Relation Between Kp and Kc

The equilibrium constant, k, is a number that describes the relationship between the number of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

K_p and K_c are equilibrium constants of ideal gas mixtures considered under reversible reactions. K_p 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 K_p K_c relation can be derived by understanding what are K_p and K_c. 

Let’s consider the general equilibrium equation:

A + B  ⇌ C + D

According to the law of mass action,

The rate at which A reacts ∝ [A]

The rate at which B reacts ∝ [B]

∴ The rate at which A and B react together ∝ [A][B]

So, the rate of the forward reaction = kf [A][B]

Where kf is the velocity constant for the forward reaction.

Now, the rate at which C and D react together ∝ [C][D] 

So, the rate of the backward reaction = k_b [C][D] 

k_b = velocity constant for the backward reaction.

At equilibrium, the rate of the forward reaction = rate of the backward reaction.

  kf [A][B] = kb [C][D]

  \[\frac{[C][D]}{[A][B]}=\frac{kf}{kb}\]

At constant temperature k_f and k_b are constant, therefore,\[\frac{kf}{kb}\]=k is also constant at a constant temperature.

Here, k is called the Equilibrium constant.

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Equilibrium Constant kc

Let us consider a general reversible reaction:

aA + bB ⇌ uU + vV

Applying the law of mass action here:

\[\frac{[U]^{u}[V]^{v}}{[A]^{a}[B]^{b}}\]= k or kc

In terms of expression of concentrations, k is written as kc.

This mathematical expression is called the law of chemical equilibrium.

Equilibrium Constant Definition

At a constant temperature, the equilibrium constant is the ratio of the product of the molar concentrations of the products, each raised to the power equal to its stoichiometric coefficient and the product of molar concentrations of the reactants, each raised to the power equal to its stoichiometric coefficient.

Do you know how to find equilibrium constant k_p for the reaction?

Let’s derive the equilibrium constant formula for gas-phase reactions:

The Equilibrium Constant k_p for the Reaction

When both reactants and products are in gaseous states, then we express equilibrium constant either in terms of concentration in moles per liter or partial pressures of the reactants and the products. 

Derivation:

The relation between Kp and Kc is given by the 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,

\[kc=\frac{C^{c}.D^{d}}{A^{a}.B^{b}}\]

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’

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’

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

                aA + bB ⇌ cC+ dD

 Then, equilibrium constant formula for 

Where pv, px, py, and pz are the partial pressures of V, X, Y, and Z, respectively.

The partial pressures are taken in the following units:

1. Atm

2. Bar

3. Pascal

Relation between  Kp and Kc

To derive a 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

Let us consider a general reversible reaction equation:

                         aA + bB ⇌ cC + dD

Substituting the above values in equation (a), and simplify:

\[kp=\frac{C^{c}.D^{d}RT^{c+d}}{A^{a}.B^{b}.RT^{a+b}}\]

Also, we may write equilibrium constant  (kc)  in terms of molar concentrations as;

So, from eq 

\[kp=\frac{C^{c}.D^{d}RT^{c+d}}{A^{a}.B^{b}.RT^{a+b}}\]

\[kp=kc \ast RT^{(c+d)-(a+b)}\]

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,

kp = kc \[(RT)^{\Delta n}\]

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.

In other terms, we have kc in molar concentration in the following manner:

\[kc=\frac{CC^{c}.CD^{d}}{CA^{a}.CB^{b}.}\]….(b) 

Ca, Cb, Cc, and Cd express the molar concentrations of A, B, C, and D, respectively. 

If we consider the gas as an ideal, then we can apply the ideal gas equation, that is:

pV = nRT or p = \[\frac{n}{v}\]RT = CRT

∵ \[\frac{n}{v}\] =\[\frac{\text{Number of moles}}{\text{Litre}}\]= C (Molar concentration)

∴ For the gases V, X, Y, and Z, we may write the equation (1) and (2) as;

            pa  = CaRT …(a)

            pb  = CbRT …(b)

            pc  = CcRT …(c)

            pd  = CdRT …(d)

Now, putting values of equations (a), (b), (c), and (d) in equation (2):

             kp  = \[\frac{(CcRT)^{c}.(CdRT)^{d}}{(CaRT)^{a}.(CbRT)^{a}}\]

                  = \[\frac{CC^{c}.CD^{d}.(RT)^{c+d}}{CA^{a}.CB^{b}.(RT)^{a+b}}\]

                  = \[\frac{CC^{c}.CD^{d}}{CA^{a}.CB^{b}}(RT)^{(c+d)-(a+b)}\]  

                  =  kc \[(RT)^{\Delta n}\]

 Where kc =\[\frac{CC^{c}.CD^{d}}{CA^{a}.CB^{b}}\]from equation (2), and

 Δn = (c + d) - (a + b) 

       = No. of moles of products - No. of moles of reactants

       = Change in the number of moles

Hence,             

Where R is the Universal Gas Constant whose value is 0.821 liter-atm per degree kelvin-mole, and

T = Temperature in degree Kelvin (°K).   

Here, 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

Equilibrium Constant Units     

For the general reaction: aA + bB ⇌ cC + dD               

kc =\[\frac{CC^{c}.CD^{d}}{CA^{a}.CB^{b}.}\] = \[\frac{(MolL^{-1})^{c+d}}{Mol L^{-1})^{a+b}}\] = \[(Mol L^{-1})^{(c+d)-(a+b)}=(Mol L^{-1})^{\Delta n}\]

So, the unit of kc is\[Mol L^{-1}\], and

kp = \[\frac{PC^{c}.PD^{d}}{PA^{a}.CP}\]=  \[\frac{(atm)^{(c+d)}}{(atm)^{(a+b)}}\]= \[(atm or bar)^{(c+d)-(a+d)} or (atm or bar)^{\Delta n}\]

So, the unit of kp = atm or bar.

Application of Equilibrium Constant

One of the applications of the equilibrium constant is to predict the extent of reaction. 

The magnitude of the equilibrium constant gives an idea of the relative amounts of the reactants and products.

For example, consider the following reversible equation and hence calculate Kp and Kc and derive the relationship between Kp and Kc:

                H2 (g) + Br2 (g) ⇌ 2 HBr(g) (kp = 3 x 1019)

                H2 (g) + Cl2 (g) ⇌ 2 HCl(g) (kp = 5 x 1029)

Here, values of kp are very high, i.e., reactions go almost to completion. 

Another Example:

  H2  + I2   ⇌ 2 HI

Solution:

Given the reversible equation,

  H2  + I2   ⇌ 2 HI

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

 \[kc= \frac{HI^{2}}{H^{2}I^{2}}\]

Solved Examples on Kp and Kc

Example 1: For the reaction,

N2O4(g) ⇌ 2 NO2(g) 

The concentration of the equilibrium mixture at 293 K of N2O4 is 5 x 10-8mol/L, and of NO2 is 2 x 10-6mol/L. Find the value of the equilibrium constant.

Applying the formula, k =\[\frac{[NO]^{2}}{[N2O4]}\]

Taking the concentrations w.r.t. standard state concentration of 1 mol/L:

k =\[\frac{(2x10^{-6})^{2}}{5x10^{-8}}=8x10^{-5}\]

Example 2: For the reaction,

N2 (g) + 3H2 (g) ⇌ 2 NH3(g)

If pN2 = 0.30 atm, pH2 =  0.20 atm, and pNH3 = 0.40 atm, then what is the value of kp?

Using the formula, kp =\[\frac{pNH3^{2}}{pN2.pN2^{2}}\]

=\[\frac{(0.4)^{2}}{(0.3).(0.2)^{2}}\]= 13.3 atm