Equilibrium Constant Calculator

For a balanced reversible reaction

$a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$

the equilibrium constant in terms of concentration is

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

• $[\ ]$ — equilibrium molar concentrations, never initial ones.
• The exponents are the balancing coefficients, so $K_c$ belongs to one specific way of writing the equation.
• Pure solids and pure liquids are omitted, along with the solvent in a dilute solution, because their activities are 1.

For gases the same reaction has a pressure-based constant $K_p$, built from partial pressures. The two are related by

$K_p = K_c (RT)^{\Delta n}, \qquad \Delta n = (c + d) - (a + b)$

with $R = 0.082057\ \mathrm{L\,atm\,mol^{-1}K^{-1}}$, $T$ in kelvin, and $\Delta n$ the change in moles of gas. When $\Delta n = 0$ the two are numerically equal.

What this assumes. The system has actually reached equilibrium at a stated temperature, and concentrations approximate activities. $K$ changes only with temperature — not with pressure, catalysts or starting amounts.

Reading K

A large $K$ means products dominate at equilibrium; a small $K$ means reactants do. $K$ is dimensionless by convention, because each concentration is really a ratio to a standard state.

Q versus K

The reaction quotient $Q$ uses the same expression with whatever concentrations exist right now. If $Q < K$ the reaction runs forward; if $Q > K$ it runs in reverse; if $Q = K$ it is at equilibrium.

The ICE table

1. Initial — write the starting concentrations.
2. Change — express every change as $\pm$ a coefficient times $x$.
3. Equilibrium — add the two rows.
4. Substitute into $K_c$ and solve for $x$.

Step 4 usually produces a quadratic. When $K$ is very small compared with the initial concentration, the approximation $C - x \approx C$ saves the algebra; check afterwards that $x$ is under about 5% of $C$. If the expression is a perfect square — as it is for $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ — take the square root of both sides instead.

Significant figures

$K$ carries the fewest significant figures of the concentrations used. Keep extra digits inside the ICE table and round only the final concentrations.

• Using initial concentrations in $K_c$. The expression takes equilibrium values; that is the entire purpose of the ICE table.
• Including a solid or a pure liquid. For $\mathrm{CaCO_3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO_2}(g)$, $K_p$ is just $p_{\mathrm{CO_2}}$.
• Forgetting the exponents. In $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$, hydrogen is cubed and ammonia squared. Dropping a power changes the answer by orders of magnitude.
• Ignoring how $K$ transforms. Reversing a reaction inverts $K$; doubling the coefficients squares it; adding reactions multiplies their constants.
• Using the wrong $\Delta n$. Only gases count in $\Delta n$ for $K_p = K_c(RT)^{\Delta n}$, and $T$ must be in kelvin.
• Keeping a negative root. A concentration cannot be negative; discard that solution of the quadratic.
• Thinking a catalyst changes $K$. It speeds up both directions equally and shifts nothing.