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# Enthalpy Calculations watch

1. Someone help me learn all the enthalpy types and their calculations thoroughly...
i don know how to calculate enthalpy of atomisations neutralisation etc...
help!!
2. (Original post by Kk1999)
Someone help me learn all the enthalpy types and their calculations thoroughly...
i don know how to calculate enthalpy of atomisations neutralisation etc...
help!!
Well first of all you need to know all the definitions off by heart
3. (Original post by SirRaza97)
Well first of all you need to know all the definitions off by heart
yup.. I know them by heart.. then??
4. Okay good. So if you have a good undertsanding of them and a good idea of what the question is asking for it should be fairly strtaight foward to work them out.
Normally they give you a data table to work with (I am assuming you do AQA). So here is a question from a paper I just randomnly picked:

Okay so you can do it 2 ways. Draw a quick Hess diagram or go through the steps in your head to form an equation. I do it the second way cause ain't nobody got time for diagrams boi. So dissociation is what we are calcluating. We have our atomisations, ionoisations, affinity anf formation. Well the formation one is a route by itself. What I mean is that from Hess' Law we know that the energy is the same no matter what route taken. So the formation is one route and the other steps is another route. Therefore the energy required must be equal. The we form a cheeky equation:

AgClFORM = AgATOM + ClATOM + AgIONISATION + ClAFFINITY + AgClLATTICE FORMATION

Left side is one route and right hand side is another route. So the toal energy must be the same. We know all values but Lattice formation. So let's call Lattice Formation = X and then sub our values in and find X

AgClFORM = AgATOM + ClATOM + AgIONISATION + ClAFFINITY + X

-127 = 289 + 121 + 732 - 364 + X
X = -905 KJ /Mol
That's our lattice formation. We want lattice dissociation which is just the reverse so our answer is +905 KJ/Mol

Honestly this might be super confusing for you. Find your own method that works for you! Then practice it. Eventually it will make sense. The same idea goes for enthalpy of solution. It's all about Hess' Law and equations to work with. Hope that helps.
5. (Original post by SirRaza97)
Okay good. So if you have a good undertsanding of them and a good idea of what the question is asking for it should be fairly strtaight foward to work them out.
Normally they give you a data table to work with (I am assuming you do AQA). So here is a question from a paper I just randomnly picked:

Okay so you can do it 2 ways. Draw a quick Hess diagram or go through the steps in your head to form an equation. I do it the second way cause ain't nobody got time for diagrams boi. So dissociation is what we are calcluating. We have our atomisations, ionoisations, affinity anf formation. Well the formation one is a route by itself. What I mean is that from Hess' Law we know that the energy is the same no matter what route taken. So the formation is one route and the other steps is another route. Therefore the energy required must be equal. The we form a cheeky equation:

AgClFORM = AgATOM + ClATOM + AgIONISATION + ClAFFINITY + AgClLATTICE FORMATION

Left side is one route and right hand side is another route. So the toal energy must be the same. We know all values but Lattice formation. So let's call Lattice Formation = X and then sub our values in and find X

AgClFORM = AgATOM + ClATOM + AgIONISATION + ClAFFINITY + X

-127 = 289 + 121 + 732 - 364 + X
X = -905 KJ /Mol
That's our lattice formation. We want lattice dissociation which is just the reverse so our answer is +905 KJ/Mol

Honestly this might be super confusing for you. Find your own method that works for you! Then practice it. Eventually it will make sense. The same idea goes for enthalpy of solution. It's all about Hess' Law and equations to work with. Hope that helps.
im sorry im doin EDEXCEL.
AND I DINT UNDERSTAND WHAT U EXPLAINED
6. (Original post by Kk1999)
im sorry im doin EDEXCEL.
AND I DINT UNDERSTAND WHAT U EXPLAINED
Oh alright. I'll try another way. Hess' Law states:

"change of enthalpy in a chemical reaction is independent of the pathway between the initial and final states."

What this means is no matter what route you take to make some sort of compund, the total energy will be the same for all routes. To make AgCl, we know that the enthalpy of formation of silver chloride is 1 route. It only takes one step.

Or the equation: AgClFORM = AgCl

But there is another route:

Or the equation:
AgATOM + ClATOM + AgIONISATION + ClAFFINITY + AgClLATTICE FORMATION = AgCl

This also forms AgCl but takes multiple steps. But the total energy rquired will be exactly the same as the energy required for the formation route, So we can set them equal to each other

AgClFORM = AgATOM + ClATOM + AgIONISATION + ClAFFINITY + AgClLATTICE FORMATION

Then with some maths we can work out the Lattice formation value. Since

AgClLATTICE FORMATION = - AgClLATTICE DISSOCIATION

We can work out what we want and gives us the answer +905 KJ / MOL

I tried my best. This is the best I can explain
7. (Original post by SirRaza97)
Oh alright. I'll try another way. Hess' Law states:

"change of enthalpy in a chemical reaction is independent of the pathway between the initial and final states."

What this means is no matter what route you take to make some sort of compund, the total energy will be the same for all routes. To make AgCl, we know that the enthalpy of formation of silver chloride is 1 route. It only takes one step.

Or the equation: AgClFORM = AgCl

But there is another route:

Or the equation:
AgATOM + ClATOM + AgIONISATION + ClAFFINITY + AgClLATTICE FORMATION = AgCl

This also forms AgCl but takes multiple steps. But the total energy rquired will be exactly the same as the energy required for the formation route, So we can set them equal to each other

AgClFORM = AgATOM + ClATOM + AgIONISATION + ClAFFINITY + AgClLATTICE FORMATION

Then with some maths we can work out the Lattice formation value. Since

AgClLATTICE FORMATION = - AgClLATTICE DISSOCIATION

We can work out what we want and gives us the answer +905 KJ / MOL

I tried my best. This is the best I can explain
Thanks..
8. (Original post by Kk1999)
Thanks..
I am sorry I can't really explain any further. In all honesty I learnt this part by just doing a lot of questions till I found it easy. Just try that. Look at mark schemes.
9. (Original post by SirRaza97)
I am sorry I can't really explain any further. In all honesty I learnt this part by just doing a lot of questions till I found it easy. Just try that. Look at mark schemes.
Hey!!no worries..
BTW can u help me in energy level diagrams..
how to do em step by step and which process comes first etc..
10. (Original post by Kk1999)
Hey!!no worries..
BTW can u help me in energy level diagrams..
how to do em step by step and which process comes first etc..
what energy level diagrams?
11. check these two pics. u ll understand.

Attachment 536461536463
Attached Images

12. Check this Pic
13. This is A2 right?
14. (Original post by Kk1999)
Check this Pic
In edexcel they will put the values in a table
So you should know when you want to calculate enthalpy of combustion ( reactants-products )
For formation (products-reactants)

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15. (Original post by Leen.A1998)
In edexcel they will put the values in a table
So you should know when you want to calculate enthalpy of combustion ( reactants-products )
For formation (products-reactants)

Posted from TSR Mobile
Thats the thing I dont know..I know how to calculate enthalpies or reaction,combustion and formation (did i miss something?) but not when..
16. Are you doing A2 or AS??
17. (Original post by femalebo55)
are you doing a2 or as??
as edexcel..(exams in summer 2016) currently going on...
18. (Original post by Kk1999)
Thats the thing I dont know..I know how to calculate enthalpies or reaction,combustion and formation (did i miss something?) but not when..
Can you post a question from a past paper about enthalpy changes that you're struggling with ? So we can help you more but I'll try to help
19. (Original post by Kk1999)
Thats the thing I dont know..I know how to calculate enthalpies or reaction,combustion and formation (did i miss something?) but not when..
Ask a question so I can figure out what you don't understand.

Posted from TSR Mobile
20. (Original post by Kk1999)
as edexcel..(exams in summer 2016) currently going on...

These are the two main enthalpy level diagram; a reaction iseither exothermic or endothermic.

Always for an exothermic reaction the products would havelower energy than its reactants, this is because when they form a bond energyis released rather than absorbed. Whereas for the endothermic enthalpy diagramyou can tell that you would need to put high amounts of energy to form thebonds of the products hence the positive enthalpy change value.

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