Unit 4 Questions MEGATHREAD!

[Milkshake]

How does adding NaOH(s) decrease the equilibrium concentration of O2 in the reaction:
2CO2 (g) <-> 2CO (g) + O2 (g), dH = +556 kJ mol-1

NaOH reacts with CO2 to form Na2CO3, but I had to google that, so I'm not sure if its something we should know

[golden]

How does adding NaOH(s) decrease the equilibrium concentration of O2 in the reaction:
2CO2 (g) <-> 2CO (g) + O2 (g), dH = +556 kJ mol-1

NaOH reacts with CO2 to form Na2CO3, but I had to google that, so I'm not sure if its something we should know

Thanks! Of course it is, that's why it's not a preferred primary standard. Why did I not realise that?

[HarveyD]

what would you guys choose for this one:

[HenryP]

Would the basicity constant be B? Although I'm sure since it doesn't have the appropriate units. I might be wrong as I am pretty terrible at chem.

[HarveyD]

Answer is D :S
I didnt think that H2O would be included though...

[thushan]

D is wrong.

Liquids and solids are NEVER included in an equilibrium expression. The reasons are to do with something called 'activity.'

You're right HenryP it's B.

[HarveyD]

awesome thanks for that

[luffy]

This discussion has made me interested into a new question.

Lets say we hypothetically have a reaction W_(l) + X_aq Y_aq + Z_aq.

Does that mean if I increase the concentration W to this system, le chateliers principle does not apply and in fact, nothing would happen?

Also, if I add water (i.e. increase the volume) of this system, would we, in fact, favor the forward reaction, despite having equal no. of particles on both sides?

Interesting...

[HenryP]

I might have to disagree with you luffy. First of all how would you increase the concentration of a liquid? The reason we don't include liquids or solids in the equilibrium constant is because the concentration cannot be changed to a great extent and so it can be assumed to be constant.
If you were to add some W into the equilibrium mixture the reaction would proceed in a forward direction as you are diluting all the other reactants.
As for your next question I believe that by adding enough water you will eventually turn W into an aqueous solution and then an even number of particles will be on both sides and no change to equilibrium would occur. If enough water is not added to sufficiently turn W into an aqueous solution, then I think you would be right in saying the forward reaction would dominate.
Please excuse me if I am wrong though. Like previously said, not that great at chem haha

[luffy]

I might have to disagree with you luffy. First of all how would you increase the concentration of a liquid? The reason we don't include liquids or solids in the equilibrium constant is because the concentration cannot be changed to a great extent and so it can be assumed to be constant.
If you were to add some W into the equilibrium mixture the reaction would proceed in a forward direction as you are diluting all the other reactants.
As for your next question I believe that by adding enough water you will eventually turn W into an aqueous solution and then an even number of particles will be on both sides and no change to equilibrium would occur. If enough water is not added to sufficiently turn W into an aqueous solution, then I think you would be right in saying the forward reaction would dominate.
Please excuse me if I am wrong though. Like previously said, not that great at chem haha

Ah true - good point. Thanks! Though, I don't think my example truly justified my question. Sigh - oh well.

So what is the "concentration" in which a "liquid" is stated as "aqueous"? Is there a set point? Or do we just have a general vague idea of whether or not its aqueous?

[thushan]

when you consider no. of particles in terms of dilution, we only talk about particles in aqueous and gaseous phases!

[kensan]

what would you guys choose for this one:

I thought it would be C?? Because it is not acidity constant, it's equilibrium constant and water is included. That's what I thought anyway haha

[thushan]

Acidity constants ARE equilibrium constants

[liuetenant]

Just a question...for le chatliers principle...

Is heat a product? Is it right to say that the system will partially oppose the decrease in temperature by favouring the forward reaction as heat is a product? Or should i said....it favours the forward reaction as it is TEMPERATURE increasing?

not sure if heat is considered a product! (my teacher said something like that)

Thanks!

[vea]

Heat is NOT a product but many teachers teach it as a way for students to learn the effect of temperature in relation to LCP. Make sure you do not write that heat is a product on the exam or you will lose marks.

I'm assuming the reaction you have above is exothermic (increasing the temperature of system by favouring forward reaction).

My response would be something like this:
When the temperature of an exothermic equilibrium system is decreased, Le Chatelier's Principle says that the system will try to oppose this change by increasing the temperature of the system. To increase the temperature of the system, the exothermic reaction- which releases heat and increases the temperature of the system- must be favoured. So in this case, the forward reaction (the exothermic reaction) is favoured.