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AS Chemistry- helping each other out!

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Original post by haemo
Using chloromethane (CClH3) as an example:

Initiation step: Cl2 -> 2Cl•

Propagation step 1: CClH3 + Cl• -> •CClH2 + HCl

Propagation step 2: •CClH2 + Cl2 -> CCl2H2 + Cl•

Termination step (1): •Cl + •Cl -> Cl2
Termination step (2): •Cl + CClH2 -> CCl2H2
Termination step (3): •CClH2 + •CClH2 -> CClH2CClH2 (NOT C2Cl2H4)

Care to explain what is happening with the propagation step?
Hey guys.
So i was doing some a level chemistry questions (http://www.a-levelchemistry.co.uk/AQ...2.1%20home.htm)
exercise 2.
We are given bond disassociation energies and are asked to calculate an enthalpy of formation of HF? I am unsure of how to go about this.. The question before asks to calculate an enthalpy of reaction where HF is a product
All help is greatly appreciated
Original post by Gladiatorsword
Hey guys.
So i was doing some a level chemistry questions (http://www.a-levelchemistry.co.uk/AQ...2.1%20home.htm)
exercise 2.
We are given bond disassociation energies and are asked to calculate an enthalpy of formation of HF? I am unsure of how to go about this.. The question before asks to calculate an enthalpy of reaction where HF is a product
All help is greatly appreciated


Total up all the bond enthalpies from all the bonds of all the reactants.
Total up all of the bond enthalpies from all the bonds of the products.

Do reactants minus products.
This equals delta H.

Bob's your uncle.


Posted from TSR Mobile
Can anyone help me pleeeease!!

Calculate the enthalpy change for thermal decomposition of lithium carbonate (li2c03) given enthalpy of formation for li2c03- -1216 kj mol-1, li02- -598 kj mol-1 and c02- -394 kj mol-1. I got 224 kj mol-1, but I don't think it's right.

Calculate the enthalpy change for the reaction; c(graphite)>> c(diamond)
Enthalpy of combustion of carbon- graphite is -393.5 kj mol-1 and carbon-diamond is -395.4 kj mol-1.
Original post by danconway
Total up all the bond enthalpies from all the bonds of all the reactants.
Total up all of the bond enthalpies from all the bonds of the products.

Do reactants minus products.
This equals delta H.

Bob's your uncle.


Posted from TSR Mobile

So is this delta h the same as enthalpy of formation?
(edited 9 years ago)
Original post by Gladiatorsword
S is this delta h the same ad enthalpy of formation?


It'd be the enthalpy of reaction. You may need to put further work in e.g. draw a Hess cycle.
Original post by Gladiatorsword
So is this delta h the same as enthalpy of formation?


Ah, my bad. Serves me right for trying to assume what the question is!

How you do (c):
Formation of HF will be H2 + F2 -> 2HF, which you have to change to
0.5H2 + 0.5F2 -> HF

Now do the bond enthalpy thing with this reaction – this reaction is the formation of HF.
1. Given the data:
4NH3(g) + 3O2(g) à 2N2(g) + 6H2O(l), DH = -1530kJmol-1
H2(g) + 1/2O2(g) à H2O(l), DH = -288 kJmol-1
Calculate the enthalpy of formation of ammonia.
Reply 768
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Jmedi
6
Original post by Super199
Care to explain what is happening with the propagation step?


The •Cl is highly reactive, so it reacts with the neutral compound. It therefore forms HCl and •CClH2, in this example. The •C(Cl)H2 is a free radical and so, is highly reactive. This reacts with Cl2, to form C(Cl2)H2 and a free radical, •Cl.

Hopefully that explained it?
Original post by Gladiatorsword
1. Given the data:
4NH3(g) + 3O2(g) à 2N2(g) + 6H2O(l), DH = -1530kJmol-1
H2(g) + 1/2O2(g) à H2O(l), DH = -288 kJmol-1
Calculate the enthalpy of formation of ammonia.


Have you tried drawing a Hess cycle?

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Original post by Dylann
Have you tried drawing a Hess cycle?

Posted from TSR Mobile


oh....that never occurred to me. thanks
how would i draaw the hess's cycle?
Need some help with 1c and 2a for now :tongue:. kind of cropped out the 2 but should be able to see which question it is. Is 2a the enthalpy of formation? Grr I hate this topic. Any help would be appreciated :smile:
image.jpg511.5KB
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Jmedi
6
Original post by Super199
Need some help with 1c and 2a for now :tongue:. kind of cropped out the 2 but should be able to see which question it is. Is 2a the enthalpy of formation? Grr I hate this topic. Any help would be appreciated :smile:


1c)
Okay, so you're trying to find out the C-H bond.
If you were given the value, and your question was to work out the value of the enthalpy change, you do the breaking bonds - forming bonds = enthalpy change.

Now, we use that here -
Lets say X = C-H, just to make it easier to write
Therefore,
(4X + 612 + 436) - (6X + 348) = -136
You can rearrange this to -2X = -836
You then divide by -2, to get X = 418
Therefore C-H = 418kJ/mol

I attached a picture of my workings, if that helps.


chem 1c answer tsr.jpg432.8KB
Original post by Jmedi
1c)
Okay, so you're trying to find out the C-H bond.
If you were given the value, and your question was to work out the value of the enthalpy change, you do the breaking bonds - forming bonds = enthalpy change.

Now, we use that here -
Lets say X = C-H, just to make it easier to write
Therefore,
(4X + 612 + 436) - (6X + 348) = -136
You can rearrange this to -2X = -836
You then divide by -2, to get X = 418
Therefore C-H = 418kJ/mol

I attached a picture of my workings, if that helps.



This is brilliant!! Thank you :smile: Any help with 2a?
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Jmedi
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Original post by Super199
This is brilliant!! Thank you :smile: Any help with 2a?


I'm pretty sure it's meant to be H2O2 -> H2O + 0.5O2
(I'm pretty sure as the one in the picture doesn't make sense unless I'm missing something?)

Therefore, using sum of bonds broken - sum of bonds formed

(2*463+146) - (2*463 + 496/2)
496/2 because the oxygen molecule is O=O
[ this can be simplified to
146-(496/2) ]
So, answer = -102 kJmol^-1
(edited 9 years ago)
Original post by Jmedi
I'm pretty sure it's meant to be H2O2 -> H2O + 0.5O2
(I'm pretty sure as the one in the picture doesn't make sense unless I'm missing something?)

Therefore, using sum of bonds broken - sum of bonds formed

(2*463+146) - (2*463 + 496/2)
496/2 because the oxygen molecule is O=O
[ this can be simplified to
146-(496/2) ]
So, answer = -102 kJmol^-1

Does that equation always show enthalpy change? Regardless if it is formation/ combustion or...?
Original post by haemo
When given mean bond enthalpies:

Enthalpy change = reactants - products


When given formation data:

Enthalpy change = products - reactants


When given combustion data:

Enthalpy change = reactants - products

Is there a specific reason why some of them are the other way round?
Need some help with an enthalpy change question

Calculate the enthalpy changes of reaction for each of the following reaction

a). C2H4(g) + H2(g) ----> C2H6(g)

Right so it gives you a ton of values. So since they are in the gas state it is Sum of bonds broken - Sum of bonds made.

There is a C=C double bond which is 612Kjmol^-1
But it says there is one H-H bond that gives the 436Kjmol^-1
I don't really understand that bit isn't there 4 C-H bonds? Can someone explain this. I get the rest of the question apart from this bit.

Thanks :smile:
Original post by Super199
Need some help with an enthalpy change question

Calculate the enthalpy changes of reaction for each of the following reaction

a). C2H4(g) + H2(g) ----> C2H6(g)

Right so it gives you a ton of values. So since they are in the gas state it is Sum of bonds broken - Sum of bonds made.

There is a C=C double bond which is 612Kjmol^-1
But it says there is one H-H bond that gives the 436Kjmol^-1
I don't really understand that bit isn't there 4 C-H bonds? Can someone explain this. I get the rest of the question apart from this bit.

Thanks :smile:


Did they not give the bond enthalpy for C-H & C-C bonds??
p.s. What is the answer they obtained?

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