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For a particular 10^{\circ} \mathrm{C} rise in temperature causes the rate of reaction to double of its initial rate of reaction. What will be the value of \mathrm{E_{a}} if the above condition applied to a reaction at \mathrm{290 \mathrm{~K}} ?

Option: 1

12 \, \mathrm{Kcal}^{}\mathrm{mol^{-1}}


Option: 2

1.2\, \mathrm{Kcal}^{}\mathrm{mol^{-1}}


Option: 3

120\, \mathrm{Kcal}^{}\, \mathrm{mol^{-1}}


Option: 4

1200\, \mathrm{Kcal}^{}\, \mathrm{mol^{-1}}


Answers (1)

best_answer

A/C  to Arrhenius Equation

\mathrm{T_{1}=290 \mathrm{~K}, T_{2}=305 \mathrm{~K} }

\mathrm{\log \left(\frac{k_{2}}{k_{1}}\right)=\frac{E_{a}}{2.3 \times 2 \times 10^{-3}} }\left[\frac{10}{290 \times 300}\right]

\mathrm{E_{a}=\frac{\log (2) \times 2.3 \times 2 \times 10^{-3} \times 290 \times 300}{10}}
\mathrm{E_{a}=\frac{0.3 \times 4.6 \times 290 \times 300}{10^{4}} }
\mathrm{E_{a}=12.006\, \mathrm{k\, cal} \, \mathrm{mol}^{-1}}
\mathrm{E_{a} \approx 12\, \mathrm{kcal} \, \mathrm{mol}^{-1}}

Posted by

Gautam harsolia

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