# Le Chatelier's principle

terryds

## Homework Statement

For the equilibrium reaction:
##Ag^+(aq) + Fe^{2+}(aq) \rightleftharpoons Ag(s) + Fe^{3+}(aq)\ \Delta H = -20 kJ##

action that can be done so more silver get dissolved is ...

A. Add FeCl3
B. Decrease the pressure
C. Stir the mixture
D. Add FeSO4
E. Decrease the temperature

## Homework Equations

Le chatelier's principle

## The Attempt at a Solution

Adding FeCl3 will increase the presence of ##Fe^{3+}## ions which means the equilibrium get shifted to left which leads to more Ag+ (more silver get dissolved).
But, I think the Cl- ion in FeCl3 will also react with Ag+ making AgCl precipitate (so instead of dissolving, it form a precipitation)

Decrease the pressure. I'm not sure if I should include the coefficient of the solid species (in this case the solid silver Ag(s)). If it counts, that means both sum of left coefficients and right coefficients are equal. But, if it doesn't, that means the reaction will shift to the more number which is to the left (more silver get dissolved)

Stir the mixture. I'm sure stiring mixture will just accelerate the speed of reaction. It has nothing to do with the reaction.

Adding FeSO4 will increase the presence of ##Fe^{2+}## ions which means that the equilibrium get shifted to the right, so more solid silver will form. And ##SO_4^{2-}## ions will react with ##Fe^{3+}## making ##Fe_2(SO_4)_3##

Decrease the temperature will make the equilibrium shift to the right in exothermic reaction. So, I'm sure this one is not the answer.

Please help..

## Answers and Replies

Science Advisor
Gold Member
Adding FeCl3 will increase the presence of ##Fe^{3+}## ions which means the equilibrium get shifted to left which leads to more Ag+ (more silver get dissolved).
Correct.
But, I think the Cl- ion in FeCl3 will also react with Ag+ making AgCl precipitate (so instead of dissolving, it form a precipitation)
This will have the effect of lowering the [Ag+]. What effect will that have on the reaction?

Decrease the pressure. I'm not sure if I should include the coefficient of the solid species (in this case the solid silver Ag(s)). If it counts, that means both sum of left coefficients and right coefficients are equal. But, if it doesn't, that means the reaction will shift to the more number which is to the left (more silver get dissolved)
When applying this principle, you should only think about the number of gas phase molecules (in a reaction with no gas phase products or reactants, the volume does not change substantially, so pressure has no effect).

Stir the mixture. I'm sure stiring mixture will just accelerate the speed of reaction. It has nothing to do with the reaction.
Correct. Stirring affects how quickly the reaction gets to equilibrium, but not the position of the equilibrium itself.

Adding FeSO4 will increase the presence of ##Fe^{2+}## ions which means that the equilibrium get shifted to the right, so more solid silver will form. And ##SO_4^{2-}## ions will react with ##Fe^{3+}## making ##Fe_2(SO_4)_3##
Correct. As in the case of AgCl precipitation, how would precipitating Fe3+ from solution affect reaction?

Decrease the temperature will make the equilibrium shift to the right in exothermic reaction. So, I'm sure this one is not the answer.
Correct.

terryds
Correct.
This will have the effect of lowering the [Ag+]. What effect will that have on the reaction?
The effect is the equilibrium shift to the left due to lowering of [Ag+], right??
So, more Ag+ will be produced?

When applying this principle, you should only think about the number of gas phase molecules (in a reaction with no gas phase products or reactants, the volume does not change substantially, so pressure has no effect).
So, this means that the reaction shift to the left...

I'm confused now.
It seems A and B are correct.

Science Advisor
Gold Member
The effect is the equilibrium shift to the left due to lowering of [Ag+], right??
So, more Ag+ will be produced?
Correct.

So, this means that the reaction shift to the left...

I'm confused now.
It seems A and B are correct.
Why do you think decreasing the pressure will shift the reaction to the left? Remember that pressure has an effect only on reactions where the number of gas phase molecules changes.

Last edited:
terryds
Why do you think decreasing the pressure will shift the reaction to the left? Remember that pressure has an effect only on reactions where the number of gas phase molecules changes.

Only gas? Aqueous doesn't count?

What about volume? Does it only effect on gas or aqueous too?

Science Advisor
Gold Member
Pressure matters for gas phase reactions because it affects the concentration of reactants and products. Decreasing the pressure (while keeping T and n constant) means that V must increase. This means that the concentration (n/V) decreases for all gas phase species, which can affect the reaction quotient (Q = ([product 1]*[product 2] * ...etc)/ ([reactant 1] * [reactant 2] * ...etc), if the number of gas molecules increases or decreases with the reaction. For example, in the reaction $$2NO_{2\text{ (g)}} \rightleftharpoons N_2O_{4\text{ (g)}}$$ The reaction quotient Q is $$Q = \frac{[N_2O_4]}{[NO_2]^2}$$ Decreasing the pressure by a factor of two will decrease the numerator by a factor of two but the denominator by a factor of four, so to re-establish equilibrium, the reaction will have to convert more ##N_2O_4## into ##NO_2##.

Will changing the pressure have any effect on the concentrations of ions in solution?

terryds
terryds
Pressure matters for gas phase reactions because it affects the concentration of reactants and products. Decreasing the pressure (while keeping T and n constant) means that V must increase. This means that the concentration (n/V) decreases for all gas phase species, which can affect the reaction quotient (Q = ([product 1]*[product 2] * ...etc)/ ([reactant 1] * [reactant 2] * ...etc), if the number of gas molecules increases or decreases with the reaction. For example, in the reaction $$2NO_{2\text{ (g)}} \rightleftharpoons N_2O_{4\text{ (g)}}$$ The reaction quotient Q is $$Q = \frac{[N_2O_4]}{[NO_2]^2}$$ Decreasing the pressure by a factor of two will decrease the numerator by a factor of two but the denominator by a factor of four, so to re-establish equilibrium, the reaction will have to convert more ##N_2O_4## into ##NO_2##.

Will changing the pressure have any effect on the concentrations of ions in solution?

Hmm.. I think it won't... It has nothing to do with the ions

Science Advisor
Gold Member
Hmm.. I think it won't... It has nothing to do with the ions
Correct. This is why you only need to consider the change in the number of gas phase molecules when pressure increases or decreases. Only gasses change volume significantly with changes in pressure. Liquids and solids (and aqueous species since they exist in liquids) are incompressible.

There are some exceptions, however. For example, some solids can exist as different allotropes that have different densities, so different allotropes can be favored under different pressures (e.g. carbon in coal vs diamond).

terryds