Question

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?

Answer 1

Answer:

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):

$ln (\frac{k_{2} }{k_{1} } ) = \frac{Ea}{R} (\frac{1}{T_{1}}- \frac{1}{T_{2} } )$

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:

$ln (\frac{0.032s^{-1} }{0.0160s^{-1} } )= (\frac{37400 J/mol}{8.31 J/mol.K}) (\frac{1}{298 K} - \frac{1}{T_{2} } )$

$ln (2) = (4,500.6 K) ( (\frac{1}{298 K})-(\frac{1}{T_{2} } ))$

$T_{2}= \frac{1}{(\frac{1}{298K})-(\frac{ln2}{4,500.6K}) }$

T₂= 312.33 K