Equilibrium Constant Calculator

Calculate the equilibrium constant (Keq) or reaction quotient (Q) for chemical reactions. Enter product and reactant concentrations along with their stoichiometric coefficients to determine equilibrium position and predict reaction direction.

Core Formulas: For general reaction: aA + bB ⇌ cC + dD Equilibrium Constant (Keq): Keq = [C]^c × [D]^d / ([A]^a × [B]^b) Reaction Quotient (Q): Q = [C]^c × [D]^d / ([A]^a × [B]^b) (uses current concentrations, not equilibrium) Predicting Reaction Direction: • Q < Keq → Reaction proceeds forward (→) • Q > Keq → Reaction proceeds backward (←) • Q = Keq → System at equilibrium (no net change) Temperature Dependence (van't Hoff equation): ln(K₂/K₁) = -ΔH°/R × (1/T₂ - 1/T₁) Relationship to ΔG°: ΔG° = -RT ln(Keq) where: • ΔG° = standard Gibbs free energy change • R = 8.314 J/(mol·K) • T = temperature (K) For Gas Reactions: Kp = Kc(RT)^Δn where Δn = (moles gas products) - (moles gas reactants)
Example 1 (Haber Process): N₂(g) + 3H₂(g) ⇌ 2NH₃(g) At equilibrium: [NH₃] = 0.20 M, [N₂] = 0.30 M, [H₂] = 0.10 M Keq = [NH₃]² / ([N₂][H₂]³) Keq = (0.20)² / [(0.30)(0.10)³] Keq = 0.04 / (0.30 × 0.001) Keq = 0.04 / 0.0003 = 133 (Products moderately favored) Example 2 (Water Ionization): H₂O ⇌ H⁺ + OH⁻ At 25°C: [H⁺] = [OH⁻] = 1.0×10⁻⁷ M Keq (Kw) = [H⁺][OH⁻] Kw = (1.0×10⁻⁷)(1.0×10⁻⁷) = 1.0×10⁻¹⁴ (Reactants strongly favored - water barely ionizes) Example 3 (Ester Formation): CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O At equilibrium: [ester] = 0.67 M, [H₂O] = 0.67 M [acid] = 0.33 M, [alcohol] = 0.33 M Keq = (0.67 × 0.67) / (0.33 × 0.33) Keq = 0.4489 / 0.1089 = 4.1 (Products favored) Example 4 (Reaction Quotient): For N₂ + 3H₂ ⇌ 2NH₃, Keq = 133 Current: [NH₃] = 0.10 M, [N₂] = 0.50 M, [H₂] = 0.20 M Q = (0.10)² / [(0.50)(0.20)³] Q = 0.01 / (0.50 × 0.008) = 0.01 / 0.004 = 2.5 Since Q (2.5) < Keq (133), reaction proceeds forward → Example 5 (Strong Acid Dissociation): HCl → H⁺ + Cl⁻ Keq ≈ 10⁶ (very large) Reaction goes essentially to completion [products] >> [reactants]

What is the equilibrium constant (Keq)?

The equilibrium constant (Keq or K) is a number that expresses the ratio of product concentrations to reactant concentrations at equilibrium, each raised to their stoichiometric coefficients. For aA + bB ⇌ cC + dD, Keq = [C]^c[D]^d / [A]^a[B]^b. A large Keq (>1) means products are favored; a small Keq (<1) means reactants are favored.

What is the reaction quotient (Q)?

The reaction quotient (Q) has the same form as Keq but uses current concentrations, not equilibrium concentrations. Q = [C]^c[D]^d / [A]^a[B]^b at any moment. Comparing Q to Keq predicts reaction direction: If Q < Keq, reaction proceeds forward (→). If Q > Keq, reaction proceeds backward (←). If Q = Keq, system is at equilibrium.

How do you calculate Keq from concentrations?

Measure equilibrium concentrations of all species. Raise each product concentration to its stoichiometric coefficient and multiply. Do the same for reactants. Divide products by reactants. Example: For N₂ + 3H₂ ⇌ 2NH₃, if [NH₃]=0.2M, [N₂]=0.3M, [H₂]=0.1M at equilibrium, Keq = (0.2)²/[(0.3)(0.1)³] = 0.04/0.0003 = 133.

What does Keq tell you about a reaction?

Keq indicates the position of equilibrium: Keq >> 1 (e.g., 10⁶): Reaction goes nearly to completion, mostly products. Keq ≈ 1: Significant amounts of both reactants and products at equilibrium. Keq << 1 (e.g., 10⁻⁶): Reaction barely proceeds, mostly reactants. Keq does NOT indicate how fast equilibrium is reached.

Does Keq change with concentration?

No! Keq is constant at a given temperature and does not change with initial concentrations. If you add more reactant, concentrations shift to maintain the same Keq value. However, Keq DOES change with temperature according to the van't Hoff equation: ln(K₂/K₁) = -ΔH°/R × (1/T₂ - 1/T₁).

What is the difference between Kc and Kp?

Kc uses molar concentrations (mol/L) and Kp uses partial pressures (atm or bar) for gases. They are related by: Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) - (moles of gaseous reactants), R = 0.0821 L·atm/(mol·K), and T is temperature in Kelvin. For liquids and solids, only Kc is used.

How do you use Q to predict reaction direction?

Calculate Q using current concentrations (same formula as Keq). Compare to Keq: Q < Keq → reaction shifts right (forward) to make more products. Q > Keq → reaction shifts left (reverse) to make more reactants. Q = Keq → system at equilibrium, no net change. This is Le Chatelier's principle in action.

What happens to Keq when you reverse a reaction?

When you reverse a reaction, the new equilibrium constant is the reciprocal of the original: K(reverse) = 1/K(forward). For example, if N₂ + 3H₂ ⇌ 2NH₃ has Keq = 100, then 2NH₃ ⇌ N₂ + 3H₂ has Keq = 1/100 = 0.01. This makes sense because products and reactants switch places.

Why are solids and pure liquids excluded from Keq?

Pure solids and pure liquids have constant "concentrations" (activity = 1) that don't change during the reaction, so they're incorporated into Keq itself and omitted from the expression. For example: CaCO₃(s) ⇌ CaO(s) + CO₂(g) has Keq = [CO₂], not [CO₂]/[CaCO₃]. Only gases and aqueous species appear in the equilibrium expression.

How does temperature affect Keq?

Temperature is the ONLY condition that changes Keq. For exothermic reactions (ΔH < 0), increasing temperature decreases Keq (shifts toward reactants). For endothermic reactions (ΔH > 0), increasing temperature increases Keq (shifts toward products). This follows the van't Hoff equation and explains why some reactions are favored at high or low temperatures.