there is shortcut way if you know how to subtract the units with the implied 1 after mol. if you have 5 mols of product minus 3 moles of reactants for example, the units would be mol 2 dm -6. Sorry dont know how to type subscripts. Just use the two exponent from 5-3, put it outside the brackets and work it into the units i.e, 2( mol dm -3).
Hi Sir, is it necessary for those doing AQA A level chemistry to memorise the details of the experiment to deduce Kc or had you used it as an example to show how we would go about it? As I'm unsure whether to add it to my notes or not as I don't see a mention of it in the specification, but know sometimes they can throw in application questions such as this. Thanks in advance!
You won't be asked to describe it, so not from that point of view. But it's a type of exam question you can get that nicely weaves together Esters, Kc and amount of Substance. It's also an interesting situation with the units cancelling.
Ahh okay, I see thank you! They really do like throwing in a few questions that tie in bits of other topics so I could definitely see it being a question haha thank you so much
Hello, it is a very informative video! I was wondering what does mol -1 dm 3 mean in terms of solubility/concentration since mol 1 dm-3 means 'this many x mols per (1)litre or dm3 ' ; how can the reverse still express the concentration of a solution? Thank you!
Kinetics and equilibrium constants are often out of the scope of a physical/chemical understanding of units. Most of the rest of the time, the units are meaningful as it represents a concept easily understandable by the brain. However, in the case of kinetic constants, the problem is inverted. In fact, you adapt the unit to match what the equation needs, in order to have the same unit on both sides of the "equal" sign. It is probably due to the fact that all the kinetic equations have different forms, and is almost always obtained after quite a lot of simplifications. Units don't have to be physically meaningful, they just have to work. The dimensional analysis of whatever problem you are attacking forces the units to be what they are. This is why a rate constant on its own doesn't give you a clue of the speed of the reaction. It is useful when you got to compare some of them (IF they are comparable).
At 25.08 why is the 4 squared and not just times two because there are two moles of Y? 😅😅😅 if there was 3 moles what would I do? Times it by 3 or do 3 cubed 😢
In a Kc expression, the coefficients from the balanced equation always become powers. So whatever the number that's the power you raise the concentration to. Think of it like this. If the equation said 3HF, then the Kc expression would contain the term [HF] x [HF] x [HF] Which we write as [HF]^3
The experiment uses algebra to deduce X moles of something used and made. Then you have to work out X from the Kc expression. You won't need a quadratic to solve it. For the factors affecting Kc, really you only need to know that it's just the temperature that affects Kc. For an increase in concentration of reactant, you'd expect Kc to go down as reactant concentration is on the bottom. However the equilibrium shifts right making more product and using up reactant. This means that the effects cancel out and no change is observed
I don't understand. If we increase temperature the enthalpy change is endothermic therefore shifting to the left: Reactants----》products But if the forward reaction is exothermic then we have to decrease temperature to shift equilibrium to the reacntans? Why???
Don't get too bogged down by what direction is which exo/endo as that can change. You need to know that there is an optimum temperature (and pressure and concentration ratio) and at this point the chemicals have the least energy and are the most stable. If we raise the temperature, the chemicals become less stable and so the reaction that reduces the temperature (endothermic reaction) speeds up briefly and becomes faster than the other reaction. This continues until the temperature returns to the original, after which both reactions happen at the same rate again
your chemistry videos are amazing and getting me through the AS module, thank you so much !!!!
31:50 thats the most perfect way of explaining why pressure doesn't affect Kc, thank you so much
your explanation of why Kc isn't affected by conc. is soo good. I was just thinking about it and you answered it perfectly. thanks!
Excellent! It's great to know that I made it clear 😃
there is shortcut way if you know how to subtract the units with the implied 1 after mol. if you have 5 mols of product minus 3 moles of reactants for example, the units would be mol 2 dm -6. Sorry dont know how to type subscripts. Just use the two exponent from 5-3, put it outside the brackets and work it into the units i.e, 2( mol dm -3).
Absolutely right! Very handy 👌
thank you so much this video was so helpful
Hi! your video really helped me understand this for my exam.
i was wondering if you have made a video on Kp too?
Hi, I'm really glad the video was useful ☺️
I've made a walkthrough video for a Kp question:
ua-cam.com/video/HaV7EZR6nhk/v-deo.html
Hi Sir, is it necessary for those doing AQA A level chemistry to memorise the details of the experiment to deduce Kc or had you used it as an example to show how we would go about it? As I'm unsure whether to add it to my notes or not as I don't see a mention of it in the specification, but know sometimes they can throw in application questions such as this. Thanks in advance!
You won't be asked to describe it, so not from that point of view. But it's a type of exam question you can get that nicely weaves together Esters, Kc and amount of Substance. It's also an interesting situation with the units cancelling.
Ahh okay, I see thank you! They really do like throwing in a few questions that tie in bits of other topics so I could definitely see it being a question haha thank you so much
@angel-lh8sh very welcome 🙏
thank you so so much your videos are so helpful please make some on other units like kinetics and bonding thank you
I'm really pleased you found it useful. I hope you have seen the Rate Equation and Arrhenius videos 😃
wish id seen this sooner, would have helped me in my paper 2 as well
Good luck with the revision 😃
THANK YOU SO MUCH!!
I'm really glad it's useful 😃
Hello, it is a very informative video! I was wondering what does mol -1 dm 3 mean in terms of solubility/concentration since mol 1 dm-3 means 'this many x mols per (1)litre or dm3 ' ; how can the reverse still express the concentration of a solution?
Thank you!
Kinetics and equilibrium constants are often out of the scope of a physical/chemical understanding of units. Most of the rest of the time, the units are meaningful as it represents a concept easily understandable by the brain.
However, in the case of kinetic constants, the problem is inverted. In fact, you adapt the unit to match what the equation needs, in order to have the same unit on both sides of the "equal" sign. It is probably due to the fact that all the kinetic equations have different forms, and is almost always obtained after quite a lot of simplifications.
Units don't have to be physically meaningful, they just have to work. The dimensional analysis of whatever problem you are attacking forces the units to be what they are.
This is why a rate constant on its own doesn't give you a clue of the speed of the reaction. It is useful when you got to compare some of them (IF they are comparable).
@@chemistrytutor I see - Thank you very much!
@@lfragonard7661 no worries at all. It was a very good question!
Hi sir, are quadratic equations in kc calculations in the spec? Would it be worth to practice?
No, not in the spec. If it looks like you need to solve a quadratic you should square root everything 😃
wow, thank you a lot
Very welcome 🙏
the c method for calculating units is so helpful!
Thanks 😀 glad it's useful!
At 25.08 why is the 4 squared and not just times two because there are two moles of Y? 😅😅😅 if there was 3 moles what would I do? Times it by 3 or do 3 cubed 😢
In a Kc expression, the coefficients from the balanced equation always become powers. So whatever the number that's the power you raise the concentration to. Think of it like this. If the equation said 3HF, then the Kc expression would contain the term
[HF] x [HF] x [HF]
Which we write as [HF]^3
what is they arent on the same state? how would it vary
So, if there was a solid present, you wouldn't include it in the eqm mixture. Typically they will all be (aq) or all (g)
@@chemistrytutor what if there’s was a liquid
@@generation6790 then that isn't the same phase. Your equilibrium expressions are only for homogeneous equilibria
so helpful
Thank you 😊
i dont really understand the experiment to determine kc and how concentration and pressure dont affect its value?
The experiment uses algebra to deduce X moles of something used and made. Then you have to work out X from the Kc expression. You won't need a quadratic to solve it.
For the factors affecting Kc, really you only need to know that it's just the temperature that affects Kc. For an increase in concentration of reactant, you'd expect Kc to go down as reactant concentration is on the bottom. However the equilibrium shifts right making more product and using up reactant. This means that the effects cancel out and no change is observed
@@chemistrytutor thank you so much!
@@alizahmohammed7751 😎
thankyou phew
Very welcome 😀
I don't understand. If we increase temperature the enthalpy change is endothermic therefore shifting to the left:
Reactants----》products
But if the forward reaction is exothermic then we have to decrease temperature to shift equilibrium to the reacntans? Why???
Don't get too bogged down by what direction is which exo/endo as that can change.
You need to know that there is an optimum temperature (and pressure and concentration ratio) and at this point the chemicals have the least energy and are the most stable.
If we raise the temperature, the chemicals become less stable and so the reaction that reduces the temperature (endothermic reaction) speeds up briefly and becomes faster than the other reaction. This continues until the temperature returns to the original, after which both reactions happen at the same rate again