[thuanthuan]

I have a question about the Haber process of making ammonia.
The reaction is N2 + 3H2 <-> 2NH3 + heat
To get the best yield we can increase the pressure to favor the forward reaction because it is the one that decrease the number of moles of gas, thus decrease pressure. At the same time, ammonia produced is dissolved into water and its concentration decreases. The forward reaction is again favored.
But if we decrease the number of moles of ammonia, the pressure will decrease, so the reaction will increase it, producing more N2 and H2. I wonder if it's right.
Another question that I'm not sure is that, Is "heat" involved in the reverse reaction ? I think it is. Then at equilibrium, the temperature is constant. Is that right ?

[thuanthuan]

I have another question.
Can you explain quantitatively or qualitatively why changes in pressure affect the equilibrium of a reversible reaction. K depends on temperature only.
For example:
- Change in concentration will increase or decrease K, so something must react with one another to make K unchanged.
- Change in temperature will change K, the concentrations of the reactants must change to fit the new K
- How about change in pressure ?

[AWK]

Change in pressure do not change concentrations only when moles of reagents on both side of equation  are the same.
This is the case, eg in the reaction H2+I2=2HI
but not in the synthesis of ammonia

[thuanthuan]

Change in pressure do not change concentrations only when moles of reagents on both side of equation  are the same.
This is the case, eg in the reaction H2+I2=2HI
but not in the synthesis of ammonia

What question do you answer ? You should quote it or say 1 or 2 !
Anyway, I don't understand your reply. My 2nd question is: Why change in pressure cause equilibrium displacement ?
Thank you.

[technologist]

your first Q in the first post.

If U absorb ammonia the pressure will decrease on the product side (RHS of Equation), So to compensate for that loss U need to generate more molecules of ammonia & hence rxn is favored in forward direction. Here U r confused about two sides of the equation - Reactant & Product.. are two different sides considered in LeChat Principle....U can assume them as two phases separated by a membrane then explanation of everything is easier...IF U reduce pressure on the other side of membrane more molecules will pass thru it to compensate for the pressure loss.....Similarly if U increase the pressure on this side (Reactant side) of membrane again more molecules can be pushed to the other side. So its an equilibrium ..

your Second Q in the first post.

Yes heat is involved in every reaction the only difference is it may be significant or negligible and may evolve heat (Exothermic) or may consume it (Endothermic). The satetment - At equilibrium the temp is constant - is not correct...At a given T equilibirum concentrations ratio (or K - Eqm Constant) is fixed.

your First Q in the second post.
AWK already gave a  good answer and when someone answers out of his busy time...Its your responsibility to understand it....Anyway here is my addition...

K doesn't depend on T only.....Though it is apparently so.....U must have read the Kp & Kc terms which r correlated by equn Kp = Kc * RT (May be wrong I dont remember exactly)..But they are related..

So your Eqm depends on both...This comes from the fact that Pressure do affect the concentration if U consider different kinds of rxn orders.

Even if Rxn is equimolar (when no of moles on both side are same) actual rxn equilibrium still depends on pressure in a plant reactor.., But U can leave it as its a practical part of it.

I Hope it answers all your queries. Need more answers....check out at the bottom of this post.

[AWK]

Even if Rxn is equimolar (when no of moles on both side are same) actual rxn equilibrium still depends on pressure in a plant reactor.., But U can leave it as its a practical part of it.

Technologist is absolutely right on this question, but on high school forum we use an ideal law for gases and solutions, hence my post is sufficient with framework of ideal laws.

[thuanthuan]

Thank you, but it seems that you are misunderstanding my question.

your Second Q in the first post.

Yes heat is involved in every reaction the only difference is it may be significant or negligible and may evolve heat (Exothermic) or may consume it (Endothermic). The satetment - At equilibrium the temp is constant - is not correct...At a given T equilibirum concentrations ratio (or K - Eqm Constant) is fixed.

I wonder why that statement is not correct.
A + B -> C + D + heat
A reacts with B to form C and D and evolving heat.
C react with D to form A and B along with absorb heat.
At equilibrium, the amount of heat evolved is equal to the amount of heat absorbed, so not heat is evolved anymore. Hence the temperature doesn't change.
What's wrong in my explanation ?

your First Q in the second post.
AWK already gave a  good answer and when someone answers out of his busy time...Its your responsibility to understand it....Anyway here is my addition...

K doesn't depend on T only.....Though it is apparently so.....U must have read the Kp & Kc terms which r correlated by equn Kp = Kc * RT (May be wrong I dont remember exactly)..But they are related..

So your Eqm depends on both...This comes from the fact that Pressure do affect the concentration if U consider different kinds of rxn orders.

Even if Rxn is equimolar (when no of moles on both side are same) actual rxn equilibrium still depends on pressure in a plant reactor.., But U can leave it as its a practical part of it.

Actually I don't understand your post and AWK's post.
Here is my question again:
A + B  ->  C  + D
forward speed: k1.[A].[B ]
reverse speed: k2.[C].[D]
equilibrium:  k1.[A].[B ] = k2.[C].[D] (*)
According to my understanding, the impact of temperature change and concentration change is:
- If temperature changes, both k1 and k2 change. Thus left-hand side and the right-hand of (*) are no longer equal, causing equilibrium displacement.
- If concentration changes, in A, for example, right side of (*) remain unchanged, but its left side changes since [A] changes. Again the left and the right are not equal, new equilibrium will be established
- If pressure change, neither k1,[A],[B ] nor k2,[C],[D] will change. (*) is still true. But why the old equilibrium is still broken ?

I know that you guys don't like anyone who keep asking such an easy question. But actually I don't understand your answer. Please help me. Thank you

[Borek]

If pressure change, neither k1,[A],[B ] nor k2,[C],[D] will change. (*) is still true. But why the old equilibrium is still broken ?

A + B -> C + D is not scheme of ammonia synthesis, as it doesn't involve volume changes. Write equation similar to k1.[A].[B ] = k2.[C].[D] but for ammonia.

[thuanthuan]

If pressure change, neither k1,[A],[B ] nor k2,[C],[D] will change. (*) is still true. But why the old equilibrium is still broken ?

A + B -> C + D is not scheme of ammonia synthesis, as it doesn't involve volume changes. Write equation similar to k1.[A].[B ] = k2.[C].[D] but for ammonia.

No, I want to know WHY in general.

Alright, for example ammonia-producing reaction, then
k1[N2][H2] = k2 [NH3]
It's the same, I mean increase in pressure will not change k1,k2 or [N2][H2] or [NH3] (Assume that volume remain unchanged (thus so do the concentration). We increase pressure by adding noble gas into the container, for example)
Thank you. Sorry for my bad English.

[Borek]

No, I want to know WHY in general.

Then you were already answered by AWK.

k1[N2][H2] = k2 [NH3]

Nope. It is not N₂ + H₂ = NH₃ but N₂ + 3H₂ = 2NH₃.

It's the same, I mean increase in pressure will not change k1,k2 or [N2][H2] or [NH3] (Assume that volume remain unchanged (thus so do the concentration). We increase pressure by adding noble gas into the container, for example)

Who told you noble gas addition moves equilibrium?

[thuanthuan]

Who told you noble gas addition moves equilibrium?

I guess. Because the pressure then increases.

It's the same, I mean increase in pressure will not change k1,k2 or [N2][H2] or [NH3] (Assume that volume remain unchanged (thus so do the concentration). We increase pressure by adding noble gas into the container, for example)

That's what I don't understand. Can you explain why ?

[technologist]

I wonder why that statement is not correct.
A + B -> C + D + heat
A reacts with B to form C and D and evolving heat.
C react with D to form A and B along with absorb heat.
At equilibrium, the amount of heat evolved is equal to the amount of heat absorbed, so not heat is evolved anymore. Hence the temperature doesn't change.
What's wrong in my explanation ?

Nothing is wrong. If U consider an static equilibrium stage (Say in a closed container there will not be any net change in the temperature provided it is practically possible. So U r right - But it is only possible when other factors are also not affected or changed.

A + B  ->  C  + D
forward speed: k1.[A].[B ]
reverse speed: k2.[C].[D]
equilibrium:  k1.[A].[B ] = k2.[C].[D] (*)
According to my understanding, the impact of temperature change and concentration change is:
- If temperature changes, both k1 and k2 change. Thus left-hand side and the right-hand of (*) are no longer equal, causing equilibrium displacement.
- If concentration changes, in A, for example, right side of (*) remain unchanged, but its left side changes since [A] changes. Again the left and the right are not equal, new equilibrium will be established
- If pressure change, neither k1,[A],[B ] nor k2,[C],[D] will change. (*) is still true. But why the old equilibrium is still broken ?

This is a very good example of confusion. Mind it confused people are the ones who are ahead of others bcoz if everything is clear to you - means U dont understand it..

Define [A] or and u will get the answer. I already expressed that Kp & Kc concept.. your last sentence

- If pressure change, neither k1,[A],[B ] nor k2,[C],[D] will change. (*) is still true. But why the old equilibrium is still broken?

is not correct. In a non equimolar equation - How do you change pressure - By increasing or decreasing no of moles of any reactant, product or inerts....This is already mentioned by you, so no of moles / unit volume is changed.

However u need to consider that in case of non equimolar reactions the preesure term is also there in the Equilibrium equation....That is the point which U have to understand..

I know that you guys don't like anyone who keep asking such an easy question. But actually I don't understand your answer. Please help me. Thank you

This is not true as u can see everyone is trying to sort out your problem....But "I dont understand" these words and the difference in " My question was different" is tremendous....bcoz ultimately its your language which leads us to some idea about your problem...This is not a face to face communication...Anyway dont worry....keep asking.

[Borek]

Who told you noble gas addition moves equilibrium?

I guess. Because the pressure then increases.

If you rewrite reaction equilibrium formula in terms of partial pressures you will see that nothing has changed.

http://www.chemicalforums.com/index.php?topic=4442.0

[thuanthuan]

If you rewrite reaction equilibrium formula in terms of partial pressures you will see that nothing has changed.

http://www.chemicalforums.com/index.php?topic=4442.0

Thanks, now I know that I'm wrong.
Another question about total vs partial pressure:
http://www.chemicalforums.com/index.php?topic=19438.0


Actually, I don't understand what technologist say, so I need to learn more to understand it later. I still don't understand the impact of pressure change to equilibrium.
Thanks anyway.