Haber Process

I need help to understand something about the Haber process ...

Why is it that text books say that a change in pressure does not change the position of equilibrium and does not affect the value of Kp ?
Yet quite plainly if you increase the pressure in the Haber process you will increase the yield of Ammonia as the system compensates for increased pressure by shifting to the side with least moles of gas thereby reducing vol/pressure effects ... ??
But how can Kp therefore stay the same ? Here is how it is worked out ...and following that I have gathered what data I can. But it is many years since I studied Physical Chemistry at University and apart from anything else I have hunted the internet for an answer and cannot find one. So barring buying a text book about gas laws, I am hoping there is someone out there who can help ! Thanks !

Total Pressure measured in kPa or atm or Nm-2 ....
(conversion chart for changing between units http://www.convert-me.com/en/convert/pressure/nmsq.html ...)

The total pressure of a mixture of gases equals the sum of the individual gas pressures.  Each partial pressure is the same percent of the total pressure as the percent each gas is of the total volume.

This table shows percentage of NH3 at various temps and pressure. At each temp you will see that the percentage of NH3 increases with increased pressure :
                                              Atmospheres of Pressure
                     10     25      50       100      200   300      400    500    1000
100ºC                    91.7    94.5    96.7     98.4              99.4
200ºC         50.7    63.6    74       81        89                 94.6              98.3
300ºC         14.7    27.4   39.5     52.5     66.7              79.7              92.6
400ºC          3.9      8.7     15.3    25.2     38.8     48     55.4    61      79.8
500ºC          1.2      2.9     5.6      10.6      18.3    26     32       38        57.5
600ºC                                                                   13     17        21
700ºC           0.2                1.1      2.2                                                12.9

at 500ºC  Kp = 1.45x10ˉ5  or 0.48 NH3
at 300ºC Kp = 4.34 x 10-3
at 400ºC Kp = 1.64 x10-4
at 450ºC Kp = 4.51 x 10 -5
at 472ºC Kp = 2.79 x 10-5  (7.38 atm H2 2.46 atm N2  0.166 atm NH3)
at 500ºC Kp = 1.45 x 10 -5
at 550ºC  Kp = 5.38 x 10 -6
at 600ºC Kp = 2.25 x 10-6

Partial pressure of a gas in a mixture of gases is defined as the pressure the gas would exert if it alone occupied the entire volume.

pV=nRT where p is partial pressure
R=0.0820575 atm dm3K-1mol-1  or 8.3142 JK-1mol-1  (the more common SI unit)

Le Chatelier
The effect of changing the pressure on a gas phase reaction depends on the stoichiometry of the reaction
In this case there are 4 on the left (reactants) and 2 moles on the right (products).
If we compress a system by a factor of 10 that is at 500ºC where Kp =1.4 x 10-5 we get the following:
Before Compression     After Compression 
PNH3=0.12 atm             PNH3 = 8.4 atm
  PN2   = 2.4 atm                PN2 = 21 atm
  PH2  = 7.3 atm                 PH2 = 62 atm

Before the system was compressed the partial pressure of NH3 was only about 1% of the total pressure. After the system is compressed the partial pressure of NH3 is almost 10% of the total. The system was under stress from the pressure and it has now shifted in the direction which minimises the effect of this stress. It shifts towards the products because this reduces the number of particles of gas, thereby decreasing the total pressure on the system.

But if we slot this data into the Kp equation how can the Kp value possibly stay the same ? Surely the number must increase using these new partial pressures ?

Boiling point of :
N2 = -196ºC
H2 = -253ºC
NH3 = -33ºC
so ammonia will liquefy first if cool mixture down.
Gas boiling points are higher at high pressure....

To get Kp from Kc
Kp = Kc(RT) to the power delta n where n is change is no of moles =2 in this occassion

Ideal Gas Law....
PV=nRT is initial pressure of one gas
pV=nRT to get partial pressure of one

This has loads of further info ...

and http://www.ausetute.com.au/partialp.html

Refs ;
  • A Case Study of the Haber Process
  • James Mungall tuition
  • Le Chatelier's Principle