VCE Chemistry Question Thread

[warya]

Thanks!

How come Le Chatelier's principle partially opposes a change and not completely? For instance, in the reestablishing of equilibrium, the concentration of a spiked reactant never reaches pre-change levels, although it does decrease

[paper-back]

A mixture of 0.01mol hydrogen gas and unknown amount of methane burn completely in a calorimeter... heats of combustion of hydrogen gas and methane are 286KJ/mol and 890KJ/mol respectively:
(a) write the thermochemical equations for the 2 combustion reactions

In the answers (Answer: 2H2+O2->2H2O, /\ = 572KJ/mol) they've multiplied the heat of combustion of hydrogen gas by 2, why?

[lzxnl]

Thanks!

How come Le Chatelier's principle partially opposes a change and not completely? For instance, in the reestablishing of equilibrium, the concentration of a spiked reactant never reaches pre-change levels, although it does decrease

It's mathematically impossible for a complete opposition in change.
Let's say we have the reaction A + B -> C
Q = [C]/([A]) = K initially
Now assume we double [C], so Q increases. Q will then decrease to get back to K. The only way the system can do this by itself is to effect a back reaction, so [C] decreases and both [A] and increase. [C] can't halve to go back to the initial concentration because [A] and have both increased in the back reaction, so [C] has to drop to some intermediate value between the spiked value and the initial value. Use similar reasoning for all other cases.
The problem with a perfect opposition to the change is that in an equilibrium, if you return one part of the system right back to where it began, some other part of the system will have changed, meaning your system won't then be at equilibrium.

[warya]

Could I have help with this question

OCl- + H2O--->HOCl + OH-  (at equilibrium)

100mL of pure water is added to a 100mL solution of 0.1M NaOCl, when the solution reaches equilibrium again, the
A) Concentration of H+ has decreased
B) pH of the solution has decreased
C) Concentration of HOCl has increased
D) the value of the equilibrium constant has halved

The answer is A. I understand how the concentration of OH- can be said to have decreased as the change is only partially offset and so  conc. at the new equilibrium is still lower than its initial value. But I don't get what this has to do with H+? Also, why are B and C wrong?

[sunshine98]

Could I have help with this question

OCl- + H2O--->HOCl + OH-  (at equilibrium)

100mL of pure water is added to a 100mL solution of 0.1M NaOCl, when the solution reaches equilibrium again, the
A) Concentration of H+ has decreased
B) pH of the solution has decreased
C) Concentration of HOCl has increased
D) the value of the equilibrium constant has halved

The answer is A. I understand how the concentration of OH- can be said to have decreased as the change is only partially offset and so  conc. at the new equilibrium is still lower than its initial value. But I don't get what this has to do with H+? Also, why are B and C wrong?

A-  Adding water = dilution so it decreases concentration of everything. Usually this change will be partially offset by  the net reaction occurring in  the direction that produces the greater number of particles. But in this q we have the same no. of particles on either side.
B- pH decreasing means increasing [H+] and the concentration of H+ is actually decreasing so this cannot be right
C- because dilution = decreasing concentration of errything not increasing it. 
Pls correct me if I'm wrong cause I just finished learning this topic .
Hope this helps 

[kayoak]

A mixture of 0.01mol hydrogen gas and unknown amount of methane burn completely in a calorimeter... heats of combustion of hydrogen gas and methane are 286KJ/mol and 890KJ/mol respectively:
(a) write the thermochemical equations for the 2 combustion reactions

In the answers (Answer: 2H2+O2->2H2O, /\ = 572KJ/mol) they've multiplied the heat of combustion of hydrogen gas by 2, why?

I believe it may be because the heat of combustion is for per mole of hydrogen gas, and in your equation you've got 2 moles oh H2 to 1 mol of oxygen. Therefore, the /\ is double that of what they provided you with. Does that make sense?

[warya]

A-  Adding water = dilution so it decreases concentration of everything. Usually this change will be partially offset by  the net reaction occurring in  the direction that produces the greater number of particles. But in this q we have the same no. of particles on either side.
B- pH decreasing means increasing [H+] and the concentration of H+ is actually decreasing so this cannot be right
C- because dilution = decreasing concentration of errything not increasing it. 
Pls correct me if I'm wrong cause I just finished learning this topic .
Hope this helps 

How come the concentration of H+ is decreasing if it isn't in the equation?

[BakedDwarf]

How come the concentration of H+ is decreasing if it isn't in the equation?

Increasing the amount of OH- consequently decreases the amount of H+

[paper-back]

I believe it may be because the heat of combustion is for per mole of hydrogen gas, and in your equation you've got 2 moles oh H2 to 1 mol of oxygen. Therefore, the /\ is double that of what they provided you with. Does that make sense?

But don't you need to balance the equation? Isn't the change in enthalpy applying to the entire combustion reaction?

[warya]

Increasing the amount of OH- consequently decreases the amount of H+

Except OH- has been decreased...

[knightrider]

For this picture attached.

How have they determined the right hand side of this equation?

[keltingmeith]

For this picture attached.

How have they determined the right hand side of this equation?

Probably by running an experiment.

[warya]

Could I have help with this question

OCl- + H2O--->HOCl + OH-  (at equilibrium)

100mL of pure water is added to a 100mL solution of 0.1M NaOCl, when the solution reaches equilibrium again, the
A) Concentration of H+ has decreased
B) pH of the solution has decreased
C) Concentration of HOCl has increased
D) the value of the equilibrium constant has halved

The answer is A. I understand how the concentration of OH- can be said to have decreased as the change is only partially offset and so  conc. at the new equilibrium is still lower than its initial value. But I don't get what this has to do with H+? Also, why are B and C wrong?

bump bump :/

[keltingmeith]

Could I have help with this question

OCl- + H2O--->HOCl + OH-  (at equilibrium)

100mL of pure water is added to a 100mL solution of 0.1M NaOCl, when the solution reaches equilibrium again, the
A) Concentration of H+ has decreased
B) pH of the solution has decreased
C) Concentration of HOCl has increased
D) the value of the equilibrium constant has halved

The answer is A. I understand how the concentration of OH- can be said to have decreased as the change is only partially offset and so  conc. at the new equilibrium is still lower than its initial value. But I don't get what this has to do with H+? Also, why are B and C wrong?

There are a couple of ways of thinking of this. I reckon the most straightforward is this:

By adding water, you do two things:
a) Decrease the concentration of everything,
b) Increase the volume.

Now, since the equation has 2 moles on either side, increasing the volume will not affect the equilibrium. However, note that water always has an effective concentration of 1 - this means that there's a larger decrease in concentration on the RHS of the reaction. So, the forwards reaction will increase. Now, remember that there was an initial dilution, so this means that [HOCl] doesn't necessarily increase, meaning that C isn't the answer.

However, we can say that there will be a greater production of hydroxide - and here's the kicker. In water, if you have hydroxide, you have protons (see the self-ionization of water, ). So, if you're producing more hydroxide, the equilibrium of the self-ionization of water is affected, and so this second equilibrium has a push for the back reaction. Realistically, with so much happening at once, we don't definitively know what's happened to the concentration of hydroxide - but, we can say with certainty that there is a decrease in [H+], since in all of this mess that's the only thing that doesn't both decrease and increase. Hence, the answer is A. This also knocks out B - because, once again, we cannot definitively say what happens to the pH.

EDIT: Just in case this scared you (because tbh, it scared me...), I'd definitely say this is a bit harder than what I'd expect from VCAA. If it does appear in a VCAA exam (or this did), it definitely wouldn't feature in an extended response question - there's too much going on.

[BakedDwarf]

Except OH- has been decreased...

Yes, it's concentration has decreased. But it's mole amounts has increased (I think)