For the reaction $$ \ce{2N2O5(g) -> 4NO2 + O2(g)} $$ the rate law is: $$ \frac{\mathrm{d}[\ce{O2}]}{\mathrm{d}t} = k[\ce{N2O5}] $$ At $\pu{300 K}$, the half-life is $\pu{2.50E4 s}$ and the activation energy is $\pu{103.3 kJ/mol}$.
What is the rate constant at $\pu{350 K}$?
I know there is something fishy about the rate law, but I can't make sense of it.
\begin{align} \frac{ln2}{k} &= \pu{2.50E4}\\ k &= \pu{2.773E-5}\\ \frac{\pu{2.773E-5}}{k_2} &= \frac{A\cdot\exp\left\{\frac{-103300}{8.314\times300}\right\} }{ A\cdot\exp\left\{\frac{-103300}{8.314\times350}\right\}}\\ \end{align}
Finding $k_2$ from this gives me a weird value: $k_2 = 0.0103$. The answer for this question is $\pu{7.47E-8 s^-1}$.
Where have I gone wrong?