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The rate of a reaction quadruples when the temperature
changes from \(293 \mathrm{~K}\) to \(313 \mathrm{~K}\). Calculate the energy of
activation of the reaction assuming that it does not change
with temperature.
changes from \(293 \mathrm{~K}\) to \(313 \mathrm{~K}\). Calculate the energy of
activation of the reaction assuming that it does not change
with temperature.
Solution:
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Verified Answer
\(\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{T_2-T_1}{T_2 T_1}\right)\);
\(\frac{k_2}{k_1}=4\) (given)
\(\log 4=\frac{E_a}{2.303 \times 8.314}\left(\frac{313-293}{313 \times 293}\right)\)
on solving \(E_a=52.854 \mathrm{kJmol}^{-1}\)
\(\frac{k_2}{k_1}=4\) (given)
\(\log 4=\frac{E_a}{2.303 \times 8.314}\left(\frac{313-293}{313 \times 293}\right)\)
on solving \(E_a=52.854 \mathrm{kJmol}^{-1}\)
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