The Haber Process Question (1 Viewer)

[hobbsy4]

Yeah that graph's right.

What u on seadonkey? I was saying that, u just have to interpret what i meant in graphical terms as in H2 N2 decreasing, NH3 increasing.

 

[Ultraviolent]

I believe it looked something like this:

edit: same deal as the one above

 

[queenie]

your all right, however, they decrease in the same ratio of their no of moles (i think thats what seadonkey was saying), eg, NH3 increased 2 cm, hydgrogen decreased 3cm, and nitrogen 1cm... does that make sense?

 

[seadonkey]

Yeah that graph's right.

What u on seadonkey? I was saying that, u just have to interpret what i meant in graphical terms as in H2 N2 decreasing, NH3 increasing.

umm u do know concentration is inversely proportional to volume

conc = moles /volume
so if volume decreases conc increases
ALL reactants increase, the u take in2 the equil shift of ammonia

 

[Captain Gh3y]

Hmm... I didn't make any measurements, I just drew it freehand. (they were nice looking curves though...)

I don't know if they can mark us down on that, because there's nothing quantitative on Equilibrium in the syllabus. (or maybe there is in Industrial.. i dunno.)

 

[queenie]

Hmm... I didn't make any measurements, I just drew it freehand. (they were nice looking curves though...)

I don't know if they can mark us down on that, because there's nothing quantitative on Equilibrium in the syllabus. (or maybe there is in Industrial.. i dunno.)

lol, im sure they were nice looking curves

no, theres nothing in industrial, but industrial kids are more exposed to that sorta thing, so they woulda done it better i suppose

 

[KFunk]

Awesome, that's exactly how my graph looked. Provided yours is right I'll be sitting pretty.

 

[console]

as volume decreases, pressure increases....so, the equalibrium shifts to the right to eliminate molecues (i.e. N2 and H3) and more ammonia is produced..

so, in de graph the presence of H2 and N2 DECREASES (i.e. \----\)

with ammonia it INCREASES (i.e. /----/)

 

[Abtari]

the only reason why the curve at the very beginning of time "slopes slowly" is because an equilibrium hasn't even been established yet. that is to say, it is in the process of forming equilibrium, with concentration of hydrogen and nitrogen decreasing and that of ammonia increasing. however, after T2, an equilibrium having been established, the decrease in volume would mean an increase in pressure, and an IMMEDIATE response, in accordance with Le Chatelier's principle, to counteract PARTIALLY the imposed change.

so, in my opinion, the graph would not have sloped after T2, rather shot straight up, partially, for ammonia and shot straight down, partially for hydrogen and nitrogen - thereafter forming equilibrium at a different concentration. And also another point i might add is the ratio of the lengths of how much you 'increase' concentration of ammonia and how much you 'decrease' concentration of hydrogen and nitrogen. they should be in the ratio 2:3:1 respectively (or at least an approximation to indicate you understand the concept of gaseous volumes relating DIRECTLY to molar concentrations)

 

[queenie]

the only reason why the curve at the very beginning of time "slopes slowly" is because an equilibrium hasn't even been established yet. that is to say, it is in the process of forming equilibrium, with concentration of hydrogen and nitrogen decreasing and that of ammonia increasing. however, after T2, an equilibrium having been established, the decrease in volume would mean an increase in pressure, and an IMMEDIATE response, in accordance with Le Chatelier's principle, to counteract PARTIALLY the imposed change.

so, in my opinion, the graph would not have sloped after T2, rather shot straight up, partially, for ammonia and shot straight down, partially for hydrogen and nitrogen - thereafter forming equilibrium at a different concentration. And also another point i might add is the ratio of the lengths of how much you 'increase' concentration of ammonia and how much you 'decrease' concentration of hydrogen and nitrogen. they should be in the ratio 2:3:1 respectively (or at least an approximation to indicate you understand the concept of gaseous volumes relating DIRECTLY to molar concentrations)

i agree with the second part of your post; that the increase/decrease is proportional to the molar concentrations, but when increasing/decreasing the pressure the concentrations don't shoot up/down. They only/shoot up or down when your increasing/decreasing the concentration. increasing/decreasing the pressure is a process that takes time, and thus, it would be sloped.

 

[Casmira]

It was a "sketch" question we don't have to make it a perfect concave

 

[jarro_2783]

The only time a graph shoots up is when you add more reactants or products such as to increase the concentration.
eg. When you are making an ester, say you have pure ethanol and 1mol/L ethanoic acid and it gets to equilibrium then you add 10mol/L ethanoic acid it will shoot up because the concentration has suddenly increased when you add the ethanoic acid like this:


But the haber process one doesn't shoot up because you haven't actually changed the concentration. You have changed the pressure, there is a difference. When you change the pressure there is still 10mol of N₂, 10mol H₂ and 4mol NH₃ (for example). But because there is a pressure increase the system reacts to oppose the pressure change, so there is no sudden jump, there is a slow reaction to oppose the pressure change which results in a curve.

 

[~ ReNcH ~]

My answer.

My explanation:

As the volume of the container is decreased, the pressure on the system is increased. N2 + 3H2 <--> 2NH3. The reaction that produces the least number of moles of gas will be favoured. Hence, more ammonia will be produced. The equilibrium moves to the right.

Ok, cool.
That's pretty much what I had...although I didn't have the bottom two intersecting: is it necessary to have them intersecting?

Btw. I pretty much had the same explanation...was it necessary to say that the equilibrium was established again? I said that it shifted to the right, but I didn't actually say that it was "established again", so I don't know if that's a sufficient explanation of the graph itself.

 

[seadonkey]

The only time a graph shoots up is when you add more reactants or products such as to increase the concentration.
eg. When you are making an ester, say you have pure ethanol and 1mol/L ethanoic acid and it gets to equilibrium then you add 10mol/L ethanoic acid it will shoot up because the concentration has suddenly increased when you add the ethanoic acid like this:


But the haber process one doesn't shoot up because you haven't actually changed the concentration. You have changed the pressure, there is a difference. When you change the pressure there is still 10mol of N₂, 10mol H₂ and 4mol NH₃ (for example). But because there is a pressure increase the system reacts to oppose the pressure change, so there is no sudden jump, there is a slow reaction to oppose the pressure change which results in a curve.

all products AND reactants DO shoot up...
do u do industrial chem?? cos if they DONT shoot up how can u maintain a constant K value if the equil shifts? only way possible is if they all rapidly increase, then shift 2 favour ammonia

that is ALL reactants rapidly increase in conc

remember...
conc = moles/volume
therefore decrease in volume = increase in conc

 

[jarro_2783]

oh yeah, derr, you're right. That means the graph should shoot up for all molecules and then reestablish equilbrium by ammonia going up and h and n going down.
So the graph everyone is taking as right is actually wrong. It should look like this:

 

[fadykozman]

hello

just for your info i believe you must have shown that the decrease of H2 was three folds the decrease of N2, as the molar ratio is 3 : 1...

I believe thats already one mark for the drawing...who knows, may be i am wrong, but it didnt hurt to show it and write it in pencile next to the graph

 

[fadykozman]

Dear SeaDonkey

all products AND reactants DO shoot up... do u do industrial chem?? cos if they DONT shoot up how can u maintain a constant K value if the equil shifts? only way possible is if they all rapidly increase, then shift 2 favour ammonia that is ALL reactants rapidly increase in conc remember... conc = moles/volume therefore decrease in volume = increase in conc

The equilib constant doesn't change with change in volume or concentration, it only varies with the temperature, thus the basis of your argument are gone!

I can understand your point that concentration of all products and reactants increase.

 

[Captain Karl]

hello

just for your info i believe you must have shown that the decrease of H2 was three folds the decrease of N2, as the molar ratio is 3 : 1...

I believe thats already one mark for the drawing...who knows, may be i am wrong, but it didnt hurt to show it and write it in pencile next to the graph

I had that change according to the molar ratio as well - forgot to say that the eqm was reestablished but. Is the bit about the molar ratio right?

 

[fadykozman]

i think so...look at the equaiton

N2 + 3H2 <...> 2NH3, i reckon, if they dont give a mark for this then i am gonna kill somebody, they should be testing us on our understanding of the process of equilibrium, it makes perfect sense.

cheers..

 

[Captain Karl]

Thanks. I put it in there with that same logic and then put in an explanation. It made sense to me like that, I was waiting for someone to comment on it.