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  • Calculate the activation energy of a reaction whose rate constant is tripled by <img alt="\mathrm{10^{\circ} \mathrm{C}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B10%5E%7B%5

Calculate the activation energy of a reaction whose rate constant is tripled by \mathrm{10^{\circ} \mathrm{C}} rise in the temperature from \mathrm{22^{\circ} \mathrm{C}}.
Take ln(3)=1.1

Option: 1

46.3 \mathrm{~kJ} \mathrm{~mol}^{-1}


Option: 2

82.2 \mathrm{~kJ} \mathrm{~mol}^{-1}


Option: 3

168 \mathrm{~J} \mathrm{~mol}^{-1}


Option: 4

2403 \mathrm{~J} \mathrm{~mol}^{-1}


Answers (1)

best_answer

The temperature rises from \mathrm{22^{\circ} \mathrm{C}} to \mathrm{32^{\circ} \mathrm{C}}.
Thus it rises from 295 K to 305 K
During this rise in temperature, the rate constant of reaction is tripled.
\mathrm{\therefore \frac{k_{305}}{k_{295}}=3 }
According to Arrhenius the relationship between temperature and rate constant is given by,
\mathrm{ k=A e^{-E_a / R T} \\ }

\mathrm{ \frac{k_{305}}{k_{295}}=\frac{A e^{-E_a / R T_{305}}}{A e^{-E_a / R T_{205}}}=3 \\ }

\mathrm{ \frac{e^{-E_a / R T_{305}}}{e^{-E_a / R T_{295}}}=3 }
Taking In on both side,
\mathrm{ \ln 3=-\frac{E_a}{R T_{305}}+\frac{E_a^{\prime}}{R T_{295}}}

\mathrm{ \ln 3=-\frac{E_a}{R}\left[\frac{1}{305}-\frac{1}{295}\right] \\}

\mathrm{ \ln 3=\frac{E_a}{R}\left[\frac{10}{305 \times 295}\right] \\ }

\mathrm{ E_a=\frac{\ln 3 \times 8.314 \times 305 \times 295}{10} \\ }

\mathrm{ E_a=82285.7 \mathrm{Jmol}^{-1} }

\mathrm{ E_a \approx 82.2 \mathrm{~kJ} \mathrm{~mol}^{-1} }

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Rishabh

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