Converting Between Kc and Kp

**luna** · 09-26-08, 12:30 AM

N2(g)+3H2(g) =2NH3(g)
Kc 9.60 at 300C. Calculate Kp for this reaction at this temperature.

UM

**reject_wolf** · 09-26-08, 01:09 AM

Sorry, this is the BF2 board, not the Chemistry board. I actually would have been able to help you had the question been a bit more specific and typed out better though...

**DJ Ms. White** · 09-26-08, 01:13 AM

Errr...by Kc and Kp, are you by chance talking about specific heats? Define your nomenclature. Different disciplines have different meanings for the same symbols.

**Cainun** · 09-26-08, 01:14 AM

Umm, your first post is a chemistry question... Whats the deal, you stopping by here to ask a question of a gaming community? Or is that some sort of code for 'HI THERE'

**DTMiglin** · 09-29-08, 07:37 AM

I think he made a wrong turn at Albequerque.

**capnemo314** · 09-30-08, 04:37 PM

Well, I'll just show you the calculations. I'd presume you've supposedly gone over it in class or something.

Given:
N2(g)+3H2(g) =2NH3(g)

Kc=9.60 at 300 C.

So, hopefully you know that PV=nRT...

Or, if you tinker with the equation a bit:
P=RT*(n/V)

...and n/V is the same thing as the concentration!

Continuing on:
Kc = ([NH3]2)/([N2][H2]3)

So, to convert the concentrations into partial pressures, each concentration must be multiplied by RT...
Kp = (([NH3]RT)2)/([N2]RT([H2]RT)3)

Simplifying that a bit, in this case you'll end up with...
Kp=Kc*((RT)2)/((RT)4)

Kp= Kc/(RT)2

Finally, plugging in the numbers, you'll end up with...
Kp= 9.60/(0.0821*573)2
Kp= 2.21x103

Of course, I could have lost some vital bit of the equation due to all of the obnoxious coding... :P

I showed you the general idea at the beginning anyway, P=RT*(n/V). Thus, you can freely convert between the concentration (n/V) and the pressure. The only thing to worry about is keeping the powers of RT straight.

EDIT: Erm, welcome to TTP. I suppose I missed this thread a couple of days ago...