Equilibrium Constant (Kc & Kp) — Formula & Calculations

The equilibrium constant tells you, at a fixed temperature, how a reversible reaction is balanced once it reaches equilibrium. In plain terms, it tells you whether products are favored, reactants are favored, or neither side is strongly favored.

For a general reaction

$$aA + bB \rightleftharpoons cC + dD$$

the usual introductory forms are

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

and, for gases written with partial pressures,

$$K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$$

Use $K_c$ when the problem gives equilibrium concentrations. Use $K_p$ when it gives equilibrium partial pressures for gases. In a full thermodynamics treatment, equilibrium constants are defined with activities, but in most first-course chemistry problems these concentration and pressure forms are the ones you use.

If $K \gg 1$, products are favored at equilibrium. If $K \ll 1$, reactants are favored. If $K$ is near $1$, neither side is strongly favored. That tells you about the equilibrium position, not the reaction speed.

What The Equilibrium Constant Formula Includes

Three rules do most of the work:

• Put products on top and reactants on the bottom.
• Turn coefficients in the balanced equation into exponents.
• Leave out pure solids and pure liquids in standard introductory equilibrium expressions.

That last rule matters. If a reaction includes a pure solid, changing how much of that solid is present does not make it appear in the usual $K$ expression.

When To Use $K_c$ Vs. $K_p$

Use $K_c$ when equilibrium values are written as concentrations, usually in $\mathrm{mol/L}$. Use $K_p$ when the equilibrium values are given as partial pressures of gases.

If a problem mixes both forms, first check what information you are actually given. Do not convert unless the problem requires it.

How $K_c$ And $K_p$ Are Related

For gas-phase equilibria in the usual introductory form,

$$K_p = K_c(RT)^{\Delta n}$$

where

$$\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}$$

Only gaseous species count in $\Delta n$. If $\Delta n = 0$, then $K_p = K_c$. If $\Delta n \neq 0$, they are generally different.

This relation is useful only when the equilibrium expression is built from gases. It is also temperature-dependent, because the value of the equilibrium constant itself depends on temperature.

Worked Example: How To Calculate $K_c$

Consider

$$\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$$

Suppose the equilibrium concentrations are

$$[\mathrm{N_2O_4}] = 0.20 \, \mathrm{M}, \qquad [\mathrm{NO_2}] = 0.40 \, \mathrm{M}$$

Write the expression first:

$$K_c = \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]}$$

Now substitute the equilibrium values:

$$K_c = \frac{(0.40)^2}{0.20} = \frac{0.16}{0.20} = 0.80$$

So for this temperature, the equilibrium is not strongly product-favored or reactant-favored. It leans slightly toward reactants because $K_c < 1$, but it is still of order $1$, not extremely small.

This same reaction also shows why $K_c$ and $K_p$ are not always equal. Here, $\Delta n = 2 - 1 = 1$, so in the usual gas-equilibrium form

$$K_p = K_c(RT)$$

If the problem gave equilibrium partial pressures instead of concentrations, you would calculate $K_p$ with the same exponents and the same product-over-reactant structure.

Common Mistakes With Equilibrium Constant Problems

• Using initial concentrations instead of equilibrium concentrations. The equilibrium constant expression uses equilibrium values.
• Forgetting exponents. In this example, the coefficient $2$ in front of $\mathrm{NO_2}$ becomes the exponent $2$.
• Including pure solids or pure liquids in the expression. In standard introductory chemistry, they are omitted.
• Assuming a large $K$ means a fast reaction. Equilibrium constants describe position at equilibrium, not how quickly equilibrium is reached.

Where The Equilibrium Constant Is Used

This idea appears anywhere reversible reactions matter: gas equilibria, acid-base chemistry, solubility, and industrial reaction design. It is especially useful when you want to compare product-favored and reactant-favored systems, or when you want to compare the reaction quotient $Q$ with $K$ to predict which way a system will move.

Try A Similar Equilibrium Constant Problem

Write the equilibrium expression for

$$\mathrm{H_2(g)} + \mathrm{I_2(g)} \rightleftharpoons 2\mathrm{HI(g)}$$

Then check $\Delta n$. Because the number of gaseous moles is the same on both sides, this is a good case for testing the shortcut $K_p = K_c$.