Percent degree of dissociation of gas by pressure

Question

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.

Answer 1

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 $\alpha \approx 3.10^{-3}$.