Chemqueries: Relationship between Kp ,Kc, Kx and Kn

## Sunday, 4 November 2018

### Relationship between Kp ,Kc, Kx and Kn

• Kp=Equilibrium constant in terms of partial pressure.
• Kc=Equilibrium constant in terms of concentration.
• Kx=Equilibrium constant in terms of  mole fraction.
• Kn=Equilibrium constant in terms of  number of moles.
•  Kp and Kc are related as  KP = KC (RT)∆n ...........................(eq 1)

#### Relationship between KP and Kn

...........................(eq 2)
From ideal gas equation,PV=nRT
=>P=n(RT/V)
Where, n is the number of moles

So,  PA =nA(RT/v), PB = nB(RT/V), PC =nC(RT/V) and PD =nD(RT/V)

Replacing equation 2 by the above value we get that,
[nC(RT/V)]c  . [nD(RT/V)]d
K=  -------------------------------------
[nA(RT/V)]a  . [nB(RT/V)]b
nCc . nDd
= > Kp =  -------------- .  ( RT/V ) (c + d) – (a + b)
nAa. nBb
= >  Kp = Kn . (RT/V) ∆n

## = >  Kp = Kn . (PT/nT) ∆n

∆n=number of gaseous moles of product – number of gaseous moles of reactant

#### Relation between  Kp and Kx

From  above (eq 2)

Partial pressure(P) = Mole fraction(x) . Total pressure(PT)
So,   PA =  xA .  PT
PB =  xB .  P
PC =  xC .  PT
PD =  xD .  PT

Putting the values in whole equation:
( xC .  P)c . ( xD .  PT )d
K= ---------------------------------
( xA .  PT )a . ( xB .  PT )b

xCc .  xDd
=  --------------   PT(c+d) - (a+b)
xAa .  xBb

## = >     KP  = Kx  (PT) ∆n

N2O4(g)=2NO2(g) ,Calculate Kand Kfor the given reaction for which Kp=0.157