Question

Classical thermodynamics with a twist.

Answer 1

Synthesis gas, a mixture of carbon monoxide and hydrogen gas is a very important industrial intermediate. Among other things, synthesis gas is used in the synthesis of methanol, ammonia and synthetic petrol. Synthesis gas is produced from coal and water at 1200 ° C:
C(s) + H2O(g) ⇌ CO(g) + H2(g)
In the attachment is found a table with all thermodynamic data you will need to do the following questions:

a) Calculate the equilibrium constant for the reaction at 25 ° C from the data in the table.

b) Calculate the equilibrium constant for the reaction at 1200 ° C from the data in the table. Consideration must be given to the standard reaction enthalpy and standard reaction entropy depends on the temperature, explain the best approximation done.

c) Calculate the equilibrium constant for the reaction at 1200 ° C from the data in the
table as in b), but use a simpler approximation.

d) Show that the fraction of H2O reacted, \(\alpha\), is given by the expression:
\[\large \alpha=\sqrt{\frac{ K }{ K+ \frac{ p }{ p^\theta }}}\] Here is \(p\) the total pressure.

e) How does this affect the fraction of water that has reacted, if we increase the pressure by 10 kPa by reducing the volume of the container? The answer must be justified.

f) The total pressure in the container is increased by 10 kPa by adding Ar (g). How does this affect the fraction of water that has reacted, if we assume that all gases behave as ideal gases? The answer must be justified.

g) How does this affect the fraction of water that has reacted, if we increase the temperature further.

Happy hunting.

Answer 2

Ok.

Answer 3

Here are my answers to the first two parts (a) and (b):

I started by using this:
\[K=e^{- \Delta G^{\ominus} / RT}\]

And got for part (a) \(K=e^{-(-228.6+137.2)*1000/(8.314*(273+25))} \approx 10^{16}\) which seems like a pretty big number, so since I don't actually know anything about this reaction, I guess that either means it goes to completion or I have messed up in converting my units somewhere.

Part (c) I can also do by the same exact method... \(K=e^{-(-228.6+137.2)*1000/(8.314*(273+1200))} \approx 1743\)

I'm not sure how to do part (b) and account for the adjustments... I'll have to look that up but tell me if I'm on the right track so far or if I'm leaving something out in my considerations.

Answer 4

Both correct.
For b) I will give a hint you are gonna love:
\[\Delta H(T)=\int\limits_{T_1}^{T_2} \Delta _rC_{p,m}^\Theta dT\]
And
\[\Delta S(T)=\int\limits_{T_1}^{T_2}\frac{ \Delta _r C_{p,m}^\Theta }{T}dT\]


Your state functions now depends on the temperature (in c. they didn't :D )

Answer 5

And if you want to be a showoff one write
We assume bla bla bla bla:
\[\large \left( \frac{ \partial C_p }{ \partial T } \right)_v=\left( \frac{ \partial ^2 H }{ \partial T^2 } \right)=0\]

Answer 6

Right no pV work so this is a really nice trick! :D

Answer 7

No wait I meant partial for p :P not v. typo :P

Answer 8

\[\large \left( \frac{ \partial C_p }{ \partial T } \right)_p=\left( \frac{ \partial ^2 H }{ \partial T^2 } \right)_p=0\]
Really fail showoff when I choose to write it wrong.

Answer 9

Ok instead of plugging in numbers I feel like it would be better to just show what equations I'll use since I'm already taking a lot of time.

\[K = \exp(-\Delta G^{\ominus}/RT_2)\]

after integrating I'll plug in for \(\Delta G^\ominus\):

\[\Delta G^\ominus = \Delta C^\ominus_{p,m} (T_2-T_1) - T_2 \Delta C^\ominus_{p,m}\ln\frac{T_2}{T_1} \]

When you put a subscript 'r' on the delta, what does that mean? \( \Delta_r C^\ominus_{p,m}\)

Answer 10

r stands for reaction.
"The change in heat capacity for the reaction"

Answer 11

So to be clear, that's going to be my answer for (b),

\(T_2 =1200 C\)
\(T_1 = 25 C\)

I think everything else is able to be looked up in the table. I was planning on calculating those delta C's from the reaction by adding the product C's and subtracting the reactant C's so I guess that makes sense that it's "r" I guess what my real question is... what would be an alternative heat capacity change?

Answer 12

I guess what I'm saying is in this context it seems clear since there's no other way it can change here... is there?

Answer 13

Well not really. also remember we use the MOLAR heat capacity, it no longer depends on the amount of substance in the system, we've ensured that by normalizing 1/n :)

Answer 14

So as the reaction move forward it doesn't really matter much as it is just per mol times it runs.

Answer 15

Kind of like the specific heat capacity, which is normalized by the mass instead of moles, right? :)

Answer 16

exactly. Also classical thermodynamics operates near equilibrium usually.

Answer 17

many of the equations has the assumption that they assume equilibrium, they just never mention it in texts.

Answer 18

True, so have you ever seen something where dealing with the intensive heat capacity was actually not a good enough approximation?

Answer 19

I mean where it was independent of temperature.

Answer 20

Yes :)

Answer 21

I would imagine as you get down smaller to a particle or molecule basis that would become quite important and interesting, could you even determine it geometrically with bond strengths and degrees of rotational freedoms?

Answer 22

I was trying to determine enthalpy of fussion one day for methanol (when I had experimental physical chemistry), we made a bet who could hit the best value compared to the literature text, I choosed to assume the heat capacity dependent on the temperature and hit a more accurate value

Answer 23

You can think about it this way:
the smaller your temperature interval is, the better is the approximation.

Answer 24

Haha awesome. That sort of stuff seems quite interesting, but I guess I'm not quite done with this problem yet.

So I am not sure how to show the expression for \(\alpha\) but I am sort of playing with the equilibrium expression. However I can do the next 3 parts I'm fairly sure:

For part (e) Increasing the pressure will make \(\alpha\) much smaller since it's in the denominator which means the fraction of water that's reacted to is small, probably close to none at this high pressure.

Part (f) is fun but should have the exact same effect as part (e) since it will just be adding partial pressure. Since it's assumed to be an ideal gas, that means there's no interaction between the particles, and since it is Argon it's an inert gas so won't react either.

(g) From \(K=e^{-\Delta G/RT}\) we can see that increase in temperature will continue to make \(K \to 1\) so there will be equal concentrations of products and reactants.

Ok how'd I do?

Answer 25

Correct for (e) and (g) but (f) is funny :)
In thermodynamics the "total pressure" we really mean "the systems total pressure"
Ag (g) is not a part of the system :)

Answer 26

Yeah I was considering it to be the same as if we had squeezed the walls in on the gas, it should be indistinguishable I think?

Answer 27

Yeah I thought so too (I admit when I answered this popquiz I got that one wrong) but because the total pressure is the sum of the partial pressures, it shouldn't make a difference.

Answer 28

Interesting. I wonder how true this is in an actual experiment or if this is more like "thermodynamic math oddity"

Answer 29

31. of august I meet my prof who did this popquiz then I'll ask for a experimental proof :)

Answer 30

Hahaha alright. So I guess there are two open things here. I'm not entirely sure how you used that second derivative of enthalpy to solve this and I still don't know how to show \(\alpha\) is true. Can you give me some hints?

Answer 31

About (d) I can help a little:
\[\Large K=\frac{ \frac{ p_{CO} }{ p^\Theta } \frac{ p_{H_2} }{ p^\Theta }}{ \frac{ p_{H_2O} }{ p^\Theta } }\]

For the amount of substance of water we can write \(n_{start}\)

1) Relate \(n_{start}\) to the amount of substance in equilibrium for H2O, CO and H2

2) Using the equations of 1) relate them to the molfraction in equilibrium

3) Using the equations in 2) relate them to the partial pressure in equilibrium

3) insert into the big fast equation I wrote.

Answer 32

Weird, I am not sure I can wrap my mind around this.

I guess since they are ideal gasses this does make sense in the model. Adding particles of argon that are ideal won't hit the other molecules so in no way does this added pressure somehow mimic the pressure of "squeezing the box".

The weird thing for me is imagining the argon in the box having a pressure and the actual space being removed to create that same pressure. But somehow because that volume of space is not pushed all against the side and it's really all floating around in there it doesn't affect it. Or something. I guess it's just the model and I shouldn't take it too seriously. Fun :P

Answer 33

I had forgotten that I could represent the equilibrium constant for gasses in terms of partial pressures. :O

Answer 34

Specifically when you write \(p^\ominus\) what does this mean?

Answer 35

Ahhh I just remembered there are these things called activity coefficients. I have a lot to study.

Answer 36

A true badass thermodynamic chemist would ALWAYS have his equilibrium constant without a unit so we use the standard pressure (1 bar) to get rid of the unit.

Answer 37

Hahaha alright. 1 bar that's about 101kPa is that right?

Answer 38

Makes sense though since equilibrium constants are truly ratios, now I at least know that much haha.

Answer 39

that is atmospheric :)
10^5 Pa. = 1 bar
Also written it in the notes you were reading :D

Answer 40

Why 1 bar became the standard pressure I don't know.... Guess we need to ask the 1800s Germans about that.

Answer 41

I recall using mmHg, pascals, and barr at different points throughout my chemistry labs depending on what the measurement was written on the barometer in which lab we were in lol.

Answer 42

Lol I only use SI units, but we all know that rockets blow up when people can't agree to units :P

Answer 43

Well I generally only use SI units but I live in the US so we have problems. I though Pascals were SI units since \(1Pa = 1 N/m^2\)

Answer 44

Pascal is the SI unit :P that is why I find it funny they use 1 bar

Answer 45

Alright you wrote out a nice thing of how to relate things together, unfortunately I don't really know what relations actually exist between those quantities.

\[K=\frac{[CO][H_2]}{[H_2O]} \]

(for some reason we don't include C because it's a solid... Which reminds me there is something called Ksp that I've forgotten as well. As far as my intuition is concerned, changes in moles of gas contribute to entropy changes I believe. I'm kind of scattered here.

Perhaps I can get the concentrations here by plugging this in:
\[ \frac{n}{V}= \frac{p}{RT}\]

And then I can try to work out partial pressures? I feel like this might not be quite right.

Answer 46

Maybe this could do the trick. see attachment :)

Answer 47

Remember Dalton's law of partial pressures? :)

Answer 48

Oh wow this is really quite basic, I haven't seen one of these in a while. I do remember Dalton's law, that the total pressure is the sum of the partial pressures, which means the pressures the gasses would have in the container if none of the other gasses were present. Actually I think understanding this was the key to understanding part (f) or whatever with the argon.

Answer 49

Exactly yes. I wrote in the message that this question was a trick to it. (in terms of math at least)

Answer 50

As I know you are superiour good with mathematics in general, I guess the question was originally meant to be hard as you no longer had numbers to support you, but is instead left alone with a theoretical situation.

Answer 51

I have completely forgotten how to do these tables, but this is great practice I appreciate it. I'm much more comfortable with abstract values at least so this kind of question isn't particularly hard thankfully.

So when I see this table I'm sort of lost at the 3rd row. Where is the mol fraction in equilibrium coming from? I think specifically that 1 is throwing me off. Thanks again for helping me out, by the way.

Answer 52

How can I explain it the best way....... Think it is best I just write the definition of the mole fraction, then I think you see why (just constantly have in mind what alpha is):
\[\Large x_i=\frac{ n_i }{ n_{total} }\]
So to take the first one:
How much do we have left....
In the start the fraction is equal to 1, and then alpha is the fraction of H2O reacted.
So \(n_i\) must be equal to 1-alpha. The total must be then 1+alpha as the reactions precedes. Maybe you can explain it better if you understand what I mean :P

Answer 53

\[\Large \sum_{i}^{N}x_i=1\]

Answer 54

Ahhhh right ok so it's essentially normalizing them so that when we add them we get 1. I guess I'm confused about why the whole first column C(s) has nothing in it except that one spot, where is this particular thing coming from? Everything else though I see clearly what you are saying, I think putting the mathematical formula was a good idea. :D

Answer 55

Well, it is because the activity of pure solids are equal to 1 :P

Answer 56

The equilibrium expression can't be applied to heterogeneous reactions, so we discard the solids and focus on the gases.

Answer 57

But what did you think of the questions in general? :)

Answer 58

About the solids, I guess it is just one of the things we must accept, I was nicely told to shut up when I asked why it was the case.

Answer 59

I thought it was fun, I really like the trickiness. I think I really like the format of it too, it's very short and I really like how it contrasts the different ways of changing the pressure or changing temperature makes you really use a kind of wider range of knowledge.

More like this any day you want!

Answer 60

Haha yeah I feel like I never was told how to derive most of the thermodynamic things I encountered before physical chemistry. Specifically I would like to understand how the equilibrium expression came to be in the first place, I don't particularly see why we should multiply the products over the reactants, it might as well just be addition to me.

Answer 61

Well the equilibrium expression is also a simplification. The REAL expression we should use is the activity of each specie.

Answer 62

Same goes for pH (I got a question for that)

Answer 63

To see how pH is approximated with the concentration instead of activity and how big the difference is,

Answer 64

can post it as the next one if it is.

Answer 65

(We just use Debye–Hückel theory to look at the difference)
https://en.wikipedia.org/wiki/Debye%E2%80%93H%C3%BCckel_theory

Answer 66

Yeah, I think I recall that generally the activities sort of aren't important (I don't even remember what they are other than some multiplicative factor of the concentrations???) But maybe they're some kind of chemical or electrical potential thing. I wouldn't mind seeing something like that for pH, sounds like you have a bunch of fun problems just lying around haha awesome

Answer 67

I do have that.
But fast so we don't forget anything!

Answer 68

Solution above so I don't cheat you for it.

Answer 69

3 mrd... before the question got answered... my soul can rest now.

Answer 70

Ahahaha woah I didn't even realize that.

Answer 71

Also thanks for the solution I think I'll just sorta try to do this real quick from scratch tomorrow and see if I still got it. I think most of this question is now pretty sturdy in my brain though, which is good haha.