Which system at equilibrium will not be influenced by a change in pressure?

$$ \begin{align} \ce{3 O2(g) &<=> 2 O3(g)}\tag{A}\\ \ce{N2(g) + 3 H2(g) &<=> 2 NH3(g)}\tag{B}\\ \ce{2 NO2(g) &<=> N2O4(g)}\tag{C}\\ \ce{H2(g) + I2(g) &<=> 2 HI(g)}\tag{D}\\ \ce{2 W(g) + X(g) &<=> 3 Y(g) + 2 Z(s)}\tag{E} \end{align} $$

My attempt: Equilibrium will be influenced if the number of moles on one side of the equation is different than the other side. Thus, I can eliminate option A,B, and C. But how to differentiate between options D and E? They both look right to me, because there are the same amount of moles of gases on both sides, and solids do not play a factor in determining equilibrium.

For reference, the answer given in the book is D.

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    $\begingroup$ The solid is slightly compressible, so there will be a change in volume. It's minimal compared to the gases, but it's not zero. $\endgroup$ – Zhe Apr 12 '17 at 3:27
  • $\begingroup$ @Zhe But doesnt the equilibrium constant only depend on gases, as they are the only ones that occupy a significant amount of volume? $\endgroup$ – Teoc Apr 12 '17 at 3:40
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    $\begingroup$ I think that the answers should be D and E for as @Zhe writes the change in volume of a solid is minuscule and can be ignored. There might be typo in E which is why the answer is D only. $\endgroup$ – porphyrin Apr 12 '17 at 8:35

As Zhe explains in the comments, the answer is D. Even though small changes in pressure are not expected to significantly affect the chemical potential of a solid, it is a function of pressure:

$$\mu(\mathrm{s}) = \mu^\circ(\mathrm{s}) + \int_{p^\circ}^{p}V_\mathrm{m}\,\mathrm dp \qquad T = \mathrm{const}$$ This means that a small change in the chemical potential of the solid also contributes to the total free energy change of the system when the pressure is altered. Usually this response is ignored as $V_\mathrm{m}$ is orders of magnitude greater for gases than for condensed phases.



Well it's about the amount of moles of gases on left and the right sides of the equal/equilibrium sign

for A, it is 3 moles vs 2, so then if we apply pressure it shifts to the right, to make this more visual, imagine a tower with 3 blocks stacked and another tower stacked 2 blocks high, if we were place them side by side and push them down with a plate, obviously the tower with the 3 blocks is going to be compressed First

for B, it is 4 vs 2 moles

for C, it is 2 vs 1 moles

for D, it is 2 vs 2 moles

for E, it is 3 vs 2=3 moles (with solid)

From the list, we can see that in choice D, we have the same amount of moles on both sides. Adding pressure or decreasing them wouldn't affect in which direction it would go to.

You said that you wanted help between choice D and E, but again if we look at the problem, for choice E, we have a tower stacked 3 blocks high vs 3+solid blocks high, so that means that the one with the 3+solid blocks is higher and thus means unbalance

to convert that into chemistry terms, because you have an imbalance of moles of gases, that means that if you apply pressure, you are going to be disrupting the side with 3+solid moles as compared to 3 moles.


you want to have choice D because there is no imbalance of moles of gases

Also I understand that you are wondering about the solid being there, I believe it is not choice D because we have the same amount of moles but the definition of an equilibrium is that it goes both forward and back so then that implicates that the Z turns into products as well. So by adding pressure you actually favor one side more because either adding or decreasing pressure means that it becomes harder for Z to become a gas or to turn into a solid depending on the imbalance created. Z will eventually become a gas at one point and the added pressure or decreased pressure might make that harder. Z turns into gas molecules at one point in time, so your going to have more than 3 moles gas on left side, it does that naturally because definition of equibrium, but if you apply pressure it becomes harder to do that because RXN shifts right

  • $\begingroup$ But only gases apply to equilibrium, answer choice E has 2 mols of solid, making it 3 vs 3 moles. $\endgroup$ – Teoc Apr 12 '17 at 4:23

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