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Le Chatelier's Principle to predict the change in position of equilibrium

Submitted by chrisf on Mon, 06/23/2008 - 21:12

Describe the effect on the position of the equilibrium if AgNO3 were added to the following equilibrium system.

FeCl3 (aq) + KSCN(aq) Fe(SCN)2+(aq) + 3Cl-(aq) + K+(aq)

The equilibrium will shift left towards reactants.

Why does the NO2 / N2O4 gas mixture initially darken upon compression and then lighten when equilibrium is established?

As pressure is increased, equilibrium shifts to favor products but as equilibrium is reached N2O4 is favored in the reaction

I hope these are right..


The equilibrium will shift left towards reactants. - I believe this is incorrect.

Adding Ag+ ion will precipitate out the Cl- ion.  A reduction in the concentration of Cl- will shift the reaction to the right as the system attempts to replace the precipitated Cl-

As pressure is increased, equilibrium shifts to favor products but as equilibrium is reached N2O4 is favored in the reaction. 

I agree that increasing the pressure would shift the reaction toward the products since this would reduce the pressure as there are fewer molecules on the product side than the reactant side:
                          2 NO2    N2O4
That would explain the initial darkening as N2O4 is brown.

I believe that since the reaction involves the formation of a bond, the forward reaction is exothermic.
                          2 NO2    N2O4  +  Heat
As the heat is given off, the temperature would rise, this would have the effect of driving the reaction to the left (toward the reactants or in the endothermic direction).  This would result in the lightening of the reaction mixture as the brown N2O4 is used up.


spock wrote:

The equilibrium will shift left towards reactants. - I believe this is incorrect.

Adding Ag+ ion will precipitate out the Cl- ion.  A reduction in the concentration of Cl- will shift the reaction to the right as the system attempts to replace the precipitated Cl-

As pressure is increased, equilibrium shifts to favor products but as equilibrium is reached N2O4 is favored in the reaction. 

I agree that increasing the pressure would shift the reaction toward the products since this would reduce the pressure as there are fewer molecules on the product side than the reactant side:
                           2 NO2      N2O4
That would explain the initial darkening as N2O4 is brown.

I believe that since the reaction involves the formation of a bond, the forward reaction is exothermic.
                           2 NO2      N2O4  +  Heat
As the heat is given off, the temperature would rise, this would have the effect of driving the reaction to the left (toward the reactants or in the endothermic direction).  This would result in the lightening of the reaction mixture as the brown N2O4 is used up.

Okay, I understand where you're coming from on the first one...

But on the second one, isn't NO2 brown while N2O4 is colorless? If so, if you were to compress the gas (decrease volume/increase pressure) shouldn't the reaction shift to favor the least molecules (stoichiometrically) which would be N2O4?

I'm confused, because this is the exact opposite of what happens in lab..


Describe the effect on the position of the equilibrium if an acid were added to the following equilibrium system:

Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)

The reaction would shift to the right because the concentration of OH- would decrease when the acid was added.

How is the addition of HCl to this reaction able to affect the position of equilibrium despite the fact that neither an H or Cl ion appears in the equilibrium reaction?

Cu2+(aq) + 4NH3(aq) Cu(NH3)42+(aq)

I'm not sure.

How is the addition of dry ice able to affect the position of equilibrium despite the fact it does not appear in the equilibrium reaction?

Ind = Indicator

HInd(aq) + H2O(l) Ind-(aq) + H3O+(aq)

I'm not sure.


How is the addition of HCl to this reaction able to affect the position of equilibrium despite the fact that neither an H or Cl ion appears in the equilibrium reaction?

I see 2 possibilities.  Either the H+ ion form with the NH3 to make NH4 (it won't stay this way long, in fact I don't know that it's safe to say this would even happen here)

Or:  Cl- will bond with Cu2+ forming CuCl2.  This decreases your Cu2+ concentration and the system will shift left to replace it.

How is the addition of dry ice able to affect the position of equilibrium despite the fact it does not appear in the equilibrium reaction?

Dry ice is solid CO2.  Not sure why this will affect it though.


SchoolBoyDJ wrote:

Or:  Cl- will bond with Cu2+ forming CuCl2.  This decreases your Cu2+ concentration and the system will shift left to replace it.

This is what I was thinking.


You are right,  NO2 is reddish brown.  I based my answer on the description that was given that an increase in pressure initially causes the gas to darken and then lighten.  Given that situation, it was logical to assume that the N2O4 must be the darker gas.  I'm familiar with the demonstration, but haven't done it in years and had forgotten which gas was colorless and which gas was reddish-brown.

If you are saying that you tested this in the lab and found that the solution lightened as pressure increased, and then darkened as time went on, then those observations would be consistent with the explanation I gave earlier.

How is the addition of HCl to this reaction able to affect the position of equilibrium despite the fact that neither an H or Cl ion appears in the equilibrium reaction?

Cu2+(aq) + 4NH3(aq) Cu(NH3)42+(aq)
The H+ reacts with the NH3 to form NH4+.  The decrease in the concentration of NH3 would cause the reaction to shift to the left, reducing the concentration of the complex Cu ion.

As for dry ice, as the solid CO2 sublimes CO2 gas reacts with water to form carbonic acid, a weak acid.  This partially ionizes to produce additional H3O+ ions which then tend to drive the Indicator equilibrium to the left, increasing the concentration of HInd while reducing the concentration of Ind-.


I'm trying to work this other problem out too, and it's not making a whole lot of sense (supposed to be related to our lab, I assume).

When 20.0mL of SO2 gas is mixed with 10.0mL of O2 gas, the resulting volume is 25.0mL. The experiment is conducted at 750 torr and a temperature of 25.0C. Determine the equilibrium constant for the reaction.

nSO2 = (0.987atm)(0.02L) / (0.0821 Latm/molK)(298.15K) = 8.06 * 10-4 mol

nO2 = (0.987atm)(0.01L) / (0.0821 Latm/molK)(298.15K) = 4.03 * 10-4 mol

nTotal = (0.987atm)(0.025L) / (0.0821 Latm/molK)(298.15K) = 1.01 * 10-3 mol

I think what I'm confused about is where to go next...

I know that for every 1 mole of SO2 there is 1 mole of SO3. This means the most SO3 I should be able to make is 8.06 * 10^-4 mol and any moles of gas extra would be excess O2 right?

If that's right, I sure don't know how to calculate it.