Chemical Equilibrium: Kp Explained & Relation to Kc | CHEMCA

Chemical Equilibrium: K_p Explained & Relation to K_c

Published by Abhishek Sengar | CHEMCA India

We already know how to write the Equilibrium Constant (K_c) using molar concentrations. But when dealing with gas-phase reactions, calculating the exact molarity of a gas inside a large vessel can be tricky.

It is significantly easier to hook up a pressure gauge and measure the Partial Pressures of the gases! Because of this, chemists defined a second equilibrium constant specifically for gases: K_p. But how does this new constant relate to our old friend K_c? Let's derive the legendary formula.

Video Tutorial: Deriving the K_p - K_c Relationship

Watch Abhishek Sengar sir from CHEMCA decode how the Ideal Gas Law forms the perfect bridge between partial pressures and molar concentrations.

1. Writing the K_p Expression

For a general gaseous reaction: aA_(g) + bB_(g) ⇌ cC_(g) + dD_(g)

We write K_p exactly the same way we write K_c, but we replace concentration brackets [ ] with Partial Pressure symbols (P):

K_p = (P_C^c × P_D^d) / (P_A^a × P_B^b)

The "Gases Only" Rule!
Partial pressure ONLY exists for gases. If a reaction contains a pure solid or a pure liquid, it does not exert a meaningful partial pressure in the equilibrium mixture. You must completely ignore solids and liquids when writing K_p (assign them a value of 1).

Example: CaCO_3(s) ⇌ CaO_(s) + CO_2(g)
Correct: K_p = P_CO2

2. The Mathematical Derivation

How do we connect Pressure (P) to Concentration (C)? We use the Ideal Gas Law: PV = nRT.

Rearranging the Ideal Gas Law:
Divide both sides by Volume (V):
P = (n / V) × RT
The Concentration Connection:
Since moles divided by Volume (n/V) is the exact definition of Molarity (Concentration, C or [ ]), we can write:
P = [Concentration] × RT.
Therefore: P_A = [A]RT, P_B = [B]RT, etc.
Substitution:
Plug these values into the K_p formula. Separate the concentration terms from the RT terms. The concentration fraction becomes K_c, and the powers on RT subtract to form Δn_g.

Fig: Visually tracking how the ideal gas law converts partial pressure fractions into concentration fractions.

What is Δn_g?
Δn_g = (Total moles of Gaseous Products) - (Total moles of Gaseous Reactants).
Do not count solids or liquids when calculating this number!

Practice Questions for JEE & NEET

Let's test your ability to apply the formula and calculate Δn_g correctly!

Question 1: For the famous Haber process: N_2(g) + 3H_2(g) ⇌ 2NH_3(g), what is the exact mathematical relationship between K_p and K_c?

Answer: K_p = K_c / (RT)²

Reasoning:

First, calculate Δn_g.
Products: 2 moles of NH₃ gas.
Reactants: 1 mole N₂ + 3 moles H₂ = 4 moles of gas.
Δn_g = 2 - 4 = -2.

Now plug it into the formula: K_p = K_c(RT)^-2.
A negative exponent means we drop it to the denominator: K_p = K_c / (RT)².

Question 2: Under what specific condition will the numerical value of K_p be exactly identical to the numerical value of K_c? Give an example of such a reaction.

Answer: When Δn_g = 0. Example: H_2(g) + I_2(g) ⇌ 2HI_(g)

Reasoning:

If we look at the formula K_p = K_c(RT)^Δn_g, the only way K_p can equal K_c is if the (RT) term turns into the number 1.

Any number raised to the power of zero equals 1. Therefore, K_p = K_c ONLY when Δn_g = 0.

In the reaction H_2(g) + I_2(g) ⇌ 2HI_(g), we have 2 moles of gaseous product and 2 moles of gaseous reactants. Δn_g = 2 - 2 = 0. For this specific reaction, K_p and K_c are numerically identical!

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