The rate of a reaction doubles, when the temperature is changed from \(300 \mathrm{~K}\) to \(310…

The rate of a reaction doubles, when the temperature is changed from \(300 \mathrm{~K}\) to \(310 \mathrm{~K}\). Activation energy of the change is....... \(\left(R=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 2=0.301\right)\)
  1. \(53.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  2. \(48.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  3. \(58.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  4. \(60.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Solution

From Arrhenius equation, \(\begin{aligned} \log \frac{K_{310}}{K_{300}} & =\frac{E_a}{2.303 R}\left(\frac{310-300}{300 \times 310}\right) \\ \Rightarrow \log 2 & =\frac{E_a}{2.303 \times 8.314 \times 10^{-3}}\left(\frac{10}{300 \times 310}\right) \\ E_a & =53.6 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\)

Asked in: AP EAMCET 2020 (18 Sep Shift 1)

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