VCE Chemistry Question Thread

[tigerclouds]

If you increase the temperature in an exothermic reaction, then the endothermic reaction will be favoured. This is because an exothermic reaction results in a net release of energy, whereas an endothermic reaction results in a net absorption of energy. When you increase temperature, you add energy, so the system will attempt to partially oppose this change by trying to remove this energy (which it will do by favouring the reverse endothermic reaction). This is why Kc changes. An increase in temperature in an exothermic reaction will favour the formation of the reactants, so Kc will decrease.

Oh ok, thank you whys! I didn't realise that it links with Le Chatelier's principle. So just to make sure I have it right:

When temperature increases, the Kc of an exothermic reaction will decrease because the system will counteract the change by absorbing heat to remove the heat from surroundings, therefore favouring the reverse reaction which is endothermic and less products are produced as a result.
On the other hand, the Kc of an endothermic reaction will increase because the system will still oppose the increase in temperature by absorbing the temperature but since the reaction is endothermic, it is already doing that and therefore does not need to favour the backwards reaction, so more products are formed. Is that all correct?

So does that mean that a system does not necessarily have to favour the reverse reaction if a change is made? Does it just depend on which direction is able to counteract the change?

[ErnieTheBirdi]

Hey, y'all I was just wondering whether someone would be able to help me find some information for my research investigation for Unit 1 Outcome 3? The topic I chose is "Is Alchemy Chemistry". Do you reckon its a good topic? im not too sure. Also If possible could someone give me some subheadings I would be able to do? Thanks in advance

[Mkiryo]

Hi,
I am a bit confused about why Kc can only be changed by temperature. Why can't it be changed by an increase in concentration?
I don't know if we're meant to know this, but could someone please help me understand this?

[whys]

Hi,
I am a bit confused about why Kc can only be changed by temperature. Why can't it be changed by an increase in concentration?
I don't know if we're meant to know this, but could someone please help me understand this?

When concentration is changed, the system will attempt to oppose this change by keeping Kc constant and changing the concentrations to achieve this. However, when the temperature is changed, then the reaction completely changes. E.g. if an exothermic reaction is heated, the endothermic reaction (reverse reaction) is favoured which completely changes Kc. Temperature is the only thing that can completely change Kc because the other factors still allow the system to bounce back and partially oppose the change.

[Mkiryo]

When concentration is changed, the system will attempt to oppose this change by keeping Kc constant and changing the concentrations to achieve this. However, when the temperature is changed, then the reaction completely changes. E.g. if an exothermic reaction is heated, the endothermic reaction (reverse reaction) is favoured which completely changes Kc. Temperature is the only thing that can completely change Kc because the other factors still allow the system to bounce back and partially oppose the change.

But doesn't the system also oppose a change in temperature in order to keep Kc constant? Why does favouring the reverse reaction completely change Kc? Also, when heat is applied to an endothermic reaction, the reverse reaction isn't favoured so why does that increase Kc?

[whys]

But doesn't the system also oppose a change in temperature in order to keep Kc constant? Why does favouring the reverse reaction completely change Kc? Also, when heat is applied to an endothermic reaction, the reverse reaction isn't favoured so why does that increase Kc?

Good questions! When temperature is added, the concentrations of the reactants/products are completely changed which changes Kc. This is because the system will oppose the change in temperature, NOT a change in concentration. On the other hand, all other factors only change concentration directly, and will work to oppose this change in concentration. Essentially, a system with changed temperature will attempt to oppose this change in temperature, whereas a system with any other changed factor will attempt to oppose this change in concentration. With your specific question, heating an endothermic reaction will cause a great increase in the formation of products as the forward reaction is favoured. This means Kc will become larger. Favouring the reverse reaction in the previous example I gave you would mean that the formation of reactants is greatly favoured, which means Kc will become smaller. A system opposes a change in temperature by changing the concentrations of reactants/products, which changes Kc because Kc is dependent on the concentration of the reactants/products.

For example, think about decreasing pressure. Decreasing pressure means there is an increase in volume, so the concentrations of all substances will decrease. The system will partially oppose this by favouring the side which produces the most particles. This does not change Kc because the system deals with the change by trying to make sure the concentrations remain in the same ratio as before. This is true for all factors excluding temperature.

[Mkiryo]

Good questions! When temperature is added, the concentrations of the reactants/products are completely changed which changes Kc. This is because the system will oppose the change in temperature, NOT a change in concentration. On the other hand, all other factors only change concentration directly, and will work to oppose this change in concentration. Essentially, a system with changed temperature will attempt to oppose this change in temperature, whereas a system with any other changed factor will attempt to oppose this change in concentration. With your specific question, heating an endothermic reaction will cause a great increase in the formation of products as the forward reaction is favoured. This means Kc will become larger. Favouring the reverse reaction in the previous example I gave you would mean that the formation of reactants is greatly favoured, which means Kc will become smaller. A system opposes a change in temperature by changing the concentrations of reactants/products, which changes Kc because Kc is dependent on the concentration of the reactants/products.

For example, think about decreasing pressure. Decreasing pressure means there is an increase in volume, so the concentrations of all substances will decrease. The system will partially oppose this by favouring the side which produces the most particles. This does not change Kc because the system deals with the change by trying to make sure the concentrations remain in the same ratio as before. This is true for all factors excluding temperature.

Oh. My. Goodness. You've clarified so much confusion, thank you SO much! So temperature is the only thing that changes Kc because it is trying to oppose the change in temperature only and as a side effect of that, concentration changes so Kc changes whereas with other factors, the system tries to bring back concentrations to what they initially were so Kc doesn't change as the ratio is the same! Makes sense. How does your brain work in this way? I've looked at so many resources online that couldn't answer these questions. You're amazing!
In saying that, I just want to clarify a little detail, because the system can only partially oppose a change, how can the ratio of the products and reactants return back to how they were originally?

[whys]

Oh. My. Goodness. You've clarified so much confusion, thank you SO much! So temperature is the only thing that changes Kc because it is trying to oppose the change in temperature only and as a side effect of that, concentration changes so Kc changes whereas with other factors, the system tries to bring back concentrations to what they initially were so Kc doesn't change as the ratio is the same! Makes sense. How does your brain work in this way? I've looked at so many resources online that couldn't answer these questions. You're amazing!
In saying that, I just want to clarify a little detail, because the system can only partially oppose a change, how can the ratio of the products and reactants return back to how they were originally?

Haha I’m flattered you think so! Im glad I could help. If you are willing to engage with the content, ask lots of questions (like you!) and look beyond the textbook, you will find that you will understand everything so much better! I also used to struggle with the whole only temperature changes Kc stuff.

I get what you mean with the partially opposes thing. My take on this is that the concentrations never do return to their original level (as seen on concentration vs time graphs) when a certain factor (other than temperature) is changed. However, since this is true for all the concentrations of all the reactants/products, the ratios will still remain relatively similar even if the concentrations may not be the exact same. This is because the reactants/products will all change concentration by the amounts dictated by their coefficients, so the original ratio of concentrations remains intact. I’m not entirely sure if this is 100% correct, but I can’t think of any alternative explanation. Perhaps someone with more knowledge in this topic could clarify. Hope this made sense!

[tigerclouds]

Hey guys, you know how as you go up an electrochemical series, the greater the tendency of the reaction to reduce and as you go down, oxidation is more likely to occur... does that flip around for electrolysis? As in, does oxidation increase going up and reduction increase going down for electrolytic cells?

[Snow Leopard]

Hi, I would really, really, really appreciate if someone could please answer my questions!

- What's oil drilling? (I understand why they use diamonds in it but I don't actually know what it is itself)
- With Question 1 (that I've attached) would I have to research it or would I be able to use the general info about diamonds to figure it out?
- With Q 1 c) (that I've attached), I got 0.001 micrometres as my answer by dividing 1nm by 1000 but I should've gotten 1000 as my answer
- Do we learn about addition reactions in alkynes in 3/4 Chem?
Thanks so much in advance!

[Mkiryo]

Hi guys, hope you're all doing well.
Could someone please help me understand the solution to this question? I don't understand the mole calculations in the third line because my understanding is that these are just finding the mole ratio which should be multiplied by another value and I also don't understand why the Faraday constant doesn't really play a role in the calculations.

[Comet striker]

Hi guys, hope you're all doing well.
Could someone please help me understand the solution to this question? I don't understand the mole calculations in the third line because my understanding is that these are just finding the mole ratio which should be multiplied by another value and I also don't understand why the Faraday constant doesn't really play a role in the calculations.

Question basically says Q = 96500
so 96500 = 96500*n(e)
n(e) = 1 mol
ratio of Cl2: e --> 1:2 so
n(Cl2) = .5 mol
then u can just find the volume using ideal gas formula

[Comet striker]

Hey there,
Question from textbook:
Lithium metal is prepared by electrolysis of a molten mixture of lithium chloride and potassium chloride.
a) Write a half-equation for the reaction occurring at the cathode in this cell
The answer uses Lithium for the reaction at the cathode, however in the presence of potassium chloride, isnt potassium a greater oxidant?

[jackmcgilligan]

Hey! Quick question, do we need to memorise the reactions at the anode/cathode for electrolysis? Or just the concepts relating to each? Thanks!

[Cassidyhogi]

Hey! Quick question, do we need to memorise the reactions at the anode/cathode for electrolysis? Or just the concepts relating to each? Thanks!

You had to remember the reactions before 2017. Now you just need to understand the concepts since in the exam you will be given the reactions to answer questions on . Sometimes you will be given the reactants and products and asked to write out the reactions yourself, but that should be doable, there is no need to memorise.