Question

What is the activation energy for a reaction it its rate doubles when the temperature is raised from $20^{\circ} C $ to $35^{\circ} C $ ? $(R = 8.314 \, J \, mol^{-1} K^{-1})$

• A. $15.1 \, kJ \, mol^{-1} $
• B. $342 \, kJ \, mol^{-1} $
• C. $269 \, kJ \, mol^{-1} $
• D. $34.7 \, kJ \, mol^{-1} $

Answer 1

Answer: (D)

$\log \frac{k_{2}}{k_{1}} = \frac{Ea}{2.303 \times R} \left(\frac{1}{293} - \frac{1}{308}\right)$
$ \log 2 = \frac{Ea}{2.303R} \left(\frac{308 - 293}{293 \times308}\right) $
$ E_{a} = 34673 J = 34.67 kJ$