A certain reaction with an activation energy of 155 kJ/mol was run at 495 K and again at 515 K . What is the ratio of f at the higher temperature to f at the lower temperature

Respuesta :

Answer:

4.32 is the ratio of f at the higher temperature to f at the lower temperature

Explanation:

Using the sum of Arrhenius equation you can obtain:

ln (f₂/f₁) = Eₐ / R ₓ (1/T₁ - 1/T₂)

Where f represents the rate constant of the reaction at T₁ and T₂ temperatures. Eₐ is the energy activation (155kJ / mol = 155000J/mol) and R is gas constant (8.314J/molK)

Replacing:

ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)

Where 2 represents the state with the higher temperature and 1 the lower temperature.

ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)

ln (f₂/f₁) = 1.4626

f₂/f₁ = 4.32

4.32 is the ratio of f at the higher temperature to f at the lower temperature