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STD 12 - 3. Chemical kinetics — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3. Chemical kinetics4 Marks
MCQ
For a reaction taking place in three steps at same temperature, overall rate constant $\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3}$. If $\mathrm{Ea}_1, \mathrm{Ea}_2$ and $\mathrm{Ea}_3$ are $40$,$50$ and $60 \mathrm{~kJ} / \mathrm{mol}$ respectively, the overall $\mathrm{Ea}$ is ________ $\mathrm{kJ} / \mathrm{mol}$.
  • A
    $20$
  • B
    $10$
  • ✓
    $30$
  • D
    $45$
✓

Answer

Correct option: C.
$30$
c
$ \mathrm{K}=\frac{\mathrm{K}_1 \cdot \mathrm{K}_2}{\mathrm{~K}_3}=\frac{\mathrm{A}_1 \cdot \mathrm{A}_2}{\mathrm{~A}_3} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}\right)}{R T}} $

$ \mathrm{~A} \cdot \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}=\frac{\mathrm{A}_1 \mathrm{~A}_2}{\mathrm{~A}_3} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}\right)}{R T}}$

$\mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}=40+50-60=30 \mathrm{~kJ} / \mathrm{mole}$

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