Equilibrium Pressure Question

[EthanS24]

Could anyone please explain to me how the answer is increase
This question appears in:
Q.25 of U3 AOS2 in Chem 3/4 Heinemann Textbook

Carbon monoxide and iodine pentoxide react to form
iodine and carbon dioxide in the equilibrium reaction.
5CO(g) + I2O5(s) = I2(g) + 5CO2(g) ΔH = –1175 kJ

If the volume doubles, does the Amount of CO2 Increase/Decrease/no change ?

Thanks

[sweetcheeks]

The key here is amount of CO2, not concentration of CO2. When you double the volume, you initially have the same amounts of each species, except that the concentration of all of them have decreased. The system will now shift to the side with more gaseous products to partially counter the change. This means that there is an increase in the amount of CO2 formed, as the right has 6 gaseous products and the left only 5. (the concentration would still be less than original).

[vcestressed]

The key here is amount of CO2, not concentration of CO2. When you double the volume, you initially have the same amounts of each species, except that the concentration of all of them have decreased. The system will now shift to the side with more gaseous products to partially counter the change. This means that there is an increase in the amount of CO2 formed, as the right has 6 gaseous products and the left only 5. (the concentration would still be less than original).

Hey, I'm a little confused here.. Wouldn't that apply only if ALL molecules were gases? In this case, iodine pentoxide isn't a gas.

[Lear]

Iodine Pentaoxide is a solid and we know that solids and liquids are not included in equilibrium expressions and thus can be ignored for the sake of predicting how a system would shift if a change was opposed. Ignore Iodine Pentaoxide we have 5 molecules on the left and 6 on the right. Increasing the volume decreases pressure and the system partially opposes this by favoring the reaction producing the greatest moles of gas. In this case the forward reaction and in turn the concentration of CO2.