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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?

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    $\begingroup$ I get the same answer you do. Reality check: the hotter the reaction, the faster it'll go. Therefore, your k value should be growing as your temperature increases. The given answer is 3 orders of magnitude smaller at 350 K than at 300 K. I think your book/solution guide is wrong. $\endgroup$ – tralston Nov 14 '14 at 3:26
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    $\begingroup$ If looks like your book divided k_1 by the ratio of k_2/k_1 instead of multiplying it (see my response below). As @tralston says, k should increase with temperature. $\endgroup$ – Yandle Nov 14 '14 at 15:51
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I arrived at the same $k_{\pu{300 K}}$ as you did. I find it a little weird that there's a 3 order of magnitude decrease in rate constant for a $\pu{50 K}$ increase in temperature.

Per the note below, the original equation I put had a mathematical error. The below should be correct:

\begin{align} \frac{k_{\pu{350 K}}}{k_{\pu{300 K}}} &= \exp\left\{\frac{E_\mathrm{a}}{R\left(\frac{1}{T_{\pu{300 K}}} - \frac{1}{T_{\pu{350 K}}}\right)}\right\}\\ \frac{k_{\pu{350 K}}}{k_{\pu{300 K}}} &= \exp\left\{\frac{103000}{8.314\left(\frac{1}{300} - \frac{1}{350}\right)}\right\}=371.14 \end{align}

If you take $k_{\pu{300 K}} = \pu{2.773E-5}$ and multiply by that factor above (as you should) you get your answer, if you take $k_{\pu{300 K}}$ and divide by the factor above, you get the book's answer, which is where I believe their mistake is.

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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.

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