MCQ
The activation energy for a reaction which doubles the rate when the temperature is raised from $298\, K$ to $308\, K$ is ........... $kJ\, mol^{-1}$
- ✓$52.89$
- B$39.2$
- C$52.9$
- D$29.5$
✓
Answer
Correct option: A.
$52.89$
a
Activation energy can be calculated from the equation.
Activation energy can be calculated from the equation.
$\frac{{\log \,\,{K_2}}}{{\log \,\,{K_1}}}\, = \,\frac{{ - \,{E_a}}}{{2.303\,R}}\left( {\frac{1}{{{T_2}}}\, - \,\frac{1}{{{T_1}}}} \right)$
Given $\frac{{\log \,\,{K_2}}}{{\log \,\,{K_1}}}\, = 2$ $T_2\,=\,308$: $T_1\,=\,298$
$\therefore \,\,\log \,\,2\,\, = \frac{{ - \,{E_a}}}{{2.303\, \times 8.314}}\left( {\frac{1}{{308}}\, - \,\frac{1}{{298}}} \right)$
${E_a}\, = \,52.9\,\,kJ\,\,mo{l^{ - 1}}$
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