T1 = 300 K K1 = 0.69340T2 = 320 KK2 = 0.69320logK2K1 = Ea2.303 RT2-T1T1T2log0.693200.69340 = Ea2.303 x 8.314320-300320 x 300log4020 = Ea2.303 x 8.31420320 x 300 = 0.3010 x 2.303 x 8.314 x 320 x 30020 = Ea = 27764 J= 27.76 KJ

" />T1 = 300 K K1 = 0.69340T2 = 320 KK2 = 0.69320logK2K1 = Ea2.303 RT2-T1T1T2log0.693200.69340 = Ea2.303 x 8.314320-300320 x 300log4020 = Ea2.303 x 8.31420320 x 300 = 0.3010 x 2.303 x 8.314 x 320 x 30020 = Ea = 27764 J= 27.76 KJ

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Chemical Kinetics

Question

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A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.
(Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK^–1 mol^–1)

Answer

$T_{1} = 300 K K_{1} = \frac{0.693}{40} T_{2} = 320 K K_{2} = \frac{0.693}{20} \log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}}) \log \frac{\frac{0.693}{20}}{\frac{0.693}{40}} = \frac{E_{a}}{2.303 x 8.314} (\frac{320 - 300}{320 x 300}) \log (\frac{40}{20}) = \frac{E_{a}}{2.303 x 8.314} (\frac{20}{320 x 300}) = \frac{0.3010 x 2.303 x 8.314 x 320 x 300}{20} = E_{a} = 27764 J = 27.76 K J$