In a reaction at $27^{\circ} \mathrm{C}, 10^{-3} \%$ reactant molecules manage to cross over the barrier of…

In a reaction at $27^{\circ} \mathrm{C}, 10^{-3} \%$ reactant molecules manage to cross over the barrier of transition state. The energy of these molecules in excess of the average value will be $\left(R=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}ight)$ :
  1. $6.91 \mathrm{kcal} \mathrm{mol}^{-1}$
  2. $3.00 \mathrm{kcal} \mathrm{mol}^{-1}$
  3. $4.15 \mathrm{kcal} \mathrm{mol}^{-1}$
  4. $5.10 \mathrm{kcal} \mathrm{mol}^{-1}$

Solution

$e^{-E_{a} / R T}=10^{-3} \%=10^{-5}$
$E_{a}=2.303 \times 2 \times 300 \times 5 \mathrm{cal}$
$=6.91 \mathrm{kcal} \mathrm{mol}^{-1}$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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