# find out the percentage of the reactant molecules crossing over the energy barrier at 325 K, given that ΔH325=0.12kcal, Ea(backward)=+0.02kcal.give the detailed solution.

Sheershak
8 Points
7 years ago
$eq=\Delta H = E_{a_f} - E_{a_b}\\ E_{a_f} = 0.12 + 0.02 \;\;kcal \;\; = 140\;cal/mol.K\\ R \; = 2\;cal/mol.K\\\\ Fraction \;of\;molecules\;with\;energy\;greater\;than\;E_a = e^{-\frac{E_a}{RT}} \\ x = e^{-\frac{140}{2*325}}\\ ln(x) = -\frac{140}{2*325}\\ log(x) = -\frac{140}{2*325*2.303}\\ log(x) = -0.0935234978\\ x = antilog({-0.0935234978}) \\ \Rightarrow x = 0.8063$

Thus, percentage of molecules with energy greater than or equal to activation energy is 80.6%