# What does Keq tells us?

1. Nov 12, 2017

### dRic2

Usually we say that if a reaction $$aA + bB ⇔ cC + dD$$ has $K_{eq} > 1$ the equilibrium is reached with more products than reagents. Otherwise if $K_{eq} < 1$ the inverse reaction is favoured.

Now let's consider the synthesis of ammonia: $$1.5H_{2} + 0.5N_{2} = NH_{3}$$

From letterature we know that @ T = 600K $ΔG_{R} > 1 → K_{eq} < 1$. Now let's write the equilibrium equation (assuming perfect gases as equation of state): $$\frac {P*y_{NH_{3}}} {(P*y_{H_{2}})^{\frac 3 2} * (P*y_{N_{2}})^{\frac 1 2}} = K_{eq} < 1$$

It seems to me that the fact of $K_{eq}$ being smaller than 1 is a meaningful information since I can increase the concentration of ammonia ($y_{NH_{3}}$) simply by increasing the Pressure. So if $P→inf$ then $y_{NH_{3}} → 1$ while $y_{H_{2}}$ and $y_{N_{2}} → 0$.

Conclusion: working at high pressure I have lots of ammonia even thought K_{eq} < 1. Why then should I worry about K_{eq}? I mean if I MUST have products no metter what I just have to work at higher pressure (it this example).

2. Nov 12, 2017

### Staff: Mentor

You just used the K value to predict the reaction will be not giving yields high enough, yet you ask what is K good for?

3. Nov 12, 2017

### dRic2

Yeah, but it's just for prediction. Working on pressure I can overcome the problem so it is like some information that can be useful but it is somehow irrelevant

4. Nov 12, 2017

### dRic2

Maybe I wasn't clear. K is vital to understand a reaction, but I don't get the point of my professor worrying about K being < 1. If you have to do something you just do it, I mean you write the equation and find values that fits your problem. You can get ammonia in high quantity even if K is < 1 so what is the point of giving so much important to the exact value of K ?

PS: I mean, in my example, working at high pressure the formation of ammonia is favoured even though K < 1

5. Nov 12, 2017

### Staff: Mentor

Actually it is not, even for really high pressures.

Typically optimizing the process on paper is much less expensive than doing the same by experimental trial and error.

6. Nov 12, 2017

### dRic2

Mhm, I found that $y_{NH_{3}} = 0.65$ @ T = 633K and P = 40kPa at equilibrium...

Does it mean is still not favoured?

7. Nov 12, 2017

### Staff: Mentor

I am under the impression amount of ammonia produced is quite low, but perhaps the numbers I recall seeing were results of the measurements of the real reaction taking place in flowing gas and not getting to the equilibrium.

8. Nov 12, 2017

### dRic2

Don't know about the whole precess yet, but I refer to this table I have (from my professor)... Sorry but I forgot where he got it

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9. Nov 13, 2017

### Lord Jestocost

The pressure P is not 40 kPa, but about 40000 kPa in this case!

10. Nov 13, 2017

### dRic2

Yes, thank you. I lost 10^-3 on the way... My point should have been clear anyway

11. Nov 13, 2017

### Lord Jestocost

In order to optimize an industrial process like the NH3 synthesis, one needs a lot more than saying: “Why should I worry about Keq. In case Keq < 1, simply increase the pressure.”

12. Nov 14, 2017

### dRic2

Perhaps I was to 'drastic' and I realized after posting mine was a silly question. Anyway what I was trying to say is that, since Keq is indipendent of pressure, I find it misleading to judge the equilibrium condition only by considering Keq.

Yes, there are lots of other problems concerning the Kinetics or the physics of the reaction, but working on the pressure to optimize a process is the very first thing that you try to do after you studied the thermodynamic of the system, I guess.

13. Nov 14, 2017

### Lord Jestocost

All you need to calculate equilibrium concentrations or equilibrium partial pressures is the equilibrium constant. Nothing more is needed for such evaluations. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. Have a look at http://www.chemguide.co.uk/physical/equilibria/change.html.

Last edited by a moderator: Nov 14, 2017
14. Nov 14, 2017

### dRic2

Yes, I know that. I don't know how can I express what I was thinking about (english is not my language). I don't question the equilibrium law, I was trying to say that, to me, it is meaningless to speculate about Keq without making calculations. Mine was a simpler question. In lot of chemistry and thermodynamics books you read that Keq "tells" you what's the equilibrium will be, yeah but only at $P=P_{ref}$. I was saying that those "considerations" (about Keq) that you read in books are meaningless (to me) because there are other factors to consider while studying the equilibrium of a system (and pressure is one example).

I realize it is a stupid question, I know, I posted it while I was anxious about an exam I had... I felt like I lost what I've learned. Now, being more focus, I acknowledge the frivolousness of this thread and, if some of you wasted time on this, I apologize.

15. Nov 15, 2017

### Lord Jestocost

Regarding this point, you are right.