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  1. I wanted to understand application of LeChatelier's principle for reversible reactions of type $\ce{A(g) <=> B(g) + C(s)}$. Because of increase in moles, you can say lower pressure will favor the reaction going right. But does this apply even if one of the products is solid, like in this reaction? If not, why? I need to understand this intuitively, if possible, with less math.

  2. Another related question I have is we talk about reactions having thermodynamic limitations when considering their practical applicability. i.e say a reaction such as above might only yield 20% conversion at industrially practical conditions. But isn't this a bit misleading? As long as you find a way to remove products as soon as they formed, you can push ANY reversible reaction to the right. So for all practical purposes you can say any reversible reaction can be designed to give high conversions (80-100%) if you devise a mechanism to remove products quickly from reaction space. So the thermodynamic limitation is not really a hard limitation at all.

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  1. I wanted to understand application of LeChatelier's principle for reversible reactions of type $\ce{A(g) <=> B(g) + C(s)}$. Because of increase in moles, you can say lower pressure will favor the reaction going right. But does this apply even if one of the products is solid, like in this reaction? If not, why? I need to understand this intuitively, if possible, with less math.

Let's get back to the fundamentals of Le Chatelier's principle :

When any system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to counteract (partially) the effect of the applied change and a new equilibrium is established.

Let's assume you have a reaction $\ce{A(g) <=> 2B(g)}$. When lowering the pressure, the system will adjust itself to counteract this, i. e. increase the pressure. This can be achieved by displacing the equilibrium towards the right, as the direct reaction increases the amount of gas molecules.

In your case, however, there is no increase in the amount of gas compounds. Regarding the solid, as it does not participate to the overall pressure in the system (only gases do), it should not be taken into consideration here. Therefore, a change in pressure wouldn't alter the equilibrium state of your reaction.

That's for the math-less part.

You can also understand this concept using the equilibrium constant. For my example, it would be $K_c = \frac{[\ce{B}]^2}{[\ce{A}]}$. Lowering the pressure corresponds to an increase in volume, thus a decrease in both concentrations. This leads to a ratio that is different from the equilibrium constant, which will force the system to a new equilibrium. $$K_c = \frac{[\ce{B}]^2}{[\ce{A}]} = \frac{\left(\frac{n_\ce{B}}{V}\right)^2}{\frac{n_\ce{A}}{V}} = \frac{n_\ce{B}^2}{n_\ce{A}}\cdot\frac{1}{V}$$ When increasing $V$ to $V'$, $$\frac{[\ce{B}]^2}{[\ce{A}]} = \frac{n_\ce{B}^2}{n_\ce{A}}\cdot\frac{1}{V'} < K_c$$ As this ratio is lower than $K_c$, the reaction will evolve towards the formation of products.

For your equation, $$K_c = \frac{[\ce{B}]}{[\ce{A}]} = \frac{\frac{n_\ce{B}}{V}}{\frac{n_\ce{A}}{V}} = \frac{n_\ce{B}}{n_\ce{A}}$$ there is no dependance on the volume, thus none either on the pressure. C is not included as it is a pure solid ($a(\ce{C}) = 1$).

  1. Another related question I have is we talk about reactions having thermodynamic limitations when considering their practical applicability. i.e say a reaction such as above might only yield 20% conversion at industrially practical conditions. But isn't this a bit misleading? As long as you find a way to remove products as soon as they formed, you can push ANY reversible reaction to the right. So for all practical purposes you can say any reversible reaction can be designed to give high conversions (80-100%) if you devise a mechanism to remove products quickly from reaction space. So the thermodynamic limitation is not really a hard limitation at all.

It is true that, if you find a way to remove products from the system, you can reach full conversion of your reagents. But that's actually the catch : it can be very hard to remove those products. Take as an example the Haber-Bosch process : $$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$ As it is a slow reaction, a high temperature is required to get the product in a reasonable time. How would you remove $\ce{NH3}$ from the system, leaving the reagents inside? We then use other ways of increasing the yield (pressure and temperature), but they do not allow a full conversion of the reagents.

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  • $\begingroup$ Thanks Thomas. I agree that it is not always practically possible to remove products easily (ie like using a simple membrane reactor and so on), but my question was to show that at least theoretically no reaction has an equilibrium limitation such as, the limitation is more about inability to devise practical means to quickly remove products. And may be this where a challenge can be converted into an opportunity, to innovate. $\endgroup$
    – daraj
    May 29 '17 at 9:08
  • $\begingroup$ Thomas and anybody else, I actually have this doubt about removing reaction products from reaction space part. In say a gas-solid catalytic reaction, or gas-liquid, isn't the product gas immediately removed from reaction space just by virtue of constant flow through the reactor? If I had a reaction say A(g)-->B(g) happening on a solid catalyst surface in a fixed-bed reactor, as soon as B is formed it is not going to stay on catalyst surface but come out of the reactor due to gas flow anyways. So are we rather talking about SELECTIVE removal of product alone from reaction space? $\endgroup$
    – daraj
    May 29 '17 at 19:39
  • $\begingroup$ @daraj To increase overall yield, you would have to circulate the reaction mixture while continuously removing products. The Haber Bosch process does that. $\endgroup$ Nov 21 '19 at 10:34
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"Because of increase in moles, you can say lower pressure will favor the reaction going right" - no, in your equation it is the same on both sides and favors neither. But if it had been, it would have applied regardless of solid products. The volumetric influence of 1 unit of gas is significantly higher than 1 unit of solid, and as such the unit of solid will never impact the partial pressure of any of the gases. The best you can argue for it is that it can, in an enclosed reaction chamber, cause a negligible increase in system pressure.

For some combinations of reactants you can see the limitations when you draw the phase diagram - if you apply Gibbs Phase rule and have a mind to check for incompatible coexistence you will find that some reactions are thermodynamically limited - the combination of reactants necessary to produce something will instead produce something else. So if you "remove all products" as you say, you end up producing something you weren't supposed to produce.

edit: An example. (Please, thermodynamic wizards clean up my notation, it is too long since I were in school to remember what was the G$^0$'s and G's and there is probably multiple formal errors in here...)

So, we would like to make Silicon from what the earth gives us, Quartz - $\ce{SiO2}$, and coal - $\ce{C}$

We put quartz and coal together and heat it up, we know something will happen because

$\ce{SiO2(s) + 2C(s) <=> Si(l) + 2CO(g)}$

has a positive $\Delta$G so heating it up will force something to happen. But, to our surprise, no matter how vigorously we (try to) remove the carbon monoxide and the silicon, we only find silicon carbide $\ce{SiC}$. Turns out that

$\ce{Si(s,l) + C(s) = SiC(s)}$

There are competing reactions happening, disturbing our equilibrium, consuming our product. But, chemists as we are, we blow oxygen into the mix, thinking - now we will get our silicon, because

$\ce{2SiC + O2(g) <=> 2Si(l) + CO}$

and that has a negative $\Delta$G no less, so we will regain our losses! Again, we are stymied, getting only Silicon monoxide gas $\ce{SiO}$.

$\ce{2Si(l) + O2(g) = 2SiO(g)}$

But, we are closer to our target. What if we can react these two different products together? Like...

$\ce{SiO(g) + SiC(s) <=> 2Si(l) + CO(g)}$

Well, now we got our precious Silicon, but the trouble we had to go through. And, we now reach our understanding - it is impossible to produce silicon without also producing the intermediaries. And we have to be careful how we treat the intermediaries because SiO and CO are quite sensitive and there are circumstances (like availability of C(s) or Si(l)) that can drive opposing reactions with the intermediaries.

$\ce{SiO(g) + C(s) <=> SiC(s) + CO(g)}$

It turns out that for this (these) specific reaction(s) you just simply cannot have 100% yield. The conditions that are optimal requires you remove some of the SiO and all of the CO, and you only get a 90-something % yield. This is one example of a reaction that is thermodynamically favorable but impossible to "drive all the way". I am just a sledgehammer chemist, there are certainly organic chemists that can tell you that a 90% yield is worthy of a celebration, many thermodynamically inhibited reactions yield less.

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  • $\begingroup$ "if you apply Gibbs Phase rule and have a mind to check for incompatible coexistence you will find that some reactions are thermodynamically limited - the combination of reactants necessary to produce something will instead produce something else." you mean the reaction will be spontaneous but in a different direction? i.e it will yield different products? can you elaborate? if this is the case it should be that way from the beginning itself,not just at equilibrium $\endgroup$
    – daraj
    May 29 '17 at 9:06
  • $\begingroup$ I'll try to edit in an elaboration $\endgroup$ May 29 '17 at 10:14
  • $\begingroup$ Stian , please also address the question I added above in response to Thomas $\endgroup$
    – daraj
    May 29 '17 at 19:43

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