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

Question

Quizplus · Accepted Answer

The Arrhenius equation can be used to relate the rate constant (k) to the temperature (T) and the activation energy (Ea):

k = A e^(-Ea/RT)

where A is a pre-exponential factor, R is the gas constant, and T is the temperature in Kelvin.

Taking the natural logarithm of both sides yields:

ln(k) = ln(A) - Ea/RT

This form of the equation can be used to determine the activation energy from two sets of experimental data at different temperatures. Taking the ratio of the two equations gives:

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

where k1 and k2 are the rate constants at temperatures T1 and T2, respectively.

Substituting the given values, we have:

ln(3.5 × 10^14/2.5 × 10^13) = Ea/8.314 [(1/268) - (1/298)]

solving for Ea gives:

Ea = -ln(3.5 × 10^14/2.5 × 10^13) × 8.314 [(1/268) - (1/298)]

Ea ≈ 58 kJ/mol

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

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