Question:medium

Calculate activation energy for a reaction if it's rate doubles when temperature is raised from $20^\circ\text{C}$ to $35^\circ\text{C} \left( \text{R} = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \right)$}

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Shortcut: Rate doubles $\Rightarrow \ln 2 = 0.693$
Updated On: May 8, 2026
  • $17.336 \text{ kJ}$
  • $26.900 \text{ kJ}$
  • $34.673 \text{ kJ}$
  • $44.236 \text{ kJ}$
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The Correct Option is C

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