `therefore` X=Antilog (-18.8323)=`1.4713xx10^(-19)`


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