- 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
**K**_{P}**= K**...........................(eq 1)_{C}(RT)^{∆n }

####
Relationship between **K**_{P} and **K**_{n}

_{P}

_{n}

...........................(eq 2)

From ideal gas equation,PV=nRT

=>P=n(RT/V)

Where,

**n**is the number of moles
So,

**P**_{A}=n_{A}(RT/v), P_{B}= n_{B}(RT/V), P_{C}=n_{C}(RT/V) and P_{D}=n_{D}**(RT/V)**

**Replacing equation 2 by the above value we get that,**

**[n**

_{C}(RT/V)]^{c }. [n_{D}(RT/V)]^{d }

**K**

_{p }**= -------------------------------------**

**[n**

_{A}(RT/V)]^{a }. [n_{B}(RT/V)]^{b }**n**

_{C}^{c }. n_{D}^{d }**= >**

**K**

_{p}**=**

**-------------- . ( RT/V )**

^{(c + d) – (a + b)}**n**

_{A}^{a}. n_{B}^{b}**= > K**

_{p}= K_{n}. (RT/V)^{∆n}##
= > K_{p} = K_{n} . (P_{T}/n_{T})
^{∆n }

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

#### Relation between Kp and Kx

_{A = }**x**

_{A . }**P**

_{T}
P

Putting the values in whole equation:

##
= > K

_{B = }**x**_{B . }**P****P**

_{C = }**x**

_{C . }**P**

_{T}**P**

_{D = }**x**

_{D . }**P**

_{T}

Putting the values in whole equation:

(

**x**_{C . }**P**_{T })**. (**^{c}**x**_{D . }**P**)_{T}^{d}
K

**= ---------------------------------**_{P }
(

**x**_{A . }**P**)_{T}**. (**^{a}**x**_{B . }**P**)_{T}^{b}**x**

_{C}

^{c}

_{ . }**x**

_{D}

^{d}**= --------------**

_{ }**P**

_{ }

_{T}

^{(c+d) - (a+b)}**x**

_{A}

^{a}

_{ . }**x**

_{B}

^{b}##
= > K_{P }= K_{x }**(P**_{T}**)**** **^{∆n }

_{P }

_{x }

^{∆n }

**Q.Check your understanding.**

**N**

_{2}O_{4}(g)=2NO_{2}(g) ,Calculate K_{c }_{and}K_{x }for the given reaction for which Kp=0.157
Nyc

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ReplyDeleteWhen number of moles in both product side and reactant side are equal,i.e. when ∆n=0, all the equilibrium constants become equal.

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