5J.5 (part d) Textbook Problem

[Elizabeth Kaplan 3I]

State whether reactants or products will be favored by an increase in the total pressure (resulting from compression) on each of the following equilibria. If there is no change, explain why that is so.
(d) 2HD(g) + H2(g)--> --< D2(g)

For part (d), I thought that the products will be favored because there is 1 mole of gas on the right (products) side, compared to 3 moles of gas on the left (reactants) side, however the answer key says that there will be no change. Does anyone know why this is?

[Bella Wachter 1A]

I went to the Workshop session this morning with Riya, and she said that there should be a shift, like you said (she also said she'd talk to Professor Lavelle about it and see if it was an error/could be added to the "Solution Guide Errors" document on the 14B website). I'm not sure why the book states otherwise, because the only way there would be no shift is if both sides had equal moles of gas, which is not the case.

[Zaid Bustami 1B]

Hey Elizabeth,
Correct me if I'm overlooking something but I think this could be the textbook's fault. The reaction they list is 2HD + H₂ <-> D₂, which is doesn't make sense because there are no regular hydrogens present on the right side of the reaction (I say regular because D stands for deuterium I think, which is an isotope of H with 2 neutrons)
I think it could be a misprint and is supposed to say 2HD + H₂ <-> D₂ + 2H₂, in which case the book's answer key is correct because there are three moles of gas on either side of the reaction.

[Ria Nawathe 1C]

I also think this might be an error in the solutions. Prof Lavelle mentioned in one of the module videos that one of the situations in which there would be no change is if the pressure was increased by adding an inert gas (which doesn't affect the volume and therefore the concentration), but in this question the volume is changing so your answer should be correct.

[EnricoArambulo3H]

In my textbook, I have this problem as "2HD <->H2 + D2". For this, it would be no change since the moles of gas are equal on both the reactants and products. Hope this helped!