Kc, Kp, and Their Relationship

There are two ways to express the equilibrium constant, and the MCAT expects you to know both. The difference is simply what you plug into the expression - concentrations or partial pressures.

K_c uses molar concentrations (mol/L) for all species. This is the version you have already seen.

K_p uses partial pressures (atm) for gaseous species. It is used when the problem gives you pressures instead of concentrations.

Both describe the same equilibrium. They just speak different “languages.”

When to Use Each

• Use K_c when the problem gives concentrations in mol/L
• Use K_p when the problem gives partial pressures in atm
• For dilute aqueous solutions, K_c = K_eq (they are interchangeable)

Example: Writing Kc and Kp

For the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

K_c = [NH₃]² / ([N₂][H₂]³)

K_p = ($P_{\text{NH}}$₃)² / ($P_{N}$₂ × ($P_{H}$₂)³)

Same structure, different units plugged in.

The Formula Connecting Kc and Kp

Key Insight: When Kp = Kc

When delta-n = 0 (the number of moles of gas is the same on both sides), then (RT)⁰ = 1, so Kp = Kc.

Example: H₂(g) + I₂(g) ⇌ 2HI(g)

delta-n = 2 - (1 + 1) = 0, so Kp = Kc for this reaction.

Worked Example

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), if Kc = 0.50 at 400°C, find Kp.

Step 1: Calculate delta-n = 2 - (1 + 3) = -2

Step 2: Convert temperature: T = 400 + 273 = 673 K

Step 3: Kp = Kc(RT)^delta-n = 0.50 × (0.0821 × 673)^(-2)

= 0.50 × (55.2)^(-2) = 0.50 × ($\frac{1}{3047}$) = 0.50 / 3047

$K_p \approx 1.6 \times 10^{-4}$

Because delta-n is negative (fewer moles of gas on the product side), Kp < Kc for this reaction.