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.

  • $\begingroup$ $K_p$ has a unit. It may be in bar, or in atm, or any other pressure unit. What is it ? $K_c$ has also a unit. Which one ? $\endgroup$
    – Maurice
    Sep 15, 2020 at 19:58
  • $\begingroup$ You do not need the sought conversion. This is not hard as you think it is. Follow the concept. $\endgroup$ Sep 15, 2020 at 20:30
  • $\begingroup$ $K_p$ should not have units, its log could never be taken if it did. In old texts units are used but this habit was changed years ago by dividing as necessary by 1 unit of concentration/pressure etc . (In practice this means just using numerical value) $\endgroup$
    – porphyrin
    Sep 16, 2020 at 10:04

2 Answers 2


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."


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.

  • 1
    $\begingroup$ I thought this might be the case when I first did the problem. But then I had second thoughts since the "test" values given were in moles and not pressures (and therefore I thought I needed to use $K_c$). This confusion is why I asked the question. It seems, from your answer, that I CAN use $K_p$ and compare it with Q. Thank you. Also, thank you M. Mahindaratne, because your answer shows that, in this case, the pressures are equal to the number of moles. $\endgroup$ Sep 16, 2020 at 13:18

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