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I stumbled across the following interesting problem:

Assume that you have started to live on a new planet where standard pressure condition is $\pu{2 bar}$, standard concentration is $\pu{1 M},$ and all types of gases behave as an ideal gas. On this planet, you are asked to determine equilibrium conditions for the reaction below:

$$\ce{XY4(g) <=> X(s) + 2 Y2(g)}$$

$K^\circ = \pu{1.1455E-9}$ (at $\pu{298 K}$).

Calculate the percent degree of dissociation for $\ce{XY4}$ at $\pu{298 K}$ where total pressure is $\pu{0.2 bar}.$

Since the equilibrium constant is very small, I approximated that the pressure of $\ce{Y2}$ at equilibrium is insignificant. Therefore I got:

$$p(\ce{Y2}) + p(\ce{XY4}) = \pu{0.2 bar}$$ $$p(\ce{XY4}) = \pu{0.2 bar}$$

I don't know how to proceed from here though and how to calculate the degree of dissociation.

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  • $\begingroup$ I think I see how to solve the problem, do you have an answer? $\endgroup$
    – MaxW
    Apr 20 '20 at 20:31
  • $\begingroup$ Unfortunately, no. Could you share any tips? $\endgroup$ Apr 20 '20 at 20:50
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    $\begingroup$ First what does the circle on the equilibrium constant $K$ mean? $\endgroup$
    – MaxW
    Apr 20 '20 at 20:56
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    $\begingroup$ The circle means standard conditions. // I didn't read the problem carefully enough. I don't see how standard conditions could be (1) pressure of 2 bar, (2) concentration of 1 molar and (3) gases behave as ideal gases (about 11 liters per mole at 2 bar and 298 K) . // I was thinking that the problem needed you to convert $K_c$ to $K_p$ at 298 K. $\endgroup$
    – MaxW
    Apr 20 '20 at 21:37
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    $\begingroup$ Yes, $\ce{(XY4_\mathrm{inital}/XY4_\mathrm{final})\times100\%}$ $\endgroup$
    – MaxW
    Apr 21 '20 at 7:40
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If $\alpha$ is the degree of dissociation, at equilibrium $1-\alpha$ of XY is present and $\alpha$ of X and $2\alpha$ of Y making a total of $1+2\alpha$. The partial pressure of XY is $\displaystyle p_{XY}=\frac{1-\alpha}{1+2\alpha}\frac{P}{P^\mathrm{o}}$. The $P^\mathrm{o}$ is the standard pressure used to make the equilibrium constant dimensionless. In our world this is 1 bar, in your new world it is 2 bar, thus $P/P^\mathrm{o}=0.1$. If you calculate the partial pressures for the other species in terms of $\alpha$ and form the equilibrium constant then $\displaystyle K=\frac{4\alpha^3}{(1+2\alpha)^2(1-\alpha)}\left(\frac{P}{P^\mathrm{o}}\right)^2$.(please check). As $K$ is small we will assume that $\alpha$ is also then $\displaystyle K=4\alpha^3\left(\frac{P}{P^\mathrm{o}}\right)^2$ which produces $\alpha \approx 3.10^{-3}$.

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  • $\begingroup$ In my solution, I approximated the pressure of $\ce{XY4}$ to be 0,2 bar and calculated the pressure of $\ce{Y2}$ from K. How should I take the standard pressure of 2 bar into account in my solution? Should the pressure of $\ce{XY4}$ be divided by 2? $\endgroup$ Apr 23 '20 at 6:18
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    $\begingroup$ All the pressures used in calculating $K$ have to be dimensionless which means dividing by the standard. $\endgroup$
    – porphyrin
    Apr 23 '20 at 7:20
  • $\begingroup$ This results in 0.0107% as the degree of dissociation (I divided the pressure of $\ce{Y2}$ by the pressure of $\ce{XY4}$, which is 0.1 bar). So is there an error in my solution? $\endgroup$ Apr 23 '20 at 7:57

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