# How to calculate the rate constant at different temperature for the decomposition of dinitrogen pentoxide?

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?

• 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. – tralston Nov 14 '14 at 3:26
• 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. – Yandle Nov 14 '14 at 15:51

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.
\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.
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.