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GCSE Chemistry Unit 3 (C3)

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The Haber process isn't kept in dynamic equilibrium i thought, as the conditions are trying to encourage the reaction to favor the production of Ammonia, unless the high pressure moving the equilibrium to the right and the high temperature moving it to the left keeps it at dynamic equilibrium? The forward reaction is extothermic right? if that is so, then the 450 degree temperature is used to speed up the reaction as it would move the equilibrium the "wrong way" (away from ammonia production) and then the 200atm pressure is used to move it to the right towards the ammonia production. So it moves to the left because of temperature, and then to the right because of pressure, the catalyst has no effect on equilibrium so maybe that keeps the Haber process in equilibrium?
the temperature 450 degrees is a compromise, because a low temperature would slow the rate of reaction down even though a low temperature favours an exothermic reaction which is the forward reaction. the catalyst (iron) speeds up the rate of the reaction without affecting the position of equilibrium and although a high pressure favours the forward reaction, in industry producing and maintaining a high pressure is expensive and dangerous which could lead to an explosion.
Original post by LeEastafrican
the temperature 450 degrees is a compromise, because a low temperature would slow the rate of reaction down even though a low temperature favours an exothermic reaction which is the forward reaction. the catalyst (iron) speeds up the rate of the reaction without affecting the position of equilibrium and although a high pressure favours the forward reaction, in industry producing and maintaining a high pressure is expensive and dangerous which could lead to an explosion.


I might be reading your post wrong, but it looks like you're trying to say the catalyst speeds up equilibrium. The catalyst only decreases the time it takes for the reaction to REACH equilibrium. You're correct in saying it has no effect on the position of equilibrium.
(edited 9 years ago)
Original post by Shadowninja107
In the haber process, when the rate of the forward reaction is increased by lowering the temperature or increasing pressure for instance, surely the reaction comes out of equilibrium, does anyone understand beyond what is written in the textbook?


Nope! My teacher explained this, as you INCREASE the pressure it shifts towards the left( reverse) and then this causes equilibrium to shift to the right to counter it!:smile: Book doesnt explain the first part :smile:
Original post by Marcusroye98
Nope! My teacher explained this, as you INCREASE the pressure it shifts towards the left( reverse) and then this causes equilibrium to shift to the right to counter it!:smile: Book doesnt explain the first part :smile:


An increase in pressure shifts to the side with the least molecules, i.e. the right side in the Haber Process.

Edit: Nevermind I read your post wrong, yes that is correct. It's called Le Chatelier's principle.
(edited 9 years ago)
how is everyone feeling?
Original post by Parallex
An increase in pressure shifts to the side with the least molecules, i.e. the right side in the Haber Process.

Edit: Nevermind I read your post wrong, yes that is correct. It's called Le Chatelier's principle.


Do we need to know that or can we just say a high pressure favours the reaction with fewer molecules to counteract the change of equilibrium?
Original post by Thomith
Do we need to know that or can we just say a high pressure favours the reaction with fewer molecules to counteract the change of equilibrium?


Yeah, that's all you need to say in GCSE. So in the Haber Process, there are less molecules on the right so an increase in pressure shifts equilibrium to the right (exothermic) side.
Original post by Parallex
An increase in pressure shifts to the side with the least molecules, i.e. the right side in the Haber Process.

Edit: Nevermind I read your post wrong, yes that is correct. It's called Le Chatelier's principle.


When you increase the pressure it shifts it to the left then equilibrium counters it to move it to the right ?:frown:
I am really confused : What products are formed in the electrolysis of sodium sulfate?
Original post by Marcusroye98
When you increase the pressure it shifts it to the left then equilibrium counters it to move it to the right ?:frown:


Don't confuse yourself. All you need to say is that equilibrium will (ultimately) shift to the side with the least molecules to counteract the change. In the Haber Process, this is the right side.
Original post by Marcusroye98
I am really confused : What products are formed in the electrolysis of sodium sulfate?


Sodium Sulfate - hydrogen at the cathode and oxygen at the anode.
Original post by Parallex
Sodium Sulfate - hydrogen at the cathode and oxygen at the anode.


thanks, i just dont understand where 4OH - came from/D:
Original post by Marcusroye98
thanks, i just dont understand where 4OH - came from/D:

the water form H+ and OH- ions, meaning the 4OH- is the ion that forms oxygen in the form 2H2O and O2 when oxidised in a solution electrolysis at the anode.
Original post by Marcusroye98
thanks, i just dont understand where 4OH - came from/D:


Learn the electrolysis rules, they're very helpful.
Do we need to know the formulas ;

Mass concentration = Mass of solute / volume

Mass concentration = Mole concentration x Mr

Mole Concentration = mass concentration / Mr
Original post by Marcusroye98
Do we need to know the formulas ;

Mass concentration = Mass of solute / volume

Mass concentration = Mole concentration x Mr

Mole Concentration = mass concentration / Mr


You need to know every single formula in the book.
Original post by Parallex
You need to know every single formula in the book.


They are in the cpg one not in the EDEXCEL textook
Original post by Marcusroye98
Do we need to know the formulas ;

Mass concentration = Mass of solute / volume

Mass concentration = Mole concentration x Mr

Mole Concentration = mass concentration / Mr

I believe this particular formula is purely for ease but i have learnt it anyway and i advise you do the same.
Original post by Thomith
I believe this particular formula is purely for ease but i have learnt it anyway and i advise you do the same.


They're just rearranged equations from the original. Just remember the original formulas.

Mass = Concentration * Volume
Mass = rfm/Mr * Number of Moles
Volume = Number of Moles * Molar Volume

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