1.

Find out the percentage of the reactant molecules crossing over the energy barrier at 325K, given that DeltaH_(325)=0.12kcal, E_(a(b))=+0.02kcal.

Answer»

Solution :Given , `DeltaH=0.12xx10^(3)cal`,
`E_(a(B))=0.02xx10^(3)cal` (`E_(a)` can NEVER be negative)
`:. DeltaH=E_(a(f))-E_(a(b))`
`:.E_(a(f))=0.12xx10^(3)+0.02xx10^(3)cal`
`=0.14xx10^(3)cal`
`%` of molecule crossing over the BARRIER
`=100xxe^(-E_(a(f))//RT)`
`=100xxe^(-140//(2xx325))`
`=100xx0.8065=80.65%`


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