1.

The activation energy for the reaction 2HI (g) to H_(2)(g) +I_(2)(g), is 209.5 kJ "mol"^(-1) at 581 K. Calculate the fraction of molecules of reactants hving energy equal to or greater than activation energy.

Answer»

Solution :Fraction of molecules having energy equal to or greater than activation energy, X is GIVEN by:
`x=(n)/(N)=e^(-E_(a)//RT) or "In"x= -(E_(a))/(RT) or " log"x= -(E_(a))/(2.303RT)`
SUBSTITUTING the values, we get `log x= -(209.5xx10^(3)"J mol"^(-1))/(2.303xx8.314JK^(-1)"mol"^(-1) XX 581K)= -18.8323`
`therefore x="Antilog" (-18.8323)="Antilog" bar(19).1677=1.471xx10^(-19)`.


Discussion

No Comment Found

Related InterviewSolutions