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31/01/2022 5:03 pm
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The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
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31/01/2022 5:11 pm
From Arrhenius equation, we obtain
log \(\frac{k_2}{k_1}\) = \(\frac{E_a}{2.303R}\) \(\Big(\frac{T_2 - T_1}{T_1T_2}\Big)\)
It is given that, k2 = 4k1
T1 = 293 K
T2 = 313 K
Therefore, log \(\frac{4k_1}{k_2}\) = \(\frac{E_a}{2.303 \times 8.314}\) \(\Big(\frac{313 - 293}{293 \times 313}\Big)\)
⇒ 0.6021 = \(\frac{20 \times E_a}{2.303 \times 8.314 \times 293 \times 313}\)
⇒ Ea = \(\frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20}\)
= 52863.33 J mol-1
= 52.86 kJ mol-1
Hence, the required energy of activation is 52.86 kJ mol-1.