Is is valid to compare reaction quotient to $K_p$ as well as $K_c$?

Question

Can I use the reaction quotient compared to $K_p$ to answer the following question?

The solid $\ce{XY}$ decomposes into gaseous $\ce{X}$ and $\ce{Y}$: $\ce{XY(s) <=> X(g) + Y(g)}$ with a $K_p = 4.1$ at $\pu{0 ^\circ C}$. If the reaction is carried out in a $\pu{22.4 L}$ container which initial amounts of $\ce{X}$ and $\ce{Y}$ will result in the formation of solid $\ce{XY}$?

a. $\pu{5.0 mol}$ of $\ce{X}$, $\pu{0.5 mol}$ of $\ce{Y}$

b. $\pu{2.0 mol}$ of $\ce{X}$, $\pu{2.0 mol}$ of $\ce{Y}$

c. $\pu{1 mol}$ of $\ce{X}$, $\pu{1 mol}$ of $\ce{Y}$

d. none of the above

If I had been given $K_c$, I could put the different answer choices into the equilibrium expression and obtain $Q$. If $Q$ turned out to be smaller than $K_c$, then that would suggest that products would be favored. Contrariwise, if $Q$ was larger than $K_c$, then reactant(s) would be favored. So, I would simply see if any of the choices resulted in a $Q$ larger than $K_c$.

But, I was not given $K_c$. Must I convert $K_p$ to $K_c$ before making the comparison with $Q$?

I suspect the answer is "yes" (i.e. I must do the conversion) because (1) the respective values are very different and (2) I get different answers to the question depending on which value I use. This leads me to believe that I cannot compare $Q$ with $K_p$. Any additional insight into why this might be the case would be appreciated.

Answer 1

The reaction of interest is:

$$\ce{XY(s) <=> X(g) + Y(g)} \tag1$$

Thus, $K_p = P_\ce{X}\cdot P_\ce{Y} = 4.1$ since $P_\ce{XY} = 1$ in given condition because it is a solid. If we assume $\ce{X}$ and $\ce{X}$ are real gases, the given conditions are such that $P_\ce{X} = \frac{n_\ce{X} RT}{V} = n_\ce{X}$ and $P_\ce{Y} = \frac{n_\ce{Y} RT}{V} = n_\ce{Y}$, no matter what units are used for $P$.For example:

$$P_\ce{X} = \frac{n_\ce{X} \ \pu{mol} \times \pu{0.082 L atm K-1 mol-1} \times \pu{273.15 K}}{\pu{22.4 L}} = n_\ce{X} \ \pu{atm}$$

Suppose, initial amounts of $\ce{X}$ and $\ce{X}$ are $x$ and $y$ and amount of $\alpha$ reacted to produce $\alpha$ amount of $\ce{XY}$ at the equilibrium (to be realistic, $\alpha \gt 0$). Thus, at the equilibrium: $P_\ce{X} = x- \alpha$ and $P_\ce{Y} = y- \alpha$. Therefore:

$$K_p = P_\ce{X}\cdot P_\ce{Y} = (x- \alpha)(y- \alpha) = \alpha^2 - (x + y) \alpha + xy = 4.1 \tag2$$

1. If $x = 5$ and $y = 0.5$: From the equation $(2)$, $$ \alpha^2 - 5.5 \alpha + 2.5 = 4.1 \ \Rightarrow \ \alpha^2 - 5.5 \alpha - 1.6 = 0 \\ \therefore \ \alpha = \frac{5.5 \pm \sqrt{5.5^2 + 4 \times 1.6}}{2} = \frac{5.5 \pm 6.05}{2} $$ Thus, $\alpha = 5.78$ (not realistic) or $\alpha \ne -0.28$ (because $\alpha \gt 0$). Therefore no reaction.
2. If $x = 2.0$ and $y = 2.0$: Again, from the equation $(2)$, $$ \alpha^2 - 4 \alpha + 4 = 4.1 \ \Rightarrow \ \alpha^2 - 4 \alpha - 0.1 = 0 \\ \therefore \ \alpha = \frac{4 \pm \sqrt{4^2 + 4 \times 0.1}}{2} = \frac{4 \pm 4.05}{2} $$ Thus, $\alpha = 4.025$ (not realistic) or $\alpha \ne -0.025$ (because $\alpha \gt 0$). Therefore no reaction.
3. If $x = 1.0$ and $y = 1.0$: Similarly, you can prove that there is no reaction.

Hence, answer is "$(d)$. none of the above."

Answer 2

Since you are given the volume, temperature and initial amount of each gas, you can compute initial partial pressures as $n_iRT/V$ and from these the product $Q_p=p_X\cdot p_Y$ and compare this to $K_p$. If $Q_p>K_p$ then solid will form:

a. $\pu{5.0 mol}$ of $\ce{X}$, $\pu{0.5 mol}$ of $\ce{Y}$

$Q_p=2.50 \rightarrow$ no solid is formed

b. $\pu{2.0 mol}$ of $\ce{X}$, $\pu{2.0 mol}$ of $\ce{Y}$

$Q_p=4.01 \rightarrow$ no solid is formed

c. $\pu{1 mol}$ of $\ce{X}$, $\pu{1 mol}$ of $\ce{Y}$

$Q_p=1.00 \rightarrow$ no solid is formed

I assume that the pressure unit is atmospheres.