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  • If the rate of reaction doubles for or <img alt="10 \mathrm{~K}" src=

If the rate of reaction doubles for 10^{\circ} \mathrm{C} or 10 \mathrm{~K} rise of temperature from 410 to 420 \mathrm{~K}, the activation energy of the reaction will be approximately:

Option: 1

2.38\, \mathrm{kcal\, mol}^{-1}


Option: 2

6.28\, \mathrm{kcal\, mol}^{-1}


Option: 3

62.8\, \mathrm{kcal\, mol}^{-1}


Option: 4

23.8\, \mathrm{kcal\, mol}^{-1}


Answers (1)

best_answer

\mathrm{\ln \frac{K_{2}}{K_{1}}=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)}

\mathrm{\ln 2 =\frac{E_{a}}{2 \text { Cal }}\left[\frac{1}{410}-\frac{1}{420}\right] }

\mathrm{E_{a} =\frac{0.693 \times 2 \times 410 \times 420}{10} \quad }
\mathrm{E_{a} =23866.9 \mathrm{cal} \Rightarrow E_{a}=23.8 \, \mathrm{kcal}}

Correct answer (d).

Posted by

shivangi.shekhar

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