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  • Which equation is true to calculate the energy of activation, if the rate of reaction is doubled by increasing temperature from <img alt="\mathrm{T_1 K}" src="https://entrancecorner.oncodecogs.com/

Which equation is true to calculate the energy of activation, if the rate of reaction is doubled by increasing temperature from \mathrm{T_1 K} to \mathrm{T_2 K} ?

Option: 1

\mathrm{\log _{10}\left(\frac{k_1}{k_2}\right)=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{2 T_2}\right]}


Option: 2

\mathrm{\log _{10} 2=\frac{2 E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]}


Option: 3

\mathrm{\log _{10} \frac{1}{2}=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]}


Option: 4

\mathrm{\log _{10} 2=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]}


Answers (1)

best_answer

The Arrhenius equation is:
                        \mathrm{ k=A e^{-\frac{k_{\mathrm{a}}}{R T}} }
Then, \mathrm{\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]}
When reaction rate becomes double then \mathrm{\frac{k_2}{k_1}} will be equal to 2
Then, \mathrm{\quad \log _{10} 2=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]}

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Rishi

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