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Q.

The rate of a reaction decreased by 3.555 times when the temperature was changed from 40°Cand 30°C .the activation energy (in KJ mol^-1) of the reaction is ___ (Take R $8.314 J {mol}^{- 1} k^{- 1}$ , $\ln 3.555 = 1.268$ )

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Detailed Solution

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The Arrhenius equation is $k = {Ae}^{\frac{E_{a}}{RT}}$
Assuming A and E_a to be independent of temperature
$\begin{array}{l} \ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \\ \ln 3 .555 = \frac{E_{a}}{8 .314} (\frac{1}{303} - \frac{1}{313}) \end{array}$
$\begin{array}{l} \Rightarrow E_{a} = \frac{1 .268 \times 8 .314 \times 303 \times 313}{10} \\ = 99980 .7 = 99 .98 kJ / mol \end{array}$

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The rate of a reaction decreased by 3.555 times when the temperature was changed from 40°Cand 30°C .the activation energy (in KJ mol-1) of the reaction is ___ (Take R 8.314Jmol−1k−1, ln⁡3.555=1.268)