Chemist 10 K rate double 25℃ 35℃ activation energy

Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction.

What must the activation energy be for this statement to be true for a temperature increase from 25 to 35℃?

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Arrhenius equation

ln(k_2 / k_1) = Ea/R (1/T_1 – 1/T_2)

( R = 8.314 J/mol•K )

> k_1 = 1, T_1 = 273 + 25 = 298 K

> k_2 = 2, T_2 = 273 + 35 = 308 K

Ea = ln(k_2 / k_1) × R / (1/T_1 – 1/T_2)

= ln(2 / 1) × 8.314 / (1/298 – 1/308)

= 52893.5 J/mol

= 53 kJ/mol

답: 53 kJ/mol

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