We have a reaction $$\ce{2N2O5 -> 4NO2 + O2},$$ which has a rate constant $k=6,2\cdot10^{-4}/\mathrm{min}$ at temperature $T=318.15\ \mathrm{K}$. How much oxygen is formed in litres when $1.00\ \mathrm{mol}$ of $\ce{N2O5}$ is at $318.15\ \mathrm{K}$ in $104\ \mathrm{kPa}$ pressure for $20$ hours. Correct answer: $6.67\ \mathrm{l}$

My attempt:

For gases we use partial pressures to measure rates. So using equation $$\ln P_t=-kt+\ln P_0$$ we get $P_t=49421.76426\ \mathrm{Pa}$. This means that oxygen has a partial pressure of $P_{\ce{O2}}= \frac{104000\ \mathrm{Pa}-49421.46426\ \mathrm{Pa}}{5}=10915.64715\ \mathrm{Pa}$.

Now we say $n_{\ce{O2}}=x$ so according to the reaction $n_{\ce{N2O5}}=1-2x$. If the total volume is $V$ we can type two equations: $$ \begin{align} P_t&=\frac{n_{\ce{N2O5}}RT}{V}\\ P_{\ce{O2}}&=\frac{n_{\ce{O2}}RT}{V} \end{align} $$ Solving for $x$ and $V$ yields $x=0.153195\ \mathrm{mol}$ and $V=0.0371657\ \mathrm{m^3}$. Now calculating the volume for oxygen at $104\ \mathrm{kPa}$ gives $$ V_{\ce{O2}}=\frac{0.153195\ \mathrm{mol}\cdot 8.3145\ \mathrm{\frac{Pa\ m^3}{K\ mol}\cdot318.15\ \mathrm{K}}}{104000\ \mathrm{Pa}}=3.900828\cdot 10^{-3}\ \mathrm{m^3}\approx3.9\ \mathrm{l} $$ My answer is somehow incorrect but what I am doing wrong here?

  • $\begingroup$ Is this reaction at constant pressure? In that case, the volume would increase over time, and the first order integrated rate law would not apply. Is this a textbook problem? $\endgroup$
    – Karsten
    Nov 24, 2019 at 18:30
  • 1
    $\begingroup$ It was from a exercise sheet. I don't know the origin of the problem and my translation might not be perfect. $\endgroup$
    – jte
    Nov 24, 2019 at 20:31

1 Answer 1


If we assume that the reaction takes place in a large amount of inert gas (like nitrogen) and that the reaction is first order in $\ce{N2O5}$, we can reach the numerical answer given by neglecting the volume change (the reaction increases the number of particles, but if there is a large amount of inert gas, the volume change will have little effect on partial pressures). The path given by Maurice is fine (no reason to figure out the partial pressure first). If you convert the first-order rate constant into a half-life (18.6 h), you can quickly estimate that a bit more than half of the reactant will be gone after 20 h.

If we assume, like the OP, that you start with pure $\ce{N2O5}$, the rate law still applies, but the integrated rate law does not (because while the reaction goes on, dilution also happens through the volume increase). You could numerically integrate, or try to find the integrated rate law that includes the dilution effect. Overall, you would expect the reaction to go slower, i.e. produce a smaller amount of oxygen. Because the final volume is larger, the partial pressure will be lower (compared to the inert gas case) by even more.

The value calculated by the OP is incorrect for either case because it assumes an integrated rate law that does not apply for the assumed conditions.


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