The activation energy of a certain reaction is 37.4 kJ/mol . At 25 ∘C , the rate constant is 0.0160s−1 . At what temperature in degrees Celsius would this reaction go twice as fast?

The correct answer is 312.3 K

Explanation:

In order to solve the problem, we use the modified Arrhenius equation which comprises two rate constants (k) and two temperatures (T):

Where Ea is the activation energy (37.4 kJ/mol= 37400 J/mol), R is the gases constant (8.31 J/K.mol), k₁ and k₂ are the rate constant. We know k₁ (0.0160 s⁻¹) and from the problem, k₂ is twice k₁, so k₂= 2 x k₁= 2 x 0.0160 s⁻¹= 0.032 s⁻¹). Finally, T₁= 25ºC + 273 = 298 K. We introduce the data in the equation and calculate T₂:

T₂= 312.33 K

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.