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

Question

Quizplus · Accepted Answer

We can use the Arrhenius equation:

$k = Ae^{-\frac{E_a}{RT}}$

Taking the natural logarithm of both sides:

$\ln k = \ln A - \frac{E_a}{RT}$

We can then use the two given values of k and temperature to create two equations:

$\ln (2.5 	imes 10^{13}) = \ln A - \frac{E_a}{R(-5+273)}$

$\ln (3.5 	imes 10^{14}) = \ln A - \frac{E_a}{R(25+273)}$

Simplifying:

$\ln A = \ln (2.5 	imes 10^{13}) + \frac{E_a}{R(278)}$

$\ln A = \ln (3.5 	imes 10^{14}) + \frac{E_a}{R(298)}$

Subtracting the second equation from the first:

$0.73 = \frac{E_a}{R}(\frac{1}{278} - \frac{1}{298})$

Solving for $E_a$:

$E_a = 58$ 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