VCE Chemistry Question Thread

[cosine]

Can someone explain what equilibirum actually means? o.O Thanks

[jyodesh.com]

The rate of the forward reaction is equal to the rate of the reverse reaction

[RazzMeTazz]

When explaining Le Chatelier's principle is it correct to say that the reaction system partially opposes the physical change with a chemical change?

In answers explaining why equilibrium positions shift, is it necessary to make this distinction between physical and chemical changes?

Thanks

[RazzMeTazz]

Would it be correct to say that the ionisation constant of water applies to not only pure water but aqueous solutions as well, because if an acid or base is added to water to form an aqueous solution, the equilibrium position will shift accordingly, to maintain the same ionisation constant?

Also I don't quite understand how buffers work.
If you have this generalised reaction of a buffer in equilibrium: (where HA= weak acid and A- = conjugate base.)

HA + H20 <==> H30+ + A-

Why is it that if we add a base (e.g. OH- ions) it will react with the H30+ ions and not the HA itself?

Thanks!

[lzxnl]

All - consider setting a corner of a piece of paper on fire. As the fire spreads, surface area increases, and the rest of the paper starts to burn more quickly. Also, the more that's burning and the hotter the paper gets, the faster it'll continue to burn. This reaction is very obviously not in equilibria.

Not the best example; work out the Gibbs energy change for this reaction and use that to calculate an equilibrium constant for combustion. Go, I dare you

Can someone explain what equilibirum actually means? o.O Thanks

When a reaction and its reverse are in equilibrium, they occur at the same rate. There is no macroscopic change. I.e. we all know that water molecules are forming H+ and OH- very quickly but we don't see that happen.

Would it be correct to say that the ionisation constant of water applies to not only pure water but aqueous solutions as well, because if an acid or base is added to water to form an aqueous solution, the equilibrium position will shift accordingly, to maintain the same ionisation constant?

Also I don't quite understand how buffers work.
If you have this generalised reaction of a buffer in equilibrium: (where HA= weak acid and A- = conjugate base.)

HA + H20 <==> H30+ + A-

Why is it that if we add a base (e.g. OH- ions) it will react with the H30+ ions and not the HA itself?

Thanks!

No one said the ionisation constant of water only held for pure water. It holds as long as water is present in a liquid form. So if there's stuff dissolved, so be it.

As for how buffers work, this is a simpler way to put it. For a buffer to work, HA has to be a weak acid. Do you expect a base would react with a stronger or weaker acid faster?
In any case, if the base reacts with HA first, your equilibrium will shift to the left to use up some H3O+ anyway so the net effect is as if the base reacted with H3O+.

[cosine]

Not the best example; work out the Gibbs energy change for this reaction and use that to calculate an equilibrium constant for combustion. Go, I dare you

When a reaction and its reverse are in equilibrium, they occur at the same rate. There is no macroscopic change. I.e. we all know that water molecules are forming H+ and OH- very quickly but we don't see that happen.

No one said the ionisation constant of water only held for pure water. It holds as long as water is present in a liquid form. So if there's stuff dissolved, so be it.

As for how buffers work, this is a simpler way to put it. For a buffer to work, HA has to be a weak acid. Do you expect a base would react with a stronger or weaker acid faster?
In any case, if the base reacts with HA first, your equilibrium will shift to the left to use up some H3O+ anyway so the net effect is as if the base reacted with H3O+.

Cheers man

So can you say that the rate that the reactants are 'turning' into their products is at the same rate that the products are 'turning' back into their reactants?

[Redoxify]

Cheers man

So can you say that the rate that the reactants are 'turning' into their products is at the same rate that the products are 'turning' back into their reactants?

Yes, hence the reactants and the products are both present

[Sundal]

If I were performing this calculation:

Ka = [(10 ^ -5.2)^2] / (0.100)

Would my final answer need to be to 1 significant figure?

My reasoning:
Since the exponent has 1 decimal place, so evaluation of the numerator would have to be to 1 significant figure, and then since this 1significant figure number  is then divided by the denominator, the final answer would also have to be to 1 sig fig?

Cheers all

[anon9884]

Hey all,
What are some ways to check whether the actual concentration of ions in a solution has changed when the solution is heated/cooled? I was thinking about pH, but then i'm pretty sure an increase/decrease in pH can be caused by just temperature alone, and not the actual change in concentrations...

Thanks in advance !

[keltingmeith]

Hey all,
What are some ways to check whether the actual concentration of ions in a solution has changed when the solution is heated/cooled? I was thinking about pH, but then i'm pretty sure an increase/decrease in pH can be caused by just temperature alone, and not the actual change in concentrations...

Thanks in advance !

Lots of ways. AAS, colourimetry, UV-vis, and so on can all calculate new concentrations. If the particular metal in the solution is coloured, a more intense colour generally means a more concentrated solution (this is actually the principle behind UV-vis).

[anon9884]

Lots of ways. AAS, colourimetry, UV-vis, and so on can all calculate new concentrations. If the particular metal in the solution is coloured, a more intense colour generally means a more concentrated solution (this is actually the principle behind UV-vis).

Oh I see! Thanks so much!
Btw, could a titration method also be used?
 For example, a titration of the solution at room temperature, and then performing a second titration of the solution after it has been heated, and then comparing the amount of titre required for the two solutions, in order to deduce the concentrations. Would this also work?

[I am a unicorn]

Hi

The TSFX notes say that a limitation of the electrochemical series is that 'it is only applicable for reactions in aqueous solutions'.

Is this right? Because I've been taught to use the electrochemical series to predict reactions involving molten electrolytes as well...

Thank you

[cosine]

The formation of hydrogen iodide from it's elements is represented by the equation:

This endothermic reaction has an activation energy of 167kj mol and the heat of reaction is +28kj mol. What is the activation energy for the reverse reaction?

May someone provide me with an explanation, please?

[Splash-Tackle-Flail]

The formation of hydrogen iodide from it's elements is represented by the equation:

This endothermic reaction has an activation energy of 167kj mol and the heat of reaction is +28kj mol. What is the activation energy for the reverse reaction?

May someone provide me with an explanation, please?

Is the answer 139 kj mol^-1 (otherwise ignore fully)!

I think for these questions it really really really really really really (you get the point) helps to draw a energy profile of the situation- but my phones gone missing so I can't take a photo of the one I've drawn. Basically the forward reaction requires 167 kj to break the activation energy barrier of one mole of the reactants, and when the new bonds form, there is 28 kj mol more chemical energy in the resultant products. So the reverse reaction would require (167-28=139) kj mol to overcome its activation energy barrier (as there is 28kj mol more energy in it already).

[cosine]

Is the answer 139 kj mol^-1 (otherwise ignore fully)!

I think for these questions it really really really really really really (you get the point) helps to draw a energy profile of the situation- but my phones gone missing so I can't take a photo of the one I've drawn. Basically the forward reaction requires 167 kj to break the activation energy barrier of one mole of the reactants, and when the new bonds form, there is 28 kj mol more chemical energy in the resultant products. So the reverse reaction would require (167-28=139) kj mol to overcome its activation energy barrier (as there is 28kj mol more energy in it already).

Your theory sort of makes sense tbh.. But the answer adds them together not minus. Any thoughts about why? Cheers