# How does increase in volume change the speed of reaction in production of NO2?

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

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.

• The reaction rate isn't necessarily proportional to the concentration in reactants: it depends on the order of the reaction. – Thomas Jungers May 29 '19 at 9:17
• @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 – Toni Antunovic May 29 '19 at 9:31
• Your $k$ is an equilibrium constant. That is entirely different from the kinetics of the reaction. – 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 $$8$$.

• 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. – andselisk May 29 '19 at 10:44
• 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. – Thomas Jungers May 29 '19 at 14:39
• 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. – Lucas Jørgensen May 30 '19 at 11:14