Both K

_{P}and K_{c}are equilibrium constants,but K_{c}is defined by molar concentration####
but K_{P} is defined by partial pressure.

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Derivation of K_{P} :

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- Consider the following reversible reaction

aA(g) + bB(g) ⇔ cC(g) + dD(g)

- If equilibrium constant for the reaction is expressed in terms of the molar concentration ,

- If the equilibrium involves gaseous species, then concentration may be expressed in terms of partial pressure of the reactants and products and it is represented as

......................... (eq1)

**P**

_{A , }**P**

_{B },**P**

_{C}and**P**are the partial pressure of A,B,C and D respectively.

_{D}####
Relationship between K_{P} and K_{c} :

We know that ,for a ideal gas pressure is directly proportional to the concentration assuming volume and temperature constant.So, PV = nRT

Divide both sides by V,

P=(n/V)RT

P=CRT

Where,C is concentration

n is the number of moles

V is the volume

R is ideal gas constant,and

n/V is the molar concentration.

Thus the partial pressures of individual gases A, B, C and D are:

**P**

_{A}= [A]RT;**P**

_{B}= [B]RT;**P**

_{C}= [C]RT;**P**

_{D}= [D]RTSubstituting these value in equilibrium constant expression equation 1, we have

**K**

_{P}**= K**

_{C}(RT)^{(c+d)-(a+b)}

^{}**K**

_{P}**= K**

_{C}(RT)^{∆n }where ∆n = (c+d) – (a+b)

=number or gaseous moles of the product-number of gaseous moles of the reactant

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