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Find out the percentage of the reactant ...

Find out the percentage of the reactant molecules crosisng over the energy barrier at `325 K`.
Given: `Delta H_(325 K) = 0.12 kcal`,
`E_(a(b)) = 0.02 kcal`

Text Solution

Verified by Experts

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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