Dinitrogen tetraoxide, N₂O₄, decomposes to nitrogen dioxide, NO₂, in a first-order process. If k = 2.5 * 10³ s¯¹ at -5$\degree$C and k = 3.5 * 10⁴ s¯¹ at 25$\degree$C, what is the activation energy for the decomposition?

Question

Quizplus · Accepted Answer

We can use the Arrhenius equation to solve for the activation energy (Ea):

ln(k2/k1) = -(Ea/R)(1/T2 - 1/T1)

where k1 and k2 are the rate constants at -5°C and 25°C, respectively; R is the gas constant (8.314 J/mol*K); T1 and T2 are the corresponding temperatures in Kelvin.

Plugging in the values given:

ln(3.5x10^10/2.5x10^10) = -(Ea/8.314)(1/298 - 1/268)

0.3365 = -(Ea/8.314)(0.00375)

Ea = -0.3365/(0.00375*8.314) = 58 kJ/mol

Therefore, the answer is B.

Dinitrogen Tetraoxide, N2O4, Decomposes to Nitrogen Dioxide, NO2, in a First-Order

Dinitrogen Tetraoxide, N₂O₄, Decomposes to Nitrogen Dioxide, NO₂, in a First-Order