# Percent degree of dissociation of gas by pressure

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.

• I think I see how to solve the problem, do you have an answer?
– MaxW
Apr 20 '20 at 20:31
• Unfortunately, no. Could you share any tips? Apr 20 '20 at 20:50
• First what does the circle on the equilibrium constant $K$ mean?
– MaxW
Apr 20 '20 at 20:56
• 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.
– MaxW
Apr 20 '20 at 21:37
• Yes, $\ce{(XY4_\mathrm{inital}/XY4_\mathrm{final})\times100\%}$
– MaxW
Apr 21 '20 at 7:40

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 $$\alpha \approx 3.10^{-3}$$.
• 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? Apr 23 '20 at 6:18
• All the pressures used in calculating $K$ have to be dimensionless which means dividing by the standard. Apr 23 '20 at 7:20
• 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? Apr 23 '20 at 7:57