$$\ce{2 NO + O2 <=> 2 NO2}$$

I understand that increasing the volume $×2$ means lower pressure so production of $\ce{NO2}$ is slowed down, but when asked how to express this mathematicaly I can't get to the correct value, which is either $4$ or $8.$ I don't see how inserting $1/2$ in $v = Δc/(2×Δt)$ can give anything close to the offered solutions.

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    $\begingroup$ The reaction rate isn't necessarily proportional to the concentration in reactants: it depends on the order of the reaction. $\endgroup$ – Thomas Jungers May 29 '19 at 9:17
  • $\begingroup$ @ThomasJungers Yes, but concentration is the only value I can change with the increase of volume. I tried placing 2V in the formula for the constant of the reaction k=([NO2]^2)/([NO]^2×[O2]) but i got different results for every option I tried and non of them are 4 nor 8 $\endgroup$ – Toni Antunovic May 29 '19 at 9:31
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    $\begingroup$ Your $k$ is an equilibrium constant. That is entirely different from the kinetics of the reaction. $\endgroup$ – Thomas Jungers May 29 '19 at 14:35

I'm not an expert but this is how I would do it.

  • Determine the reaction rate formula from the reaction equation

    $$r = K[\ce{NO}]^2[\ce{O2}]$$

  • Changing the volume by a factor $2$ means the concentration of $\ce{NO}$ and $\ce{O2}$ will become half

    $$r' = K\cdot\frac{[\ce{NO}]}{2}^2\cdot\frac{[\ce{O2}]}{2}$$

  • Compare the reaction rates

    $$r' = K[\ce{NO}]^2\cdot\frac{1}{4}\cdot [\ce{O2}]\cdot\frac{1}{2}$$

    $$\frac{r}{r'} = \frac{K[\ce{NO}]^2[\ce{O2}]}{K[\ce{NO}]^2[\ce{O2}]\cdot\frac{1}{8}} = 8$$

Conclusion: increasing the volume by $2$ will decrease the concentration (or pressure) of $\ce{NO}$ and $\ce{O2}.$ This causes the reaction rate to decrease by a factor of $8$.

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    $\begingroup$ Welcome to Chemistry.SE! Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Briefly, one can use \ce{…} for chemical formulas and $$…$$ instead of $\displaystyle …$. \displaystyle is pretty much only useful for the inline mode. $\endgroup$ – andselisk May 29 '19 at 10:44
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    $\begingroup$ The partial orders in the rate law are equal to the stoichiometric coefficients of a reaction only when that reaction is an elementary reaction. I sincerely doubt that $\ce{2 NO + O2 <=> 2 NO2}$ is an elementary reaction. As it happens, the rate law for this reaction is indeed $r = k [\ce{NO}]^2 [\ce{O2}]$, but it is a coincidence. $\endgroup$ – Thomas Jungers May 29 '19 at 14:39
  • $\begingroup$ thanks for the feedback. Thomas I agree with you on that. It's impossible for the reaction to be elementary because the reactant contains 3 molecules, I didn't think about that. I searched for the reaction mechanism on google and this: resources.saylor.org/wwwresources/archived/site/wp-content/… page 6 suggested that it's a two step reaction forming an intermediate N2O2. And it happens to match the observed rate law. $\endgroup$ – Lucas Jørgensen May 30 '19 at 11:14

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