Equilibrium Constant (Kc & Kp) — Formula & Calculations

At a fixed temperature, the equilibrium constant tells you whether a reversible reaction settles toward products, toward reactants, or neither: $K \gg 1$ means products are favored, $K \ll 1$ means reactants are favored, and $K$ near $1$ means neither side strongly wins. That is a statement about position, not speed.

$K_c$ vs. $K_p$ side by side

For a general reaction $aA + bB \rightleftharpoons cC + dD$:

                K_c                          K_p
Built from       equilibrium concentrations   equilibrium partial
                 [C]^c[D]^d / [A]^a[B]^b       pressures (P_C)^c(P_D)^d
                                               / (P_A)^a(P_B)^b
Use when         problem gives concentrations  problem gives gas
                 (usually mol/L)               partial pressures
Species counted  all (omit pure solids/liquids) gases
Related by        K_p = K_c (RT)^Δn,  Δn = (gas product mol) − (gas reactant mol)

A full thermodynamics treatment defines these with activities, but in a first course the concentration and pressure forms are the ones you use. Three rules build either expression: products on top and reactants on the bottom, balanced-equation coefficients become exponents, and pure solids and pure liquids are left out.

When to use $K_c$ vs. $K_p$, and how they relate

Use $K_c$ when equilibrium values are concentrations, $K_p$ when they are gas partial pressures. If a problem mixes forms, check what you are actually given before converting. For gas-phase equilibria,

Only gaseous species count in $\Delta n$. If $\Delta n = 0$ then $K_p = K_c$; otherwise they generally differ. The relation applies only when the expression is built from gases, and the constant itself is temperature-dependent.

Worked example: calculating $K_c$

Consider

with equilibrium concentrations $[\mathrm{N_2O_4}] = 0.20\,\mathrm{M}$ and $[\mathrm{NO_2}] = 0.40\,\mathrm{M}$. Write the expression first,

then substitute:

At this temperature the equilibrium is neither strongly product- nor reactant-favored. It leans slightly toward reactants since $K_c < 1$, but stays of order $1$. The same reaction shows why $K_c$ and $K_p$ can differ: here $\Delta n = 2 - 1 = 1$, so $K_p = K_c(RT)$. Given partial pressures instead, you would build $K_p$ with the same exponents and the same product-over-reactant structure.

Choosing and computing without the usual slips

This idea applies wherever reversible reactions matter: gas equilibria, acid-base chemistry, solubility, and industrial reaction design. It is most useful for comparing product-favored and reactant-favored systems, or for comparing the reaction quotient $Q$ with $K$ to predict which way a system moves. The recurring slips:

• Using initial concentrations instead of equilibrium concentrations.
• Forgetting exponents. The coefficient $2$ on $\mathrm{NO_2}$ becomes the exponent $2$.
• Including pure solids or liquids, which are omitted in standard introductory expressions.
• Reading a large $K$ as a fast reaction. $K$ describes position, not how quickly equilibrium is reached.

A no-conversion case worth testing

Write the expression for

and count $\Delta n$. The gaseous moles match on both sides, so $\Delta n = 0$ and $K_p = K_c$, making this the cleanest case to confirm when the conversion factor drops out entirely.