Chemical Equilibrium and Le Chatelier's Principle

• kirsten_2009
In summary, to manipulate the given equation to produce more CrO42- ions without adding any Chromium based compounds, you can alter the concentration of one of the products on the right by adding a base such as sodium hydroxide. This will shift the equilibrium to the left, resulting in more CrO42- ions. The resulting equation will have 2CrO42- + 2H3O+ + NaOH <--> Cr2O72- + 4H2O. It is important to balance the equation in either acidic, neutral, or basic media to take into account the redox reactions involved.
kirsten_2009

Homework Statement

Without adding any Chromium based compounds, how would you manipulate the above equation to produce more CrO42- ions? Be sure to use chemical equations to show the answer.

Homework Equations

2CrO42- (aq) + 2H3O+ <--> Cr2O72-(aq) + 3H2O (l)

The Attempt at a Solution

So my initial thought here is that we want the reaction to shift left so we must alter the concentration of one of the products on the right. Since adding more Chromium is "not allowed" then it would have to be the water? If this is correct though, I don't know what the equation would look like because if I add more moles of water won't I have to balance all the other species as well to have a balanced chemical equation? Thanks in advance for all the time and help.

kirsten_2009 said:
shift left so we must alter the concentration of one of the products on the right.
... or, on the left. Think about it. You are allowed to add, or remove reactants and products to push an equilibrium around to where you want it.

Chestermiller
Ah! :) good point! So then, my thoughts becomes; since hydronium is one of the reactants, if we lower the concentration of hydronium by adding sodium hydroxide to neutralize some of the acid then the equilibrium would shift left?

Very good.

Thank-you for the help. So, would this reaction equation be correct?

2CrO42- + 4H3O + 2NaOH <--> Cr2O7
2- + 7H2O + 2Na

Sodium is just a spectator ion, so you don't have to show it formally. All the "action" is in the acid-base-water bookkeeping. You have to add and remove the same amounts to and from both sides of the equation just like algebra.

O.k I understand thanks. Does this look better?

2CrO42- + 2H3O+ + NaOH <--> Cr2O72- + 4H2O

Try adding OH- to both sides as necessary to "neutralize" one side, then subtract excess water. Forget the sodium ion.

Hello,
I'm having problems with the same question...is this right?

2CrO42- + 2H3O + 2OH- <--> Cr2O72- + 3H2O + 2OH-

Good first step. 2H3O+ + 2OH- gives you what?

4 H2O?

Good. Since you've got water on both sides of the equation, you can subtract water from both sides until it disappears from one side or the other.

2CrO42- <--> Cr2O72- + 3H2O + 2OH-

thanks, ...so I could just write the equation like this?

Wait...but then it wouldn't balance...

Bystander said:
subtract water from both sides

Also: if you see the same molecule/ion/element on both sides of the equation, cancel it out. You have

Paulham said:
2CrO42- + 2H3O + 2OH- <--> Cr2O72- + 3H2O + 2OH-

You have 2OH- on both sides of the equation, they don't change - so you should remove them. They are just spectators.

kirsten_2009
Hello,

Thanks for all the help. I don't know why Paulham and I are so confused but I don't seem to see the matter clearly either. I understand the theory behind it and how NaOH will remove some H+ from the reactants side shifting the reaction to the left but it's the actual chemical equation and it's accounting that seems to be the problem. I see what Mr. Borek is saying about the OH- being spectator ions but when you remove them you would get

2CrO42- + 2H3O <---> Cr2O72- + 3H2O

which is what the original equation is...so it looks like we did nothing? Thanks again for the time...

I'm tired of this: We want to drive it to the left, so add OH-; that gives us water on the left, and water and OH- on the right; subtract three waters from both sides; 2CrO4-2 + H2O ↔ Cr2O7-2 + 2OH-. It's one of life's little headaches when playing with redox reactions deciding whether you're going to balance them in acidic, neutral, or basic media.

kirsten_2009
Oh my goodness, that makes complete sense...to be honest I didn't even clue in that this was a redox reaction. Thank you.

The chromate-dichromate equilibrium isn't redox in and of itself, but both species are used as oxidizers in quite a number of other reactions, and the pH question is an endless headache as far as keeping things balanced, as well as dragging in the oxygen hydrogen half-cell potentials for either acid or base conditions.

1. What is chemical equilibrium?

Chemical equilibrium is a state in a chemical reaction where the forward and reverse reactions occur at the same rate, resulting in no net change in the concentrations of reactants and products.

2. How is equilibrium affected by changes in concentration, pressure, and temperature?

According to Le Chatelier's Principle, when a change is made to a system at equilibrium, the system will shift in a direction that minimizes the effect of that change. For example, if the concentration of a reactant is increased, the equilibrium will shift towards the products to consume the excess reactant. If the pressure is increased, the equilibrium will shift towards the side with fewer moles of gas. And if the temperature is increased, the equilibrium will shift in the endothermic direction to absorb the excess heat.

3. Can equilibrium be reached in both exothermic and endothermic reactions?

Yes, equilibrium can be reached in both types of reactions. In an exothermic reaction, the equilibrium will favor the products as they are formed with the release of heat. In an endothermic reaction, the equilibrium will favor the reactants as they are consumed to produce heat.

4. How do catalysts affect chemical equilibrium?

Catalysts do not affect the position of equilibrium, but they do increase the rate at which equilibrium is reached by lowering the activation energy of the forward and reverse reactions. This allows the system to reach equilibrium faster, but the concentrations of reactants and products at equilibrium will remain the same.

5. What are some real-life examples of Le Chatelier's Principle?

Some examples include the production of ammonia, where the equilibrium is shifted towards the products by using high pressure and low temperature, and the production of carbon monoxide, where the equilibrium is shifted towards the products by using a catalyst. The principle also applies to everyday situations, such as the addition of salt to a pot of boiling water, where the equilibrium between the liquid and gas phases of water is shifted towards the liquid phase due to the increase in pressure.

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