# Relation Between Kp and Kc

Kp and Kc are equilibrium constants of ideal gas mixtures considered under reversible reactions. Kp is an equilibrium constant written with respect to the atmospheric pressure and the Kc is the equilibrium constant used with respect to the concentrations expressed in molarity. The Kp Kc relation can be derived by understanding what are Kp and Kc.

## Derivation:

The relation between Kp and Kc is given by following simple derivation. To derive the relation between Kp and Kc, consider the following reversible reaction:

‘a’ mole of reactant A is reacted with ‘b’ mole of reactant B to give ‘c’ moles of product C and ‘d’ moles of product D,

aA + bB ⇌ cC + dD

Where a,b,c, and d are the Stoichiometric coefficients of reactants A, B and products C, D.

### What is Kc?

Kc is the equilibrium constant for a reversible reaction and it is given by,

K$_{c}$ = [C]$^{c}$[D]$^{d}$/[A]$^{a}$[B]$^{b}$........(1)

Where,

[C] - The molar concentration of product ‘C’

[D] - The molar concentration of product ‘D’

[A] - The molar concentration of reactant ‘A’

[B] - The molar concentration of reactant ‘B’

Similarly,  Kp is the equilibrium constant in terms of atmospheric pressure and is given by the expression:

K$_{p}$ = [PC]$^{c}$[PD]$^{d}$/[PA]$^{a}$[PB]$^{b}$............(2)

Where,

[PC] - Partial pressure of product ‘C’

[PD] - Partial pressure of product ‘D’

[PA] - Partial pressure of reactant ‘A’

[PB] - Partial pressure of reactant ‘B’

To derive relation between  Kp and Kc, consider the ideal gas equation,

PV = nRT

Where,

P - Pressure of the ideal gas

V - Volume of the ideal gas

n - Number of moles

R - Universal gas constant

T - Temperature

On rearranging the above equation for P,

P = nRT/V…………..(3)

We know that the ratio number of moles per unit volume is the molar concentration of the substance, hence we can write the pressure equation as:

P = [molar concentration] RT ………………..(4)

Therefore the partial pressures of A, B, C and D can be calculated by using equation (4):

⇒ [PA] = [A] RT

⇒ [PB] = [B] RT

⇒ [PC] = [C] RT

⇒ [PD] = [D] RT

Substituting above values in equation (2) and simplify:

⇒ K$_{p}$ = [C]$^[c]$[D]$^[d]$(RT)$^[c+d]$/[A]$^[a]$[B]$^[b]$(RT)$^[a+b]$

⇒ K$_{p}$ = K$^[c]$(RT)$^{(c+d)-(a+b)}$

Where,

c + d - Number of moles of product = np

a + b - Number of moles of reactant = nr

Therefore,

(c + d) - (a + b) = np - nr = Δng

Thus we get the relation between Kp and Kc,

⇒ K$_{p}$ = K$_{c}$ (RT)$^{\Delta ng}$

Where,

Δng - Change in gaseous moles of reactant and the product.

This is the required expression that gives the relation between the two equilibrium constants. The relation between Kp and Kc Pdf can be downloaded. Depending on the change in the number of moles of gas molecules, Kp and Kc relation will be changing.

Case-1:

If Δng = 0, i.e., if the change in the number of moles gas molecules in the equation is zero.

Then Kp=Kc.

Case-2:

If the change in the number of moles of gas molecules is positive, i.e., if Δng > 0 then,

Kp>Kc.

Case-3:

If the change in the number of moles of gas molecules is negative, i.e., if Δng < 0 then,

Kp<Kc.

### Example:

Consider the following reversible equation and hence calculate Kp and Kc and derive the relationship between Kp and Kc:

H₂ + I₂ ⇌ 2HI

Solution:

Given the reversible equation,

H₂ + I₂ ⇌ 2HI

The change in the number of moles of gas molecules for the given equation is,

⇒Δn = number of moles of product - number of moles of reactant

⇒Δn = 2 - 2 = 0

Therefore, Kp=Kc

Then, Kc and Kp of the equation is calculated as follows,

⇒ K$_{c}$ = [HI]$^{2}$/[H]$^{2}$[I]$^{2}$