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

Question

Quizplus · Accepted Answer

We can use the Arrhenius equation to calculate the activation energy (Ea):
k = A*e^(-Ea/RT)
Taking the natural logarithm of both sides:
ln k = ln A - (Ea/RT)
Plotting ln k vs 1/T will give a straight line with slope = -(Ea/R), where R is the gas constant.
By using the given rate constants at two different temperatures, we can calculate the activation energy as follows:
ln (k2/k1) = Ea/R * (1/T1 - 1/T2)
ln (3.5/2.5) = Ea/8.314 * (1/298 - 1/268)
Ea = -0.73 * 8.314 * 10^3 / (1/298 - 1/268)
Ea = 57.6 kJ/mol
The answer is B, 58 kJ/mol.

Dinitrogen Tetraoxide,N2O4,decomposes to Nitrogen Dioxide,NO2,in a First-Order Process ×\times× 103 S-1 at -5°C and K = 3

Dinitrogen Tetraoxide,N₂O₄,decomposes to Nitrogen Dioxide,NO₂,in a First-Order Process $\times$ 10³ S^-1 at -5°C and K = 3