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October 30, 2020, 06:15:03 pm

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#### Harrycc3000

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##### Re: VCE Chemistry Question Thread
« Reply #8790 on: September 20, 2020, 12:53:18 pm »
+1
Hi guys this is my first ATAR notes post so apologies if I do anything wrong in terms of etiquette.

1. Why don't catalysts influence equilibrium when temperature does. Even though people say catalysts lowers the activation energy equally doesn't a change in temperature also affect the proportion of successful collisions equally too since exothermic reaction still need an input of activation energy to undergo their reactions.
If it is because that a higher proportion of the reactants undergoing the endothermic reaction gain enough kinetic energy to undergo the reaction than the reactants on the exothermic side since it is likely that the exothermic have already reached the required energy due to lower activation energy then doesn't the same principle apply for catalysts? where decreasing the activation equally still proportionately results in more endothermic reactants gaining the required energy than the exothermic reactants.
2. Why does la chateliers principle hold? The Heinemann textbook and online resources just say what it is but its kind of like they want us to blindly follow this principle of 'opposing the change' without actually knowing why.

Thanks a lot appreciate any replies.
VCE 2020: Biology
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#### keltingmeith

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##### Re: VCE Chemistry Question Thread
« Reply #8791 on: September 20, 2020, 07:16:52 pm »
+9
Hi guys this is my first ATAR notes post so apologies if I do anything wrong in terms of etiquette.

No stress - etiquette here is just like etiquette everywhere Don't be rude, use your pleases and thank yous, and we'll all get along swimmingly! (it's also worth upvoting posts you find helpful - basically counts as a thank you in and of itself. Only other special thing is remember that people are voluntarily answering your questions - we're not paid or forced to do this, we do it purely for giddy feeling of helping others. )

1. Why don't catalysts influence equilibrium when temperature does. Even though people say catalysts lowers the activation energy equally doesn't a change in temperature also affect the proportion of successful collisions equally too since exothermic reaction still need an input of activation energy to undergo their reactions.
If it is because that a higher proportion of the reactants undergoing the endothermic reaction gain enough kinetic energy to undergo the reaction than the reactants on the exothermic side since it is likely that the exothermic have already reached the required energy due to lower activation energy then doesn't the same principle apply for catalysts? where decreasing the activation equally still proportionately results in more endothermic reactants gaining the required energy than the exothermic reactants.
2. Why does la chateliers principle hold? The Heinemann textbook and online resources just say what it is but its kind of like they want us to blindly follow this principle of 'opposing the change' without actually knowing why.

Thanks a lot appreciate any replies.

I'm actually going to start by answering 2, because it's easier that way. Let's work with the simple chemical equation:

$A\leftrightharpoons B, K=\frac{[B]}{[A]}$

So, we have some chemical, A, and it turns into another chemical, B. For simplicity let's assume this equation is at equilibrium, with concentrations of 0.5 M and 0.5 M - which means our equilibrium constant is K=1. For this explanation to work, you need to accept that all reactions have some K value they want to return to. The theoretical explanation for as to why requires you understand chemical kinetics and how reaction rates are calculated - if you think this is something you want to know for understanding purposes, I can do so, but it is not required at all. Instead, just know that you can prove experimentally that, over time, the backwards reaction rate will eventually equal the forwards reaction rate, and that point happens to be at equilibrium - when K equals some special number.

So, we've accepted that this equation is going to constantly change such that K=1. Well, I'm going to add enough of B to make the concentration of B now 1.0 M. Our new Q value (that is, the reaction quotient when the reaction is NOT at equilibrium) is:

$Q=\frac{1}{0.5}=2$

Well, this is far bigger than 1! So, we want to decrease the value of Q - how do we decrease a number? We can either lower the numerator, or raise the denominator. So, I (the reaction) am going to lower the numerator by getting rid of B - but to get rid of B, I -HAVE- to make more A, because that's the only way to remove B from the equation as a chemical reaction. So, if I reduce the concentration of B to 0.9 M, then the concentration of A must also raise by 0.1 M, and so now Q is:

$Q = \frac{0.9}{0.6}=1.5$

Well, that's a little better! But we need to remove more of B. Now, the reaction will never make so much A all at once such that we have MORE A than B, because it will reach equilibrium first. But, for the sake of learning, let's pretend that the reaction removes 0.2 M of B (again, by reacting B, so now we have another 0.2 M of A), this would now give us a Q value of:

$Q=\frac{0.7}{0.8}=0.875$

Well, this is now LOWER than 1! Okay, so how do we increase a number? Raise the numerator, or reduce the denominator. So, let's reduce B by making more A... And we can do this forever, but eventually (and you may have guessed this) we're going to end up with 0.75 M of A and B, and we'll be back at equilibrium.

Okay, so what does this have to do with le Chatelier? Well, notice that by increasing the products, to get back to equilibrium, we had to "partially oppose" the reaction by making more reactants - and then, when we had more reactants, we had to "partially oppose" the reaction by making more products. All to get back to that magic K value. THAT is why le Chatelier's principle works - all it does is explain what you have to do to the chemicals to make the maths work. You can through similar explanations for everything else, not just adding more product/reactants, to explain changes in pressure, volume, etc. Let me know if this explanation didn't work for you (/what bits don't work), and I can try to explain in more detail.

So, why don't catalysts affect the constant? Well, look at the above - at no point did the rate of the reaction affect affect the maths above. The rate will certainly affect how fast the reaction reaches equilibrium, but a catalyst will increase the backwards reaction AS WELL AS the forwards reaction. So, why would the catalyst affect the equilibrium constant? It shouldn't at all!

Okay, so if catalysts don't affect equilibrium, why does temperature? If rate doesn't affect equilibrium, what does temperature do that ISN'T change the rate of reaction? Well, there's the basic explanation of, "well, if you imagine that heat is a product or reactant, then in an exothermic reaction, heat is a product, and in an endothermic reaction, heat is a reactant" - which, if that works for you, is great. No need to complicate things unnecessarily. But, if it doesn't work for you, you're right in being a little bit hesitant - heat isn't a chemical species, it's not in our equilibrium equation, so why does heat play a part?

Well, this is beyond VCE, but it all comes down to thermodynamics. Let's call the amount of energy in a system "G" (for "Gibbs' free energy". The reason is because some guy with the last name Gibbs came up with this stuff). The value of delta G depends on two different types of energy - the amount of energy from enthalpy, or heat, of the chemicals in question (the delta H value), as well as the amount of energy from the way in which the molecules arrange themselves (known as entropy, or the delta S value). Well, the Gibbs' free energy can be calculated like so:

$\Delta G=\Delta H-T\Delta S$

Notice that delta G changes with temperature? Well, this isn't the ONLY way to calculate delta G. We can also use this equation:

$\Delta G=-RT\ln K$

You may not know what ln is - it's just a logarithm. The logarithms you're familiar with from calculating pH are logarithms to the base 10, this one is to the base e, which is approximately the base 2.7. Don't think too hard about it - it's just a fancy maths thing. All of these you already know - R is just the gas constant, T is temperature, K is the equilibrium constant.

So, from this equation, it isn't obvious that temperature also affects the equilibrium. But, what if you substitute the two above equations into each other? You get:

$\Delta H-T\Delta S=-RT\ln K\\
\ln K=-\frac{\Delta H}{RT}+\frac{\Delta S}{R}$

So from this equation, you can see that K depends on temperature! It's really obvious now - if you increase temperature, then K HAS to change - but whether it goes up or down isn't quite obvious... So, let's work at this step-by-step so focusing on the delta H bit. If T increases, then:

$A=\frac{\Delta H}{RT}$

will get smaller. If T decreases, then A will get bigger. Okay, what about the following equation:

$B=-A=-\frac{\Delta H}{RT}$

Well, now we need to know what delta H does. Firstly, if B is positive (and A is negative), then the equilibrium constant will get bigger - otherwise, it will get smaller. This is because:

$\ln K=-\frac{\Delta H}{RT}+\frac{\Delta S}{R}=B+\frac{\Delta S}{R}$

So, let's say we have an endothermic reaction. Then, if you increase the temperature, A decreases, which means B will get LESS negative and MORE positive - so decreasing the temperature in an endothermic reaction causes the value of K to get bigger, so we need to make more reactants to make Q smaller. If you decrease the temperature, A increases, which means B will get MORE negative and LESS positive - so increasing the temperature in an endothermic reactions causes the value of K to get smaller, so we need to make more reactants to make Q bigger.

Exothermic reactions obey the same logic - however, since there's a negative sign outside the delta H, the two cancel out, and so increasing temperature will still cause B to get closer to 0 - but now it will become more negative instead of less negative.

And if I lost you in all that maths, all you need to know is this: temperature changes because it affects the amount of energy that single chemicals hold, it doesn't JUST change how fast they move around. (that is, it increases energy that isn't just the kinetic energy) A catalyst doesn't affect energy at all, and so the equilibrium constant doesn't change - in fact, a catalyst doesn't change the kinetic energy. In hindsight, maybe all I needed to explain to you was that a catalyst doesn't affect the kinetic energy of a system, just the pathway a system takes to get to the products, whereas temperature changes do change the energy, and the equilibrium constant is negatively proportional to energy... But oh well, you got the full thermodynamics explanation as well

So, the elephant in the room - why should you trust those equations? Well, that requires a full-on course in thermodynamics, so I can't exactly explain that to you in a short period of time... So, you kinda just have to trust those equations are true. If that explanation isn't good enough for you, then I have good news - you'd make a great scientist! Google is your friend to learn more.
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#### Harrycc3000

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##### Re: VCE Chemistry Question Thread
« Reply #8792 on: September 21, 2020, 01:41:15 am »
0
Thank you so much for the explanation I really didn't expect an answer this detailed and in depth. Probably going to use atarnotes more because this was really useful  .
Okay, so what does this have to do with le Chatelier? Well, notice that by increasing the products, to get back to equilibrium, we had to "partially oppose" the reaction by making more reactants - and then, when we had more reactants, we had to "partially oppose" the reaction by making more products. All to get back to that magic K value. THAT is why le Chatelier's principle works - all it does is explain what you have to do to the chemicals to make the maths work. You can through similar explanations for everything else, not just adding more product/reactants, to explain changes in pressure, volume, etc. Let me know if this explanation didn't work for you (/what bits don't work), and I can try to explain in more detail.
So essentially we keep Kc constant and simply due to the mathematics and because Kc needs to be this constant number we oppose any change that occurs to this constant? It does make sense and I'm assuming the same works for pressure because doubling the pressure is essentially increasing the concentration of both the reactants and products by 2. Also is it 'partially' oppose because once you've increased or decreased a certain concentration its mathematically impossible to reach the same constant with the initial 'numerator'/product? Thanks the answer really helped
Well, this is beyond VCE, but it all comes down to thermodynamics. Let's call the amount of energy in a system "G" (for "Gibbs' free energy". The reason is because some guy with the last name Gibbs came up with this stuff). The value of delta G depends on two different types of energy - the amount of energy from enthalpy, or heat, of the chemicals in question (the delta H value), as well as the amount of energy from the way in which the molecules arrange themselves (known as entropy, or the delta S value). Well, the Gibbs' free energy can be calculated like so:
This answer was really complicated and it has taken me a while to kind of understand it but I'll try and explain what I think you meant by this answer.

I did a bit of google searching and for the formula that you gave about gibbs energy being proportional to the ln (k), I found out that it was derived from the formula here:ΔG = ΔG° + RT ln Q
I think that this formula makes intuitive sense because if you increase the equilibrium quotient it means that the reaction favors the forward reaction more at the given temperature (ΔG) than at standard conditions (ΔG°) which implies that the reaction is more spontaneous and therefore has a higher free energy difference. Vice versa would occur for a lower equilibrium quotient and the same applies for temperature where increasing the temperature increases G for a constant Q (though at this moment the other formula with enthalpy and entropy with temperature makes more sense).
And yep I got to your formula by making ΔG at equilibrium thereby making it equal to zero and then resulting in your formula.

I got to your formula and I simplified it significantly to e^-H/T (is proportional to) K so I could visualise it more easily. and yeah it makes sense now because if it was exothermic then H would be negative and therefore overall e would be positive and if you wanted to make K larger then you'd reduce T which it is inversely proportional. If it was endothermic then H would be positive making e's exponent negative and to minimise the 'negativity' as you said, you can increase the temperature.

Another way I looked at it is for G=-RTln(k) G has to be constant because its the free energy under certain conditions, R has to be constant because its a constant number but T and ln(K) are proportional to each other in that if you increase the temperature, then ln(K) has to be a smaller number and vice versa. And if G was a positive value (implying endothermic if entropy equals 0) that means ln k has to be negative and if you increased the temperature then ln k would proportionally become 'less negative' and more positive and the opposite would occur with an exothermic reaction where instead ln k would have to be positive and increasing temperature would then lead to ln k being a smaller value.

Also I just realised that because the formula has H automatically negative, does that mean endothermic reactions generally have a lower equilibrium constant than exothermic reactions?

Was a huge effort to kind of understand this and idek if it will even help me on the chem exam next year but if it doesn't help me on it it's definitely taught me a lot about how all this equilibrium stuff works and I'm really grateful for that. Thanks a lot!

(oops i didn't read your last paragraph) by 'temperature changes the amount of energy chemicals hold' do you mean that a higher temperature means that chemicals have more energy and therefore are closer to their activation energy? Or do you mean that the temperature affects the way chemicals absorb energy and that somehow affects the tendency for the forward reaction to occur/the equilibrium constant. If its the first reason its still a bit confusing to me because I don't understand the difference between certain reactants gaining energy and the lowering of a reactions activation energy, To me it seems like they are just ways to increase the rate of reaction and I guess I'd need an explanation that linked the energy to the K value (thermodynamic explanation may have been the explanation I needed). Btw I watched a khan academy video and it said that the equilibrium constant was equal to the probability that the forward reaction would occur given that the required reactants were in a close enough proximity to react over the same but for the backward reaction (paraphrased) I was wondering if you were ever taught that definition and if you were, if it related to how kinetic energy changes the constant while the catalyst doesn't? Apologies if I asked too many questions usually when I don't understand something I become a bit obsessive about it.

« Last Edit: September 21, 2020, 02:12:31 am by Harrycc3000 »
VCE 2020: Biology
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#### keltingmeith

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##### Re: VCE Chemistry Question Thread
« Reply #8793 on: September 21, 2020, 04:09:29 am »
+6
Holy crap, you're not scared by maths, but also happy to look up more maths to explain things?? This is such a refreshing take, most chemistry students are scared, and run from, maths. (nothing against you if you don't like maths and still do chem - you can still succeed in chemistry and even be a scientist doing research without being good at maths. But when it comes to this field, it's just really nice to be able to explain things from first principles)

I also want to add for any readers that are bad at maths (and a little for you, Harry, in regards to your comment on if this is necessary or not) - no, you don't need to know any of this to apply the knowledge required in VCE. You also don't need to understand this to have an intuitive understanding of the material. This is all extra, bonus stuff.

Thank you so much for the explanation I really didn't expect an answer this detailed and in depth. Probably going to use atarnotes more because this was really useful  . So essentially we keep Kc constant and simply due to the mathematics and because Kc needs to be this constant number we oppose any change that occurs to this constant? It does make sense and I'm assuming the same works for pressure because doubling the pressure is essentially increasing the concentration of both the reactants and products by 2. Also is it 'partially' oppose because once you've increased or decreased a certain concentration its mathematically impossible to reach the same constant with the initial 'numerator'/product? Thanks the answer really helped

Exactly - pressure is basically concentration if temperature doesn't change (just like concentration is basically moles if volume doesn't change, if you noticed that I was using concentration that way lmao). The reason it's "partially" oppose is because "directly" oppose would be to just remove what I added. Eg, add 5 moles of B, then just remove 5 moles of B without changing A would be directly oppose. But this isn't physically possible, so we "partially" oppose by slightly changing B while changing A. I think this is what you've said, just that you've phrased it more mathematically?

This answer was really complicated and it has taken me a while to kind of understand it but I'll try and explain what I think you meant by this answer.

I did a bit of google searching and for the formula that you gave about gibbs energy being proportional to the ln (k), I found out that it was derived from the formula here:ΔG = ΔG° + RT ln Q

I'm gonna be honest, the history of which one was derived first is beyond me - I've always been taught that this form is derived from the one I gave you (in which RT ln Q is a correction factor that allows us to work in non-standard conditions). But whatever way makes sense for you is worth following - thermodynamics is confusing at the best of times lmfao

I think that this formula makes intuitive sense because if you increase the equilibrium quotient it means that the reaction favors the forward reaction more at the given temperature (ΔG) than at standard conditions (ΔG°) which implies that the reaction is more spontaneous and therefore has a higher free energy difference. Vice versa would occur for a lower equilibrium quotient and the same applies for temperature where increasing the temperature increases G for a constant Q (though at this moment the other formula with enthalpy and entropy with temperature makes more sense).
And yep I got to your formula by making ΔG at equilibrium thereby making it equal to zero and then resulting in your formula.

Yeah, this basically sounds correct.

I got to your formula and I simplified it significantly to e^-H/T (is proportional to) K so I could visualise it more easily. and yeah it makes sense now because if it was exothermic then H would be negative and therefore overall e would be positive and if you wanted to make K larger then you'd reduce T which it is inversely proportional. If it was endothermic then H would be positive making e's exponent negative and to minimise the 'negativity' as you said, you can increase the temperature.

Just as correct as what I did, of course, but if this makes more sense to you, then power to you - education is all about finding the explanations that work for you, there are usually multiple interpretations that are all correct.

Another way I looked at it is for G=-RTln(k) G has to be constant because its the free energy under certain conditions, R has to be constant because its a constant number but T and ln(K) are proportional to each other in that if you increase the temperature, then ln(K) has to be a smaller number and vice versa. And if G was a positive value (implying endothermic if entropy equals 0) that means ln k has to be negative and if you increased the temperature then ln k would proportionally become 'less negative' and more positive and the opposite would occur with an exothermic reaction where instead ln k would have to be positive and increasing temperature would then lead to ln k being a smaller value.

More great logic! It might seem weird I didn't go for this simplistic explanation, but you can arrive at it because I explained what Gibbs free energy and entropy is, and you can account for them in your own mind - I didn't want to be hiding truths from you because that just raises more questions, like "but what does G have to do with any of this, and how can energy be constant?" Tbh I'm also slightly hesitant to say that delta G has to be constant here, because it doesn't necessarily have to be, but if are very careful about how you increase temperature (eg, making sure pressure doesn't change), then you are right that it shouldn't change.

Also I just realised that because the formula has H automatically negative, does that mean endothermic reactions generally have a lower equilibrium constant than exothermic reactions?

I would not say this at all, no, and tbh in the real-world, maximising endothermic reactions is actually really easy to do. In the real world, usually, we want a reaction to make the most amount of product as quickly as possible. Increasing rate is easy - just increase the temperature, right? Hell, industry isn't scared of thousand degree reactions, because all they need to do is just make sure the reaction vessel is insulated, then nobody gets hurt. But, if your reaction is exothermic, this will decrease yield (as per our discussion - in an exothermic reaction, the equilibrium constant decreases as temperature increases), so they need to find a way to increase rate without increasing temperature, as well as finding the highest temperature where they still get "good enough" yield from the reaction. This isn't an issue with an endothermic reaction - if a reaction is endothermic, you can increase the temperature as much as you want, because not only will you increase the rate, but the yield will go up, too!

Was a huge effort to kind of understand this and idek if it will even help me on the chem exam next year but if it doesn't help me on it it's definitely taught me a lot about how all this equilibrium stuff works and I'm really grateful for that. Thanks a lot!

I mean look, all this has done has allowed you to make sense of these new ideas in a way that is more satisfying than just me telling you, "it's complicated and you don't need to know it". If you're only interested in learning things that will help you in VCE, then I can answer that way for you - but if you're asking this because you want a proper, intuitive understanding of everything, then I love explaining that kind of stuff and am happy to help.

And if nothing else, it's given you a head-start of university chemistry if you end up taking it - this is like, 60% of what you learn in first year uni chem thermodynamics.

(oops i didn't read your last paragraph) by 'temperature changes the amount of energy chemicals hold' do you mean that a higher temperature means that chemicals have more energy and therefore are closer to their activation energy? Or do you mean that the temperature affects the way chemicals absorb energy and that somehow affects the tendency for the forward reaction to occur/the equilibrium constant. If its the first reason its still a bit confusing to me because I don't understand the difference between certain reactants gaining energy and the lowering of a reactions activation energy, To me it seems like they are just ways to increase the rate of reaction and I guess I'd need an explanation that linked the energy to the K value (thermodynamic explanation may have been the explanation I needed). Btw I watched a khan academy video and it said that the equilibrium constant was equal to the probability that the forward reaction would occur given that the required reactants were in a close enough proximity to react over the same but for the backward reaction (paraphrased) I was wondering if you were ever taught that definition and if you were, if it related to how kinetic energy changes the constant while the catalyst doesn't? Apologies if I asked too many questions usually when I don't understand something I become a bit obsessive about it.

Yeah, so "amount of energy chemicals hold" is a very flippant way of describing Gibbs' free energy. Gibbs' free energy is just a culmination of the amount of energy held in its bonds (enthalpy), as well as the amount of different ways it can orient itself (entropy), so in some sense you can think of it as just the "amount of energy the chemical has". As we know, heat won't affect the enthalpy factor, but it WILL affect the entropy factor, hence why I said that energy can change with temperature. The final paragraph is just me bastardising all the complicated stuff you just managed to understand all by yourself - so you don't need to pay it much mind.

I mean, don't know how I feel about that Khan Academy definition - not something I've heard before, and I don't know if I put much stock in it. Even if true, I don't think it's an interpretation that is beneficial to answering any questions you'll come across in VCE /anyway/. Likely not a line I'd think too much about, though I am interested in hearing the exact line now so I can hear the context and try to see what ideas they're getting across.
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#### Coolgalbornin03Lo

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##### Re: VCE Chemistry Question Thread
« Reply #8794 on: September 21, 2020, 09:01:56 pm »
0
Is it likely the acid-base reaction stuff (beyond acid+base= salter+water) will be examined.

I’m asking because in 2009 VCE Chem exam it says we have to write

NH4+ + H20—-> H3O+ and NH3+

Now I know this isn’t in 3/4 on the study design but I learnt in units 1/2
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#### keltingmeith

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##### Re: VCE Chemistry Question Thread
« Reply #8795 on: September 22, 2020, 07:15:50 am »
+2
Is it likely the acid-base reaction stuff (beyond acid+base= salter+water) will be examined.

I’m asking because in 2009 VCE Chem exam it says we have to write

NH4+ + H20—-> H3O+ and NH3+

Now I know this isn’t in 3/4 on the study design but I learnt in units 1/2

The VCAA have gone on record saying that you can't be tested on empirical formulas because it's a unit 1 skill. Presumably, the same holds in this case - because acid/base calculations are not taught in units 3/4, these questions are not assessable.
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#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #8796 on: September 22, 2020, 09:38:35 pm »
0
Hello
i need help with this question

would this be right

#### keltingmeith

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##### Re: VCE Chemistry Question Thread
« Reply #8797 on: September 22, 2020, 11:03:39 pm »
+1
Hello
i need help with this question

would this be right

It may be worth asking friends if they answered the same as you, but anyway:

a. (i), correct, incorrect - both should increase. Remember that at equilibrium, the rates of both will be the same - so if one increases, they both do. If one decreases, they both do.
a. (ii), CO will increase, O2 will decrease (remember, it will get consumed), CO2 will increase
a. (iii), correct

b. (i) correct
b. (ii) correct
b. (iii) correct

c. (i) again - both will increase. We're talking about rates here, not concentrations.
c. (ii) CO and O2 we actually don't have enough information - decreasing volume will increase concentration, but equilibrium will shift to the right, so the concentration will then decrease, CO2 will definitely increase.
c. (iii) Equilibrium will shift to the right, not the left. Decrease in volume = increase in pressure, so equilibrium shifts to make less particles. There are less particles on the right (products), so we shift to the right

Someone may need to double check me, that was a lot to go through at 11 pm lol, but I think that's all good.
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#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #8798 on: September 22, 2020, 11:30:51 pm »
0
Thank you!

#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #8799 on: September 23, 2020, 02:59:29 pm »
0
Hello

would this be right
Thanks

#### tigerclouds

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##### Re: VCE Chemistry Question Thread
« Reply #8800 on: September 23, 2020, 04:50:28 pm »
0
Under the current study design, are we expected to know how to calculate the pH of a solution (using acidity constant calculations?)
Also, do we have to know how to find the percentage ionisation of a solution?
« Last Edit: September 23, 2020, 04:54:38 pm by tigerclouds »

#### keltingmeith

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##### Re: VCE Chemistry Question Thread
« Reply #8801 on: September 23, 2020, 05:38:01 pm »
+4
Hello

would this be right
Thanks

Looks fine to me.

Under the current study design, are we expected to know how to calculate the pH of a solution (using acidity constant calculations?)
Also, do we have to know how to find the percentage ionisation of a solution?

You do not need to know how to answer pH calculation questions. This question is getting asked so often tho, I should put it in my sig

Can you give context for when you've been asked to calculate percentage ionisation? Because that could mean a few different things
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#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #8802 on: September 23, 2020, 05:42:28 pm »
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i need help with this question

and is this reaction exothermic or endothermic because i thought it was exothermic

Thanks!

#### Unknown-111

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##### Re: VCE Chemistry Question Thread
« Reply #8803 on: September 23, 2020, 06:32:41 pm »
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Hey guys I need help with redox reaction questions, I still find it quite hard.

#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #8804 on: September 23, 2020, 07:11:35 pm »
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i need help urgently. pls

how do you work out solution 5 has the highest pH