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  • Given that the enthalpy of reaction is <img alt="\mathrm{\left ( \Delta rH \right )}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%20%28%20%5CDelta%20rH%20%5Cright%20%29

Given that the enthalpy of reaction is \mathrm{\left ( \Delta rH \right )} at \mathrm{350 k} is \mathrm{0.24\, kcal} & activation energy for the backward reaction is \mathrm{0.04\, kcal}.Calculate the percentage of the reactant molecules crossing over the energy barrier at \mathrm{350 \mathrm{~K}}.

Option: 1

88 \cdot 8 \%


Option: 2

90.5 \%


Option: 3

80.33 %


Option: 4

84.11 \%


Answers (1)

best_answer

\mathrm{\Delta_{r} H=E_{a(f)}-E_{a(b)} }
\mathrm{E_{ (f)}=\Delta_{r} H+E_{a(b)}=0.24+0.04=0.28\, \mathrm{kcal} }.

fraction of molecules crossing over the barrier
\mathrm{=\frac{n}{N}=x =e^{-E a / R T} }
\mathrm{\log (x) =\frac{-E_{a}(f)}{2.303 R T}=\frac{-0.28 \times 10^{3}}{2.303 \times 2 \times 350} }
\mathrm{x =0.8411 }

\mathrm{% } of molecules crossing over the barrier
\mathrm{=0.8411 \times 100=84.11 \% }.

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Riya

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