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Chemistry: Post your doubts here!

Metanoia

Metanoia

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My Name said:
I'm getting a bit confused I think.
This is what I understood so far....

When the temperature increases both( A and B side) have an increase in energies.
So both the forward and backward reaction rate increases. But they increase to different extents (depending on which direction is exo or end0).

OR

When the temperature is increased and the forward reaction is exothermic the backward reaction increases but since the system needs to be in Equilibrium the forward reaction also increases(to retain the equilibrium).

What's right and what's not?
Please and Thank you!
Click to expand...

Your first part is correct.

Screen Shot 2015-04-20 at 12.46.28 PM.png
 
Xylferion

Xylferion

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Metanoia said:
Nicely done diagrams and explanation. :)

I'll just like to add on that the 2nd product of the 3 compound would eventually be oxidized to HOOCCOOH and then to CO2.
Click to expand...

Ah yes thanks for reminding me :)

I just want to thank you Sir, as your videos have really helped me out a lot! I've learned more from your videos than I have throughout the past couple of months, thanks a ton!
 
Metanoia

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Xylferion

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Mahnoorfatima said:
https://docs.google.com/viewerng/vi...wp-content/uploads/2012/11/9701_s13_qp_23.pdf
Question 5 part d please? I can't solve questions like these at all!!!
https://docs.google.com/viewerng/vi...wp-content/uploads/2012/11/9701_s12_qp_21.pdf
Question 5, biii, cii. How are we supposed to know questions like cii????
Click to expand...

In organic chemistry, oxidation and reduction are defined differently.

Oxidation is the addition of oxygen atoms to a molecule, you often see them as [O]. It is also the removal of Hydrogen atoms.

Reduction is the removal of oxygen atoms, [O], from a molecule. It is also the addition of Hydrogen atoms.

Now of you consider the reaction in Question 5 (d),

CH2=CHCH2OH --------> CH3CH2CHO

You can see that there's an alkene group and an alcohol group in the original compound, hence why it is called an allyl alcohol.

In the product, which is propanal, you can see the -CHO group denoting an aldehyde, and the alkene group is no longer there.

What you can conclude from this is,

The -OH group was oxidized to -CHO.
The double bond was broken, as two hydrogen atoms were added across it.

Based off of the conditions for reduction and oxidation above,
Adding the hydrogen atoms across the double bond means reduction has taken place. Consequently, the OH group has been oxidized to CHO.

Both oxidation and reduction have taken place, hence you may call this a redox reaction. Which is the only thing unusual about it.

-------------------------------

I've already explained the other question here: https://www.xtremepapers.com/community/threads/chemistry-post-your-doubts-here.9859/page-608

Hope that helps!
 
Mahnoorfatima

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Xylferion said:
In organic chemistry, oxidation and reduction are defined differently.

Oxidation is the addition of oxygen atoms to a molecule, you often see them as [O]. It is also the removal of Hydrogen atoms.

Reduction is the removal of oxygen atoms, [O], from a molecule. It is also the addition of Hydrogen atoms.

Now of you consider the reaction in Question 5 (d),

CH2=CHCH2OH --------> CH3CH2CHO

You can see that there's an alkene group and an alcohol group in the original compound, hence why it is called an allyl alcohol.

In the product, which is propanal, you can see the -CHO group denoting an aldehyde, and the alkene group is no longer there.

What you can conclude from this is,

The -OH group was oxidized to -CHO.
The double bond was broken, as two hydrogen atoms were added across it.

Based off of the conditions for reduction and oxidation above,
Adding the hydrogen atoms across the double bond means reduction has taken place. Consequently, the OH group has been oxidized to CHO.

Both oxidation and reduction have taken place, hence you may call this a redox reaction. Which is the only thing unusual about it.

-------------------------------

I've already explained the other question here: https://www.xtremepapers.com/community/threads/chemistry-post-your-doubts-here.9859/page-608

Hope that helps!
Click to expand...
I'm extremely sorry Ive given you the wrong question! http://maxpapers.com/syllabus-materials/chemistry-9701-a-level/attachment/9701_s12_qp_23/
It's this question Q5 d.

https://docs.google.com/viewerng/vi...wp-content/uploads/2012/11/9701_s12_qp_21.pdf
Question Q5 part d.
 
qwertypoiu

qwertypoiu

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Just out of interest.
N2 + 3H2 = 2NH3
This is the equilibrium equation for Contact Process. We learnt that if pressure is increased, the amount of product also increases. (because of no. of moles of gases and bla bla)
In that case, why do pressure changes NOT affect Kc or Kp?
Thanks.

EDIT: sorry I mean Haber Process
 
Last edited:
Xylferion

Xylferion

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qwertypoiu said:
Just out of interest.
N2 + 3H2 = 2NH3
This is the equilibrium equation for Contact Process. We learnt that if pressure is increased, the amount of product also increases. (because of no. of moles of gases and bla bla)
In that case, why do pressure changes NOT affect Kc or Kp?
Thanks.
Click to expand...

Kc and Kp both define the rate at which a reaction proceeds, that is to assess the frequency of collisions between the reacting molecules.

When you alter the pressure of a "gas", the equilibrium " position " shifts to the side with a higher/lower amount of moles. With that in mind, the equilibrium position is devoting itself to counteract the change, everything is attempting to balance out.

To match the increase/decrease, the position of equilibrium has to adjust in order decrease/increase the change accordingly. So all that the change in pressure is doing, is telling the Position of equilibrium to shift back to cancel out this change, it is not increasing the rate of the reaction.

On the other hand, temperature affects the rate at which the reaction proceeds. Since molecules with more energy, will raise the frequency of collisions.

For a change in pressure, it's not trying to get the reaction to move faster, it's like a jug of water, when you fill it past its capacity, the rest will spill. What that simply means is, to deal with all this compressed high pressure gas, it attempts to spread the gas molecules out evenly so as to achieve a pressure that's suitable to both sides. It is not making molecules collide more.

Hope that explains it.
 
The Sarcastic Retard

The Sarcastic Retard

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qwertypoiu said:
Just out of interest.
N2 + 3H2 = 2NH3
This is the equilibrium equation for Contact Process. We learnt that if pressure is increased, the amount of product also increases. (because of no. of moles of gases and bla bla)
In that case, why do pressure changes NOT affect Kc or Kp?
Thanks.
Click to expand...
*HABER PROCESS* CONTACT PROCESS IS THE PRODUCTION OF SULFUR TRIOXIDE!!
Explanation in detail can be found here : http://www.chemguide.co.uk/physical/equilibria/change.html
POSITION OF EQUILIBRIUM MAY CHANGE ACCORDING TO LE CHATELIERS' PRINCIPLE BUT EQUILIBRIUM CONSTANTS ARE NOT AFFECTED BY CHANGE IN PRESSURE. ONLY CHANGE IN TEMPRATURE AFFECTS THE EQUILIBRIUM CONSTANTS. AND USING THAT YOU CAN FIND WHETHER THE REACTION IS EXOTHERMIC OR ENDOTHERMIC.
 
qwertypoiu

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Mahnoorfatima said:
Reaaction bw CH3CO2CH(CH30)2 with KOH(aq) warm? What type of reacton is this and whats the product?
Click to expand...
I think you made a typing mistake, I assume you meant: CH3CO2CH(CH3)2
This is an ester. You can hydrolyse this ester, and break it down into the carboxylic acid and the alcohol it was formed from. This requires reacting it with water, with heat, and you MUST have a catalyst along with it (either an acid or an alkali)
So you'd expect this to happen:
CH3CO2CH(CH3)2 + H2O -----(with catalyst)---> CH3CO2H + HOCH(CH3)2
However, since your catalyst happens to be KOH(aq), which is a base, your acid formed will react with it.
CH3CO2H + KOH ----> CH3CO2K + H2O
So the two organic products you got were:
CH3CO2K and HOCH(CH3)2
 
qwertypoiu

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The Sarcastic Retard said:
*HABER PROCESS* CONTACT PROCESS IS THE PRODUCTION OF SULFUR TRIOXIDE!!
Explanation in detail can be found here : http://www.chemguide.co.uk/physical/equilibria/change.html
POSITION OF EQUILIBRIUM MAY CHANGE ACCORDING TO LE CHATELIERS' PRINCIPLE BUT EQUILIBRIUM CONSTANTS ARE NOT AFFECTED BY CHANGE IN PRESSURE. ONLY CHANGE IN TEMPRATURE AFFECTS THE EQUILIBRIUM CONSTANTS. AND USING THAT YOU CAN FIND WHETHER THE REACTION IS EXOTHERMIC OR ENDOTHERMIC.
Click to expand...
Sorry yeah I meant Haber Process how silly of me
Your link was very helpful thanks :)
 
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qwertypoiu

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Xylferion said:
Kc and Kp both define the rate at which a reaction proceeds, that is to assess the frequency of collisions between the reacting molecules.

When you alter the pressure of a "gas", the equilibrium " position " shifts to the side with a higher/lower amount of moles. With that in mind, the equilibrium position is devoting itself to counteract the change, everything is attempting to balance out.

To match the increase/decrease, the position of equilibrium has to adjust in order decrease/increase the change accordingly. So all that the change in pressure is doing, is telling the Position of equilibrium to shift back to cancel out this change, it is not increasing the rate of the reaction.

On the other hand, temperature affects the rate at which the reaction proceeds. Since molecules with more energy, will raise the frequency of collisions.

For a change in pressure, it's not trying to get the reaction to move faster, it's like a jug of water, when you fill it past its capacity, the rest will spill. What that simply means is, to deal with all this compressed high pressure gas, it attempts to spread the gas molecules out evenly so as to achieve a pressure that's suitable to both sides. It is not making molecules collide more.

Hope that explains it.
Click to expand...
Thanks man :)
 
Xylferion

Xylferion

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Mahnoorfatima said:
Reaaction bw CH3CO2CH(CH30)2 with KOH(aq) warm? What type of reacton is this and whats the product?
Click to expand...

Hydrolysis using an alkaline catalyst. The catalyst being KOH in this reaction. The ester splits up into [CH3CO2] belonging to the carboxylic acid, and [CH(CH3)CH3] belonging to the alcohol.

First step,

CH3CO2CH(CH3)CH3 reacts with the OH- ion to form a carboxylate ion and an alcohol.

The carboxylate ion, which is negative, bonds with K.

CH3CO2K + HOCH(CH3)CH3

Second step,

Have CH3CO2K react with a hydrogen halide, like HCl.

CH3CO2K + HCl ------> CH3COOH + KCl

The products can be either CH3COOH and HOCH(CH3)CH3, OR CH3CO2K and HOCH(CH3)CH3.

Do you have a source for this question?

Hope this helped though..
 
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