VCE Chemistry Question Thread

[-_-zzz]

One more question - in the Heinemann Chem textbook, it says that the "dilution of an aqueous equilibrium system has no effect on the value of Kc for the reaction". I don't understand this - isn't the system shifting to the side of the reaction with the greater mol in order to increase the concentration of the solution (which reduced bc of the dilution). In that case, wouldn't Kc be affected since a shift has occurred and one side of the reaction is being favoured more than the other compared to before the dilution occurred? Thanks

Edit: Is it just temperature that affects the Kc value? If so, why? I thought that adding/removing reactants/products and changing pressure by changing volume would also change the Kc value since in both scenarios a shift has occurred.

In terms of the Kc value remaining constant upon dilution, adding/removing reactants/products or changing pressure, what you will find online is people explaining it in terms of mathematics and Le Chatlier's principle. However, if you want a comprehensive scientific answer as to why only temperature changes the Kc value I'm afraid that'll involve delving deep into kinetics and thermodyamics. So as of right now I wouldn't worry too much about the 'why' behind this concept

[dream chaser]

Hi Guys,

For the reaction: CH3OH(g) + HBr(g) -> CH3Br(g) + H2O(g)
                                                           <-
Considering the system at equilibrium, what would happen to the percentage yield of bromomethane when you are continuously removing the product CH3Br.

My understanding: I know that by according to Le Chatelier's principle, the system will partially oppose the effect of removing CH3Br by adding CH3Br. In doing so, the reaction would shift to the product side as the concentration of the products would increase and the concentration of the reactants would decrease. Hence, the position of equilibrium shifts to the right.

What I don't understand is even though CH3Br is being added due to Le Chaterlier's principle, its concentration would never reach the same as it was at first equilibrium. If that is the case, wouldn't you say the percentage yield of bromomethane would have decreased rather than increased?

Thanks. All help will be much appreciated.

[peachxmh]

In terms of the Kc value remaining constant upon dilution, adding/removing reactants/products or changing pressure, what you will find online is people explaining it in terms of mathematics and Le Chatlier's principle. However, if you want a comprehensive scientific answer as to why only temperature changes the Kc value I'm afraid that'll involve delving deep into kinetics and thermodyamics. So as of right now I wouldn't worry too much about the 'why' behind this concept

I've had a look online and I think I now understand why the Kc value remains constant when adding/removing reactants/products, but I still don't understand why it does so upon dilution or changing pressure.

So if you dilute or decrease the pressure of a system at equilibrium, the concentration of all the species decreases. But, according to Le Chatelier's principle, this means the system shifts towards the side of the reaction with the higher number of moles to increase the concentration. However, wouldn't this mean that whilst the side of the reaction with higher moles increases, the other side of the reaction (with lower moles) decreases as a result - so one part of the concentration fraction increases but the other part decreases even more... wouldn't the Kc value then change? (as both parts of the concentration fraction are not both increasing to try to get back to the original Kc value, despite both parts being decreased as a result of the initial dilution/decrease in pressure)

[xxxjss]

I've had a look online and I think I now understand why the Kc value remains constant when adding/removing reactants/products, but I still don't understand why it does so upon dilution or changing pressure.

So if you dilute or decrease the pressure of a system at equilibrium, the concentration of all the species decreases. But, according to Le Chatelier's principle, this means the system shifts towards the side of the reaction with the higher number of moles to increase the concentration. However, wouldn't this mean that whilst the side of the reaction with higher moles increases, the other side of the reaction (with lower moles) decreases as a result - so one part of the concentration fraction increases but the other part decreases even more... wouldn't the Kc value then change? (as both parts of the concentration fraction are not both increasing to try to get back to the original Kc value, despite both parts being decreased as a result of the initial dilution/decrease in pressure)

Here I would suggest remembering exactly what K_c is, that it is a fraction and a constant.  You can get K_c of 2 with a concentration fraction of 6/3 and 3/1.5. 3M and 1.5M are both lower but still yield the same number 2.

Yes ultimately dilution lowers the concentration of all species in a solution, and according to the classic Le chat  one part of the fraction will oppose this change by increasing and the other decreasing, but it does this to  establish, or even return to K_c not create a new Kc value. The underlying theory of Kc is that it is a constant at a particular temperature. Since diluting a solution doesn't change the temperature of the system it remains the same value
Straight from the OP heinemann textbook, I quote "For the general equation aW + bX cY + dZ at equilibrium at a particular
temperature, the equilibrium expression can be written as [insert equation] where Kc is the equilibrium constant."

This also explains why temperature does change the value of Kc as it is dependent on temperature 

Following this sweet knowledge, lets spice the math up and reinforce this concept with 3 species in a solution (1 product and 2 reactants) with still a K_c value of 2
X(aq) + Y(aq) <-> Z(aq)
Lets say [X] = 2M, [Y]= 1.5 and [Z]=6 in a 500mL solution
I dilute it by adding 500mL of water which halves the concentrations so now [X]=1. [Y]= 0.75 and [Z] = 3
Now the concentration fraction gives a value of 4, which is not equilibrium
Since Q_c>K_c it will favour a net reverse reaction which endorses Le chat's view of shifting towards the side with more particles.
So the concentration of Z will decrease and X and Y will increase.
Hypothetically [Z] will decrease from 3 to 2.5M and [X] will increase to 1.5M and [Y] will increase to roughly 0.833M which will yield a concentration ratio that is equal to equilibrium. this shows that despite the numerator decreasing and denominator of the fraction increasing it can still achieve the same value of equilibrium.

Hopefully this ungodly length of an answer helps 

[Rameen]

Hi!
Why does the shielding effect remain constant across the periods of the periodic table?

[-_-zzz]

Hi!
Why does the shielding effect remain constant across the periods of the periodic table?

Because the number of occupied electron shells which 'shield' the valence electrons from the attractive force of the positive nucleus remains constant.

[Rameen]

Because the number of occupied electron shells which 'shield' the valence electrons from the attractive force of the positive nucleus remains constant.

ok thank you!!
Am I required to memorise the polyatomic ions given in the textbook, for unit 1 chemistry AOS1?
I have a test on the topics: atomic structure, periodic table, metallic bonding, ionic bonding and quantifying chemistry when school starts.

[-_-zzz]

ok thank you!!
Am I required to memorise the polyatomic ions given in the textbook, for unit 1 chemistry AOS1?
I have a test on the topics: atomic structure, periodic table, metallic bonding, ionic bonding and quantifying chemistry when school starts.

Yes you will need to know basic polyatomic ions throughout chemistry as their formulae won't always be given to you in SACs/exams.

[zuijinde]

Hi there! I have a precipitation reaction that I noticed I had forgotten to balance when I wrote it out. However, upon closer inspection, I'm actually not sure about what the best way to go about balancing this would be.

You may have already picked where the imbalance is - we started with two sodium (Na) atoms, but for some bizarre reason, my instinct wrote what you see above. I was about to simply put a subscript 2 after the sodium in the precipitate (BaSO₄(s)) but we were just told that the golden rule is to put any balancing number before the compound. I think I've confused myself. Anyone mind un-confusing me?
Thank you

You got the Barium sulfate correct, however the other compound is wrong

The Na has a +1 charge, and since it's Na₂ (or 2 mols of Na), the overall charge is +2. Same deal with NO_3^-

Therefore, the overall equation should be Ba(No3)2(aq) + Na2So4(aq) = BaSo4(s) + 2 NaNo3(aq)

[zuijinde]

Thank you for the answer! I'm really, really stuck on this for some reason.
Are there rules for balancing these kinds of equations? I just can't seem to get it.

Not that I know of.

Write out the compounds into its constituent ions, that way it's easier to see the charges and whether they need balancing or not.

[-_-zzz]

Thank you for the answer! I'm really, really stuck on this for some reason.
Are there rules for balancing these kinds of equations? I just can't seem to get it.

There is really no shortcut to balancing these types of equations. After a decent amount of practice it should just become second nature to you. That being said I wouldn't worry too much about balancing complex equations (such as those for precipitation) as they don't really come up in 3/4.

[yplee0926]

hey guys!
In galvanic cells, do the half cells have to contain redox conjugate pairs?
I was thinkinf about a cell (that I made up) where one half cell has Pt electrode with Cu2+ solution, and another half cell with Mg electrode and Mg2+ solution. Would the reaction still proceed to have Cu deposit on Pt electrode? I’m just confused whether half cells with conjugate redox pairs are the only ones in the textbook because they’re conventional and common or because they’re the only possible combo.

[-_-zzz]

hey guys!
In galvanic cells, do the half cells have to contain redox conjugate pairs?
I was thinkinf about a cell (that I made up) where one half cell has Pt electrode with Cu2+ solution, and another half cell with Mg electrode and Mg2+ solution. Would the reaction still proceed to have Cu deposit on Pt electrode? I’m just confused whether half cells with conjugate redox pairs are the only ones in the textbook because they’re conventional and common or because they’re the only possible combo.

They most certainly don't need to. Perhaps the reason why we often use conjugate redox pairs is because that's what the ECS is based off and thus they allow us to make predicitions about particular cell reactions. Hope that helps.

[yplee0926]

They most certainly don't need to. Perhaps the reason why we often use conjugate redox pairs is because that's what the ECS is based off and thus they allow us to make predicitions about particular cell reactions. Hope that helps.

Thanks for the response! Would it matter though in the cell that I proposed? The reaction would still be quite predictable usinf ECS (from my knowledge)

[colline]

Hey everyone! Got two questions regarding significant figures:

1. If I'm using data from the data book, and the sig figs are lower than the ones in the question, which one should I follow?
(The question I'm on has 3 sig figs as the lowest number, but I'm also required to use the molar mass of hydrogen which is given in the data book as 1.0 which is 2 sig figs)

2. Can someone confirm if this is correct: (this is part of the working out for a larger question)
There's been a temperature change from 18.5 to 13.3 degrees Celsius. So delta T is 5.2 (one decimal place). The smallest number of significant figures given in the question is 3, but 5.2 has two significant figures. I'm now required to leave my answers to two significant figures instead of three because that's the lowest number of s.f in the working out, even if the question had 3 s.f.
Is this correct? This was a question on a practice SAC and I left my answer to 3 s.f., and lost a mark.

Thanks in advance!