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InterviewSolution
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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%` |
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Answer» Correct Answer - 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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