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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`

A

`80.65%`

B

`70.65%`

C

`60.65%`

D

`50.65%`

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta H = E_(a(f)) - E_(a(b))`
`E_(a(f)) = 0.12 xx 10^(3) + 0.02 xx 10^(3) cal`
`= 0.14 xx 10^(3) cal`
[The value of `R` in calories `= 1.98 cal ~~ 2.0 cal`]
Fraction of molecules crosisng over the barrier
`= (n)/(N) = x = e^(-E_(a(f)//RT))`
`:. log x = (-E_(a(f)))/(2.303 RT) = (0.14 xx 10^(3) cal)/(2 xx 325)`
`:. x = 0.8065`
`%` of molecules crosisng over the barrier
`= 0.8065 xx 100 = 80.65%`
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