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### Expansion

Posted: Tue Jan 14, 2020 11:18 am
Does expanding the volume affect the equilibrium?

### Re: Expansion

Posted: Tue Jan 14, 2020 11:27 am
Hi! To answer your question, yes increasing volume affects a reaction at equilibrium. It does not, however, affect the value of Keq as long as the system is under constant temperature as you expand volume. Expanding the volume changes the molar concentrations of the reactants and products by reducing them. This is because Molarity (M) = mol/L; and increase in L would result in a lower M. Because all the concentrations are different, the reaction quotient of the system is changed and therefore will proceed accordingly to the direction in which more moles of gas can form. If the reaction has equal number of moles on both sides, then the concentrations of such gases change by the same factor which would not disrupt the system at equilibrium. However, if there is a difference in the moles of gas between the reactants and products, then the concentrations of the species on the side with more moles of gas decrease significantly compared to the side that has less moles of gas. Thus, because the concentrations on the side with more moles of gas decrease significantly, the reaction will proceed to that side to account for that significant decrease in gas concentration. I hope this helps! (It might be a little wordy)

### Re: Expansion

Posted: Tue Jan 14, 2020 12:18 pm
Yes I think it does since when they talk about expansion they are referring to the volume of the container. Since volume affects the reaction at equilibrium, so will expansion.

### Re: Expansion

Posted: Tue Jan 14, 2020 2:23 pm
Expanding the container would shift the reaction based on the amount of product or reactant, but with overall change, the Keq would remain unchanged.

### Re: Expansion

Posted: Tue Jan 14, 2020 2:32 pm
It would make the rxn shifts to one way or another but K is remains unchanged

### Re: Expansion

Posted: Tue Jan 14, 2020 10:01 pm
^^ as the previous replies stated, expansion would affect Q (the reaction quotient) however it would not affect K because K is a fixed value.