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JEE MainChemistryToughChemical Kinetics

Find out the percentage of the reactant molecules crossing over
the activation energy barrier at 325 K, given that DH325 = 0.12 kcal,
Ea(b) = + 0.02 kcal.

Show answer and solution
Given that, ΔH=0.12×103cal\Delta H = 0.12 \times 10^{3}cal
Ea(b)=0.02×103calE_{a(b)} = 0.02 \times 10^{3}cal (EaNever negative)(E_{a}Never\ negative)
∵ΔH=Ea(f)−Ea(b)\because\Delta H = E_{a(f)} - E_{a(b)}
∴Ea(f)=0.12×103+0.02×103cal=0.14×103cal\therefore E_{a(f)} = 0.12 \times 10^{3} + 0.02 \times 10^{3}cal = 0.14 \times 10^{3}cal of molecule crossing over the barrier
= 100 \times e^{- E_{a}(f)/RT} = 100 \times \times 325} = 100 \times 0.8065 = 80.65\

Source: Grass-G / CH-15 Chemistry Paper 11 Nov..docx

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