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equilibrium question help a level chem

Hi, on question 4.4 on this paper, it confused me. I get that the concentration of all substances will decrease, but what I dont get is how this results in the equilibrium shifting to the side with the most number of moles? how and why does this occur? thanks

https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2018/june/AQA-74052-QP-JUN18.PDF
I think it might be due to the le chantelier principle.... it opposes changes in concentration..
Original post by Bertybassett
Hi, on question 4.4 on this paper, it confused me. I get that the concentration of all substances will decrease, but what I dont get is how this results in the equilibrium shifting to the side with the most number of moles? how and why does this occur? thanks

https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2018/june/AQA-74052-QP-JUN18.PDF
well yeah I get that, I just dont get how it applies to this situation. why would the position shift to the side with more moles?
Original post by Lailaaaa123
I think it might be due to the le chantelier principle.... it opposes changes in concentration..
Thats a bit confusing usually u refer to number of moles when talking abt the effect of pressure....
Original post by Bertybassett
Hi, on question 4.4 on this paper, it confused me. I get that the concentration of all substances will decrease, but what I dont get is how this results in the equilibrium shifting to the side with the most number of moles? how and why does this occur? thanks

https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2018/june/AQA-74052-QP-JUN18.PDF


Original post by Bertybassett
well yeah I get that, I just dont get how it applies to this situation. why would the position shift to the side with more moles?
The concentrations of all of the species in the eqm decrease, but as there are more moles on the RHS, the concentration of the products will decrease more than the concentration of the reactants. Therefore, in order to restore eqm, the position of eqm has to shift right in order to increase [C] and [D] and oppose the change in concentration. This, therefore, decreases [A]


Original post by Bertybassett
well yeah I get that, I just dont get how it applies to this situation. why would the position shift to the side with more moles?
thanks a lot
Original post by TODTEMPLE01
The concentrations of all of the species in the eqm decrease, but as there are more moles on the RHS, the concentration of the products will decrease more than the concentration of the reactants. Therefore, in order to restore eqm, the position of eqm has to shift right in order to increase [C] and [D] and oppose the change in concentration. This, therefore, decreases [A]

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