Chemical equilibrium

K_c and K_p: A Clear Guide to Chemical Equilibrium Constants

Audience: Class 11–12 students, chemistry teachers, and learners preparing competitive exams.

Introduction

Chemical equilibrium describes a dynamic situation where the rates of the forward and reverse reactions are equal. At equilibrium, macroscopic properties (concentrations, pressure) remain constant. Two commonly used equilibrium constants are K_c (concentration-based) and K_p (pressure-based). Understanding these helps solve many equilibrium problems in physical chemistry.

What is K_c?

K_c is the equilibrium constant expressed in terms of molar concentrations (mol·L⁻¹). For a general reaction

aA + bB ⇌ cC + dD

the expression for K_c is

Kc = [C]c[D]d / [A]a[B]b

Only species in the gas or aqueous phase (soluble) appear in this expression. Pure solids and liquids are omitted because their concentrations do not change during the reaction.

What is K_p?

K_p is the equilibrium constant expressed using partial pressures (usually in atm) for gaseous reactions. For the same stoichiometric reaction where all species are gases:

Kp = (PC)c(PD)d / (PA)a(PB)b

Partial pressures are used because gases in a mixture contribute independently to the total pressure (Dalton's law).

Relation between K_p and K_c

K_p and K_c are related when gases are involved. The formula is:

Kp = Kc (RT)Δn

Here:

• R = 0.082057 L·atm·K⁻¹·mol⁻¹
• T = temperature in Kelvin
• Δn = moles of gaseous products − moles of gaseous reactants

If Δn = 0 (no net change in the number of gas moles), then Kp = Kc.

Why does the (RT)^Δn factor appear?

Convert concentration [A] (mol·L⁻¹) to partial pressure using the ideal gas law: P = [A]RT. Substituting pressures into the K_p expression and rearranging leads to the (RT)^Δn factor. Thus the relationship is a direct consequence of the ideal gas law and stoichiometry.

Worked Example 1 — Converting K_c to K_p

For the reaction:

2SO2(g) + O2(g) ⇌ 2SO3(g)

Δn = (2) − (2 + 1) = −1.

So Kp = Kc (RT)−1 = Kc / (RT).

At T = 298 K, using R = 0.082057, compute (RT) and divide K_c by this value to get K_p.

Worked Example 2 — Using K_p to find equilibrium partial pressures

Consider:

N2(g) + 3H2(g) ⇌ 2NH3(g)

Suppose K_p at the given temperature is 4.0 × 10−3. If initial partial pressures are P_N2 = 1.0 atm and P_H2 = 3.0 atm, and no NH₃ initially, let x be the change (atm) of N₂ consumed. At equilibrium:

PN2 = 1.0 − x, PH2 = 3.0 − 3x, PNH3 = 0 + 2x.

Write K_p = (P_NH3)² / (P_N2)(P_H2)³ and solve for x (usually by approximation or numerically when algebra becomes complex). Check that pressures remain positive.

Practical Tips & Common Mistakes

• Always check the phases — include only gases and aqueous species in K expressions; omit pure liquids and solids.
• Keep units consistent. K_c and K_p are usually reported as dimensionless in thermodynamic treatments by dividing by standard states, but in problems you may see units; follow your textbook or exam convention.
• Calculate Δn carefully: count only gaseous coefficients.
• If initial amounts include products, set up an ICE (Initial–Change–Equilibrium) table to track changes systematically.
• When K is very large or very small, approximations (neglecting x compared to initial values) are often valid — but check the approximation for consistency (x should be <5% of initial concentration or pressure for safe use).

Summary

K_c and K_p give a quantitative measure of where an equilibrium lies. K_c uses concentrations; K_p uses partial pressures. Their relationship, K_p = K_c(RT)^Δn, provides a bridge between concentration- and pressure-based descriptions. Mastering ICE tables and the K_p/K_c relation will make equilibrium problems straightforward and exam-ready.

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