Author Topic: ALL CIE CHEMISTRY DOUBTS HERE !!  (Read 119180 times)

Offline HUSH1994

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #390 on: March 08, 2011, 03:38:40 am »
how do u calculate the enthalpy change of combustion, and what do u use the values given at the data booklet for,i need help in:

C8H18 + 12.5 O2 ------- 8CO2 + 9 H2O
I need full detailed explanation please,i know the defenition but how to calculate it

elemis

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #391 on: March 08, 2011, 11:13:58 am »
how do u calculate the enthalpy change of combustion, and what do u use the values given at the data booklet for,i need help in:

C8H18 + 12.5 O2 ------- 8CO2 + 9 H2O
I need full detailed explanation please,i know the defenition but how to calculate it

Calculate the energy needed to break the bonds of the reactants and subtract the energy needed to form the products.

Just use the bond dissociation energies given in the Data Booklet.

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #392 on: March 08, 2011, 11:55:18 am »
thx a lot, i knew it was something really easy... What confused me was that Mr H2=2... Dumb, I know! Thx, again!

Don't worry dear........it happens. You'll get used to it with time and more practice ;)

Offline xlane

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #393 on: March 08, 2011, 11:57:44 am »
i need explanation of partition coefficient and two immiscible layers...its for p5.......so i need the complete explanation.....some 1 help me out please....and does any1 know any p5 question based on this two topics ?

Amelia

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #394 on: March 08, 2011, 02:22:12 pm »
i need explanation of partition coefficient and two immiscible layers...its for p5.......so i need the complete explanation.....some 1 help me out please....and does any1 know any p5 question based on this two topics ?


 - The word 'partition' means a substance X is distributed between two phases in a dynamic equilibrium.

 - It is a heterogeneous equilibrium since the 'solute' is distributed between two distinct phases.

 - The two phases may be a gas and liquid-solution or, more likely, two immiscible liquids.

The basic expression is:

Kpartition = [X(phase 1)] / [X(phase 2)]

Here the K is called the partition coefficient or distribution coefficient. If it involves two immiscible liquids, K has no units.
The partition will involve the distribution of a solute between two immiscible liquid phases, which is a more likely and simpler situation to deal with.

If the solute is in the same molecular state in both liquid-phases, the following simple partition equilibrium expression will apply:

Kpartition =
 [X(liquid 1)] 
----------------
[X(liquid 2)]

K is called the partition/distribution coefficient and has no units  and is temperature dependent.

Both concentrations must be in the same units e.g. molarity mol dm-3, g dm-3, mg cm-3 or whatever.

If a substance is added to a mixture which is soluble to a greater or lesser extent in both immiscible liquids, on shaking and then allowing the mixture to settle, the concentrations in each layer become constant. However, there is continual interchange of solute between the liquid layers via the interface i.e. a dynamic equilibrium is formed.

If more of the substance X is added to the system, the solute will distribute itself between the immiscible liquids so that the ratio of the solute concentrations remains the same at constant temperature independently of the total quantity of X in the same molecular state, and that is essentially the partition equilibrium law.


Amelia

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #395 on: March 08, 2011, 02:33:12 pm »
Example question - 20g of butanedioic acid (BDA) was shaken with a mixture of 100 cm3 ether and 100 cm3 water at 25oC. After titration with standard sodium hydroxide the concentration of the acid was found to be 0.024 mol dm-3 in ether and 0.16 mol dm-3 in water.

(a) Calculate the partition coefficient Kp for butanedioic acid between ether/water.

Kp = [BDA(ether)] / [BDA(water)] = 0.024 / 0.16 = 0.15

(b) If 10g of BDA had been shaken with 50 cm3 of each solvent at 25oC, what value would expect for Kp if the layers were again analysed?

0.15, since the partition law states the ratio of concentrations remains constant at constant temperature.

(c) If 10g of butanedioic acid was dissolved in 50 cm3 of ether at 25oC, calculate how much of the acid can be extracted with 50 cm3 of water.

If we call x the mass in g extracted into water, V the liquid volumes, and expressing the concentrations in gcm-3, substitution into the partition expression gives ...

Kp =
 [(10-x)/Vether]       (10-x)
---------------------- = ------- (since V's cancel out)
   [x/Vwater]            x

Rearranging: 0.15 x = 10 - x, so 1.15x = 10
therefore x = 10/1.15 = 8.7g of BDA extracted.

 --- Check the past papers for more questions... or perhaps, the application booklet.
« Last Edit: March 08, 2011, 02:40:28 pm by Amelia »

Offline iluvme

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #396 on: March 09, 2011, 11:42:48 am »
Please explain why Kc does not change with a change in concentration.
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Offline Master_Key

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #397 on: March 09, 2011, 01:08:47 pm »
Please explain why Kc does not change with a change in concentration.

If you are changing the concentration of both the reactants then it will change.

If you are changing the concentration of one substance then it seems obvious that one of the reactant is in excess and does not react.

Offline iluvme

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #398 on: March 12, 2011, 12:32:24 pm »
If you are changing the concentration of both the reactants then it will change.

If you are changing the concentration of one substance then it seems obvious that one of the reactant is in excess and does not react.

Thanks :)
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Offline Master_Key

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #399 on: March 12, 2011, 01:08:50 pm »

Offline iluvme

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #400 on: March 20, 2011, 03:07:38 pm »
This question please,

When 1 mole of HI is allowed to dissociate, in a 1.0 dm-3 vessel at 4500C, only 0.78 mol of HI are present at equilibrium. What is the Kc value for this reaction?
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Offline Amii

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #401 on: March 20, 2011, 03:28:26 pm »
                          2HI (g)        ----------->       H2 (g) +     I2 (g)
                                           <-----------

Intial moles               1                                    0              0
Change                 (1-0.78)                          (0.5X0.22)    (0.5X0.22)
                             -0.22                             +0.11           +0.11
Moles at equilibrium   0.78                               0.11            0.11

Conc at equilibrium   (0.78/1 )                         (0.11/1)      (0.11/1)
mol dm-3                 0.78                                0.11            0.11

Kc = [H2][I2]
      ---------
        [HI]2

Kc= 0.11mol dm-3 X0.11 mol dm-3

      ---------------------------

        (0.78 mol dm-3 ) 2

=0.0199 no units

I hope it's right
« Last Edit: March 20, 2011, 03:51:35 pm by Amii »

By my sweeett nephew ;D

Offline iluvme

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #402 on: March 20, 2011, 03:49:02 pm »
Thank You! <3

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Offline Amii

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #403 on: March 20, 2011, 03:52:42 pm »
Thank You! <3


No problem  :D
Check it again.I modified it.


By my sweeett nephew ;D

Offline iluvme

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #404 on: March 20, 2011, 04:14:53 pm »
No problem  :D
Check it again.I modified it.



Thanks again.

Didn't notice anything wrong. Was just going through the steps. :)
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