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- Equilibrium shifts for slightly soluble compounds The reaction for the formation of a saturated solution of silver bromide (AgBr) can be represented as AgBr(s)?Ag+(aq)+Br?(aq)

Consider that this reaction is an endothermic process and that the AgBr solution is saturated.

Classify the following changes based on whether they will favor the formation of reactant, whether they will favor the formation of products, or whether they will have no effect on the equilibrium.

1. Adding hydrobromic acid (HBr)

2. Lowering the concentration of bromide (Br-) ions

3. Heating the concentrations

4. Adding solid silver bromide (AgBr)

5. Removing silver (Ag+) ions

Respuesta :

Answer:

1) Favor formation of reactant

2) Favor formation of product

3) Favor formation of product

4) Favor formation of product

5) Favor formation of product

Explanation:

The reaction is [tex]AgBr(s)--->Ag^{+}(aq)+Br^{-}(aq)[/tex]

The reaction is endothermic

1. Adding hydrobromic acid (HBr)

It will increase the amount of bromide ion. Bromide ion is one of the product so it will favor formation of reactant.

2. Lowering the concentration of bromide (Br-) ions

As we are decreasing the the concentration of product it will favor formation of more products

3. Heating the concentrations

If we are heating the mixture, it will increase the temperature and it will favor endothermic reaction. Thus will favor formation of product.

4. Adding solid silver bromide (AgBr)

If we are adding silver bromide it means we are increasing the concentration of reactant, it will favor formation of products

5. Removing silver (Ag+) ions

Removing silver ions means we are decreasing the concentration of product thus it will favor formation of products.

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