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  • For completion of a <img alt="1^{\text {st }}" src="https://entrancecorner.oncodecogs.com/gif.latex?1%5E%7B

For 50\, % completion of a 1^{\text {st }} order reaction, it takes 40 min at 27^{\circ} \mathrm{C} and 5 minutes at 47^{\circ} \mathrm{C}. The energy of activation of the reaction is =?

Option: 1

73 \, \mathrm{kJ\, mol}^{-1}


Option: 2

83 \, \mathrm{kJ\, mol}^{-1}


Option: 3

93 \, \mathrm{kJ\, mol}^{-1}


Option: 4

98 \, \mathrm{kJ\, mol}^{-1}


Answers (1)

\mathrm{k_{1}=\frac{0.693}{40}\, \& \, k_{2}=\frac{0.693}{5}}

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

\mathrm{\ln 8=\frac{E_{a}}{R}\left(\frac{1}{300}-\frac{1}{320}\right) }

\mathrm{E_{a}=\frac{300 \times 320 \times 2.303 \times 3 \times \log 2 \times 8.314}{20}=82999\, \mathrm{Jmol}^{-1} }
\mathrm{E_{a}=83 \, \mathrm{kJmol}^{-1} }

Posted by

Kshitij

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