Chemistry 3/4 2013 Thread

[lzxnl]

That's what I've been told.

[Alwin]

Yes, I've been told similar things. However, we were also told by our chem teacher that vcaa can put generic qs on the exam, eg:
N₂ + 3 H₂ → 2 NH₃   (ΔH = −92.4 kJ·mol⁻¹)
and ask for the conditions that will produce greatest yeild. If it was a multi choice q, the options would be something like: low temp + low pressure, low temp + high pressure, high temp + low pressure, high temp + high pressure.

Technically, the equations comes form the Haber Process, but its taken out of context and allowed. (you should be able to do that question easily)

[lzxnl]

lol react this over a copper solution. Even more yield as the ammonia complexes with the copper ions

[Aurelian]

lol react this over a copper solution. Even more yield as the ammonia complexes with the copper ions

Yeah this probably won't work as well as you might think it would. The standard operating temperature for the Haber process is around 300 - 500^oC, and at these temperatures the solubility of NH₃(g) in water is extremely low. As a result I wouldn't suspect that NH₃(g) would coordinate very stably to Cu²⁺(aq), or whatever other metal ions you had floating around.

Even ignoring this practical consideration, there are others. For instance, if you coordinate most of your ammonia to a metal cation, it isn't going to be very much use to you, is it? You're going to need at least one additional step later on to free up the ammonia anyway, which is just a pain.

[lzxnl]

I was joking initially. If we try to take this seriously, the water would probably evaporate under such high temperatures and we wouldn't have a copper solution. That, I think, is the main issue. I don't think the solubility of ammonia will be a problem as metal complexes have massive stability constants, 10^12 for tetraamine copper(II), so any ammonia that does dissolve will react, allowing more ammonia to dissolve. Less ammonia left in the air would mean the forming of more ammonia.

As for hypothetically trying to get the ammonia out of solution, we could precipitate the copper with something like hydroxide ions as the solubility product for copper hydroxide is on the order of 10^-20.
But really, I was kidding when I made the initial statement.

[Aurelian]

I was joking initially. If we try to take this seriously, the water would probably evaporate under such high temperatures and we wouldn't have a copper solution. That, I think, is the main issue.

I actually don't see why this has to be an issue... It's the gas which will necessarily be at the high temperature, not the water. If you have a sufficiently large body of water into which you're feeding the NH₃ and appropriate cooling systems in place, I don't see why the water would necessarily get too hot and evaporate. That said, of course, I'm not an engineer so there may be other practical considerations which I'm overlooking.

I don't think the solubility of ammonia will be a problem as metal complexes have massive stability constants, 10^12 for tetraamine copper(II), so any ammonia that does dissolve will react, allowing more ammonia to dissolve. Less ammonia left in the air would mean the forming of more ammonia.

I don't imagine you could use that stability constant in this sort of situation given that the NH₃(g) molecules would have unusually high kinetic energies (comparatively speaking); it's a more complicated system than those for which stability constants are typically stated.

As for hypothetically trying to get the ammonia out of solution, we could precipitate the copper with something like hydroxide ions as the solubility product for copper hydroxide is on the order of 10^-20.
But really, I was kidding when I made the initial statement.

I never said you couldn't do it, or that it would be overly complicated to do it. My point simply concerned why you'd do it given that at least one additional step would be required (which would likely decrease the economic efficiency of the process as a whole). Moreover, with your hydroxide treatment you'd end up with a heap of waste Cu(OH)₂!

Anyway, I'm glad you were just kidding! Really good to see you thinking about this stuff though =)

[lzxnl]

I actually don't see why this has to be an issue... It's the gas which will necessarily be at the high temperature, not the water. If you have a sufficiently large body of water into which you're feeding the NH₃ and appropriate cooling systems in place, I don't see why the water would necessarily get too hot and evaporate. That said, of course, I'm not an engineer so there may be other practical considerations which I'm overlooking.

But if you have the gas at 300 degrees trying to dissolve into the water, the gas will cool down and the water will heat up. I was thinking about having a large body of water as well, although if you have heaps of gas, it could be a problem.

I don't imagine you could use that stability constant in this sort of situation given that the NH₃(g) molecules would have unusually high kinetic energies (comparatively speaking); it's a more complicated system than those for which stability constants are typically stated.

Isn't the main difference just the temperature though? The Van't Hoff equation would relate the new stability constant to the old and I really doubt the constant will drop by so many orders of magnitude. Besides, when the ammonia meets the water, its thermal energy will be spread out through the volume of the water and if we have heaps of water, the water won't heat up as much, so ideally we won't even have water that's too hot.

I don't have an answer for the hydroxide problem though. Is there a use for copper(II) hydroxide somewhere?

I never said you couldn't do it, or that it would be overly complicated to do it. My point simply concerned why you'd do it given that at least one additional step would be required (which would likely decrease the economic efficiency of the process as a whole). Moreover, with your hydroxide treatment you'd end up with a heap of waste Cu(OH)₂!

Anyway, I'm glad you were just kidding! Really good to see you thinking about this stuff though =)

And yeah, economics would be a major issue for this entire process. Still, a question on this would be more fun that what VCE asks us to do

[Stick]

How important is Chapter 18 of the Heinemann textbook (the chemical industry: risks and rewards)? I don't want to spend too much time on it, but I realise that it might creep into the exam in a couple of places.

[lolipopper]

How important is Chapter 18 of the Heinemann textbook (the chemical industry: risks and rewards)? I don't want to spend too much time on it, but I realise that it might creep into the exam in a couple of places.

It only talks about general things about increasing the yeilds and some compromises that a scientist must make to get that optimum yield. The health and safety stuff is likely not of a real concern. From what my teacher told me, read over it and just pay attention to the yeild stuff. They often seem to ask questions about how to improve yield etc. in exams.

[itsdanny]

I'm aware that changes in concentration (resulting from i.e. an addition of a reactant) doesn't change the K_eq, although does it affect the reaction quotient?

[lzxnl]

ONLY temperature affects the equilibrium constant. Adding reactants or products does affect reactant quotients, yes.

[Edward21]

What is the point of adding an inert gas may i ask?? It doesn't change the concentration of the mixture in equilibrium, and although it affects pressure inside the container, it can't shift the equilibrium position according to the textbook, what is the point/? Wouldn't it slow the rate of reaction down as reactants collide with some of the wrong inert gas particles than each other?

[lzxnl]

The point is to trick poor souls like you into believing that adding an inert gas does something. The concentrations of the gases do not change, and the partial pressures of the gases do not change either (same number of moles, same volume, same temperature, so the pressure the gases exert is identical to before)

As for slowing rates of reaction down, how do you know that the inert gas won't slow down the reverse reaction as well?

[Edward21]

The point is to trick poor souls like you into believing that adding an inert gas does something. The concentrations of the gases do not change, and the partial pressures of the gases do not change either (same number of moles, same volume, same temperature, so the pressure the gases exert is identical to before)

As for slowing rates of reaction down, how do you know that the inert gas won't slow down the reverse reaction as well?

Well how about overall rates of reaction?
"trick poor souls"? I shall not be tricked by this VCAA 

[lzxnl]

I don't think you need to worry about inert gases. For the purposes of VCE, they have no effect.