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In the case of the reaction $$\ce{2NO2 <=> N2O4}$$ at equilibrium and considering Le Chatelier's principle, if we add more $\ce{NO2}$, the pressure of $\ce{NO2}$ will increase acutely and then begin to decrease until the system reaches its new equilibrium. Why is the new equilibrium have a higher pressure of $\ce{NO2}$ than the original equilibrium? How can we tell this just by looking at it? Will it always be larger?

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The total pressure of the system is, according to Dalton's law, $p_\text{T} = p_\ce{NO2} + p_\ce{N2O4}$. From here, $p_\ce{NO2} = p_\text{T} - p_\ce{N2O4}$.The equilibrium expression for this system can be written as:

$$K_\text{p} = \frac{p_\ce{N2O4}}{p_\ce{NO2}^2} = \frac{p_\ce{N2O4}}{(p_\text{T} - p_\ce{N2O4})^2}$$

So we get:

$$p_\ce{T} = \sqrt{\frac{p_\ce{N2O4}}{K_\text{p}}} + p_\ce{N2O4}$$

Assuming the process is isothermic ($K_\text{p}$ is constant) and isochoric, then introducing moles of a gas that participates in this reaction will increase the pressure of the system, hence $p_T$ will increase. Since $p_T$ increases, the previous equation shows that $p_\ce{N2O4}$ must increase. Since $K_\text{p}$ is a constant, the ratio $\displaystyle \frac{p_\ce{N2O4}}{p_\ce{NO2}^2}$ is constant. Hence, if $p_\ce{N2O4}$ increases, so does $p_\ce{NO2}$.

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  • $\begingroup$ Sweet! I am just linking my attempt as well, so the ease of getting tangled in the maths (like I got :P) is clear, and as well the brevity of this marvelous solution is more visible! $\endgroup$ – Gaurang Tandon Mar 20 '18 at 6:29
  • $\begingroup$ Yeah I started there, but once I got to the double quadratic with four unknowns, it became apparent it was very likely to lead nowhere. PS. There's a mistake on your quadratic formula, I think $\endgroup$ – ralk912 Mar 20 '18 at 7:41
  • $\begingroup$ I solved it using WolframAlpha, where's the mistake? :/ $\endgroup$ – Gaurang Tandon Mar 20 '18 at 7:42
  • $\begingroup$ Oh well, then it probably it's more efficient, I was just trying to solve by hand. I trust WA :) $\endgroup$ – ralk912 Mar 20 '18 at 7:52
  • $\begingroup$ Thank you! This makes a lot of sense! I did not realize that pressure and concentration themselves did not affect K, so K is constant in an isothermic process and therefore the pressure of NO2 must be higher than its original value. $\endgroup$ – Cassie Bishop Mar 28 '18 at 19:37
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The higher pressure of $\ce{NO2}$ even after reaching equilibrium after adding some amount of $\ce{NO2}$ can be shown mathematically.

Suppose initially before adding $\ce{NO2}$,the pressure of $\ce{NO2}$ was $p^0_{\ce{NO2}}$, and that of $\ce{N2O4}$ is $p^0_{\ce{N2O4}}$.So, the equilibrium constant is $$K_\mathrm{p} = \frac{p^0_{\ce{N2O4}}}{(p^0_{\ce{NO2}})^2}$$ Now assume that the addition of $\ce{NO2}$ just increases the pressure instantaneously to $p_x$ but immediately after that due to dissociation to achieve equilibrium, the pressure decreases by $\ce{p_y}$ atm.

$$\ce{2NO2 <=> N2O4}$$ $$p^0_{\ce{NO2}} + p_x - p_y \ce{<=>} p^0_{\ce{N2O4}} + \frac{p_y}{2}$$

So, the increase of pressure $\ce{N2O4}$ will be $p_y/2$ atm. So, now a new equilibrium is achieved where the equilibrium constant $K_\mathrm{p}$ will be still the same assuming that during addition of the gases, temperature is not changed( or the change is very much negligible in real cases). Thus, equating $K_\mathrm{p}$ on both the sides we have,

$$\frac{p^0_{\ce{N2O4}}}{(p^0_{\ce{NO2}})^2}= \frac{p^0_{\ce{N2O4}} + \frac{p_y}{2}}{(p^0_{\ce{\ce{NO2}}} + p_x - p_y)^2}$$

Taking the squares on numerator and denominator and taking square root we will find:

$$1 + \frac{p_x -p_y}{p^0_{\ce{NO2}}}= \sqrt{1+\frac{p_y/2}{p^0_{\ce{N2O4}}}} > 1$$

So, we see that $1 + \frac{p_x -p_y}{p^0_{\ce{NO2}}} > 1$. Hence it is proved that $\ce{p_x - p_y > 0}$. So, the net pressure of $\ce{NO2}$ after achieving equilibrium is $p^0_\ce{NO2} + p_x - p_y > p^0_{\ce{NO2}}$. So, its pressure has increased than the previous conditions even after reaching equilibrium.

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