« on: 25/05/2020 02:06:33 »
The National Parks Service has virtual tours of many of the parks, including the Grand Canyon, which may be useful: https://www.nps.gov/grca/learn/photosmultimedia/virtualtour.htm
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
Those, on the other hand, are not equilibrium systems.Indeed, but neither are most systems of interest. (systems at equilibrium are rather useless compared to their non-equilibrated counterparts)
There will always be some small fraction that have enough energy.Not necessarily. It is relatively common for molecules to have excited electronic states on the order 2 eV above the ground state (about 45 kcal/mol, if you prefer). At a temperature of 20 °C, we would expect the ratio of molecules in this excited state to molecules in the ground state to be on the order of 10–35 (requiring about 100 billion moles of sample to have one single excited molecule), so if we limit ourselves to samples that can fit inside an ocean liner, there will be absolutely zero molecules in this excited state at any given point in time, as long as the sample remains in the dark. Yet, as soon as you expose it to sunlight (or a lamp), this excited state can be populated.
(By analogy, LEDs can be used as photodiodes- they are lousy at it, but have the advantage of differing spectral response.)Precisely!!! in principle these things are reversible, but somehow, in reality, they aren't... I wonder why... it's almost as if the world is not made of ideal systems that readily achieve equilibrium, and that machines we make (macroscopic or molecular) can be inefficient at doing things, but more inefficient one way than in the reverse (think of pushing or pulling a mass with a string). LEDs make lousy photodiodes and PV cells make lousy LEDs, because there are ways of inherently tipping where the inefficiencies lie.
Similarly, an electrochemical cell is reversible (in principle)
Not if you have to push it both ways...The point is that you don't.Catalytic cycles with multiple elementary steps don't have to work equally well in both directions.Yes they do.
Do you accept that, if you have an equilibrium (and the principle of microscopic reversibility says you always have), then the position of the equilibrium is the point where the backward and forward reactions have the same rate?
Do you accept that if the catalyst increased the rate in one direction more than the other, it would change the equilibrium constant?
you have to push it both ways
(plus most catalysts don't actually work just as well in both directions)Yes they do.
It was noted that Ir/C has poor ORR activity, but good OER activity whereas Pt/C exhibited excellent ORR activity, but very low OER activity.https://www.sciencedirect.com/science/article/abs/pii/S092633731730735X
If you had a catalyst that favoured one reaction, but not the other, it would change the position of equilibrium and you could make a "free energy" machine.
Whoops, sorry, typo! Meant 27 °C (I'll change it).and 300 K (23 °C),300K = 26,85 °C.
My question contains 3 substances, but I can omit the nitrogen to simplify the calculation.
Can you tell me how much helium will be in the top half of the pipe, how much SO2 in bottom half of the pipe after reaching equilibrium?
How much g-force is required to get 75% SO2 in the bottom half of the pipe?
Even in a sealed container with no temperature gradient, you won't get a useful degree of separation.What do you think will happen if I put a mixture of Helium, Nitrogen, and SO2 with equal volume and pressure inside a 10 meters vertical pipe. Will we get the same composition between top and bottom part?