How to calculate the rate constant at different temperature for the decomposition of dinitrogen pentoxide? For the reaction
  $$ \ce{2N2O5(g) -&gt; 4NO2 + O2(g)} $$
  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} &amp;= \pu{2.50E4}\\
k &amp;= \pu{2.773E-5}\\
\frac{\pu{2.773E-5}}{k_2} &amp;= \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?
When using Arrhenius equation: you have to multiply the activation energy by $1000$, because it must be in $\pu{J}$ and not $\pu{kJ}$.
you must also divide by $RT$ and not by $T$ as you did.