A first order reaction is 50%. Complete in 30 minutes at 27° and in 10 minutes at 47°C. Calculate the reaction rate constant at 27°C and the energy of activation of the reaction in kJ/mol.

Solution

by using half life equation, we get

K = \frac{0.693}{t_{1 / 2}} K = \frac{0.693}{30 \min} = 0.0231 at 27 ° C or 300 K K = \frac{0.693}{10 \min} = 0.0693 at 47 ° C or 320 K

From the Arrhenius equation

\log \frac{k_{2}}{k_{1}} = \frac{Ea}{2.303} [\frac{1}{T_{1} - T_{2}}] Ea = \frac{2.303 \times R \times T_{1} \times T_{2}}{T_{2} - T_{1}} \log \frac{k_{2}}{k_{1}} = \frac{2.303 \times 8.314 J K^{- 1} {mol}^{- 1} \times 300 \times 320 K}{320 K - 300 K} \log \frac{0 . 06935^{- 1}}{0 . 02315^{- 1}} = 43.848 kJ {mol}^{- 1} .

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