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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 300 K, the half-life is 2.50·104 seconds and the activation energy is 103.3 kJ/mol.

What is the rate constant at 350 K?


My attempt at solving: I know there is something fishy about the rate law but I can't make sense of it.

ln(2)/k = 2.50 · 104

k = 2.773 . 10-5

2.773 . 10-5/k2 = Ae-103300/8.314 . 300 / Ae-103300/8.314 . 350

Finding k2 from this gives me a weird value. k2 = 0.0103

The answer for this question is 7.47 · 10-8 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 $k_{300K}$ as you did. I find it a little weird that there's a 3 order of magnitude decrease in rate constant for a 50K increase in temperature.

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

$k_{350K}/k_{300K}=e^{E_a/R(1/T_{300K}-1/T_{350K})}$

$k_{350K}/k_{300K}=e^{103000/8.314(1/300-1/350)}=371.14$

If you take $k_{300K}=2.773x10^{-5}$ and multiply by that factor above (as you should) you get your answer, if you take $k_{300K}$ and divide by the factor above, you get the book's answer, which is where I believe their mistake is.

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    $\begingroup$ The equation you're using to find k350 is incorrect. Taking the ratio of k350/k300 would not allow you to cancel the activation energy or gas constant terms, and the temperatures would be found as subtracted reciprocals: k350 = k300 * e^(E/R * (1/T300 - 1/T350)) $\endgroup$ – tralston Nov 14 '14 at 4:57
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    $\begingroup$ @tralston Thank you for pointing that out, my brain was not functioning yesterday. I will correct the answer. $\endgroup$ – Yandle Nov 14 '14 at 15:43
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When using Arrhenius equation: you have to multiply the activation energy by 1000, because it must be in $J$ and not $kJ$. you must also divide by $RT$ and not by $T$ as you did.

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  • $\begingroup$ Thanks for your answer. I edited my answer according but I'm still unable to get the correct value $\endgroup$ – Kenny Nov 13 '14 at 15:34

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