Question

For the reaction Ag2S(s) ⇌ 2Ag+(aq) + S2−(aq), what happens to the equilibrium position if the amount of silver ion is halved?For the reaction S(s) ⇌ 2(aq) + (aq), what happens to the equilibrium position if the amount of silver ion is halved?

Answer 1

More silver and sulfide ions will be yielded as the reaction proceeds.

Step-by-step explanation:

Hello,

In this case, with the given reaction at equilibrium:

`Ag_2S(s) \rightleftharpoons 2Ag^+(aq) + S^(2-)(aq)`

We take into account the Le Chatelier's principle in order to explain what would happen when the amount of solver ion is halved. Thus, it explains that when the amount of one product is diminished as in this case, since by halving its amount is reduced to the half, the direct reaction is favored (reaction will shift rightwards), it means the formation of products, so more silver and sulfide ions will be yielded as the reaction proceeds.

Best regards.