Equilibrium Constant Calculator
Calculate the equilibrium constant (Kc) from product and reactant concentrations with stoichiometric coefficients. Also converts Kc to Kp and determines the standard Gibbs free energy change. See also our Gibbs Free Energy Calculator and Activation Energy Calculator for related thermodynamics computations.
Reaction: aA + bB ⇌ cC + dD
How to Calculate the Equilibrium Constant
The equilibrium constant (K) is a fundamental quantity in chemistry that describes the ratio of product concentrations to reactant concentrations at chemical equilibrium, each raised to the power of their stoichiometric coefficients. It was first formulated by Cato Guldberg and Peter Waage in 1864 as part of the law of mass action, which states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants.
At equilibrium, the forward and reverse reaction rates are equal, and the concentrations of all species remain constant over time. The equilibrium constant quantifies the position of this equilibrium — a large K indicates that products are favored, while a small K indicates that reactants are favored. The value of K depends only on temperature and is independent of the initial concentrations of reactants or products.
- Write the balanced chemical equation for the reaction.
- Identify the equilibrium concentrations of all products and reactants.
- Raise each concentration to the power of its stoichiometric coefficient.
- Calculate Kc = [products]^coefficients / [reactants]^coefficients.
- For gas-phase reactions, convert to Kp using Kp = Kc(RT)^Δn.
- Calculate ΔG° = -RT ln(K) to find the standard free energy change.
The distinction between Kc (concentration-based) and Kp (pressure-based) is important for gas-phase reactions. Kp uses partial pressures instead of molar concentrations. The relationship Kp = Kc(RT)^Δn connects the two, where Δn is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants), R is the gas constant (0.0821 L·atm/(mol·K)), and T is the absolute temperature in Kelvin.
Equilibrium Constant Formula
For aA + bB ⇌ cC + dD:
Kc = [C]^c × [D]^d / ([A]^a × [B]^b)
Kp = Kc × (RT)^Δn
Δn = (c + d) − (a + b) [for gaseous species only]
ΔG° = −RT ln(K)
K = e^(−ΔG°/RT)
Where:
[X] = molar concentration of species X at equilibrium
a, b, c, d = stoichiometric coefficients
R = 8.314 J/(mol·K) or 0.0821 L·atm/(mol·K)
T = temperature in Kelvin
ΔG° = standard Gibbs free energy change
Pure solids and pure liquids are excluded from the equilibrium expression because their concentrations (activities) are defined as 1. For example, in the decomposition of calcium carbonate: CaCO₃(s) ⇌ CaO(s) + CO₂(g), the equilibrium expression is simply Kp = P(CO₂). This is because the "concentration" of a pure solid or liquid is constant and is incorporated into the equilibrium constant itself.
Example Calculation
Problem: For the reaction A + B ⇌ C + D at 298 K, the equilibrium concentrations are [A] = 0.5 M, [B] = 0.5 M, [C] = 1.0 M, [D] = 1.0 M. Calculate Kc, Kp (Δn = 0), and ΔG°.
Given:
• [A] = 0.5 M, [B] = 0.5 M (reactants)
• [C] = 1.0 M, [D] = 1.0 M (products)
• All coefficients = 1, T = 298 K, Δn = 0
Solution:
Kc = [C]¹[D]¹ / ([A]¹[B]¹)
Kc = (1.0)(1.0) / (0.5)(0.5)
Kc = 1.0 / 0.25 = 4.0
Kp = Kc × (RT)^Δn = 4.0 × (0.0821 × 298)⁰ = 4.0 × 1 = 4.0
ΔG° = −RT ln(K) = −(8.314)(298) ln(4.0)
ΔG° = −(2477.6)(1.386) = −3435 J/mol = −3.44 kJ/mol
Answer: Kc = 4.0, Kp = 4.0, ΔG° = −3.44 kJ/mol (reaction favors products, spontaneous)
Equilibrium Constant Reference Table
| Reaction | K (at 25°C) | Favors |
|---|---|---|
| 2H₂(g) + O₂(g) ⇌ 2H₂O(g) | 3.3 × 10⁸¹ | Products (strongly) |
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | 6.0 × 10⁵ | Products |
| H₂(g) + I₂(g) ⇌ 2HI(g) | 54.3 | Products (moderately) |
| N₂O₄(g) ⇌ 2NO₂(g) | 4.63 × 10⁻³ | Reactants |
| 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) | 6.9 × 10⁻²⁵ | Reactants (strongly) |
| AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) | 1.8 × 10⁻¹⁰ | Reactants (Ksp) |
| CH₃COOH ⇌ CH₃COO⁻ + H⁺ | 1.8 × 10⁻⁵ | Reactants (Ka) |
| NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ | 1.8 × 10⁻⁵ | Reactants (Kb) |
| CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) | 3.4 × 10⁻² | Reactants |
| Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺ | 1.4 × 10³ | Products |
Frequently Asked Questions
What does the equilibrium constant tell us about a reaction?
The equilibrium constant indicates the extent to which a reaction proceeds to completion. K > 1 means products are favored at equilibrium (more products than reactants). K < 1 means reactants are favored. K = 1 means roughly equal amounts of products and reactants. Very large K values (like 10⁸¹ for water formation) indicate reactions that go essentially to completion, while very small K values indicate reactions that barely proceed under standard conditions.
How does temperature affect the equilibrium constant?
Temperature changes shift the equilibrium constant according to the van't Hoff equation: ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁). For exothermic reactions (ΔH < 0), increasing temperature decreases K, shifting equilibrium toward reactants. For endothermic reactions (ΔH > 0), increasing temperature increases K, shifting equilibrium toward products. This is consistent with Le Chatelier's principle — the system responds to oppose the imposed change.
What is the reaction quotient Q and how does it relate to K?
The reaction quotient Q has the same mathematical form as K but uses instantaneous concentrations rather than equilibrium concentrations. Comparing Q to K predicts the direction a reaction will shift: if Q < K, the reaction proceeds forward (toward products); if Q > K, the reaction proceeds in reverse (toward reactants); if Q = K, the system is at equilibrium. Q is a powerful tool for predicting whether a precipitate will form or whether a reaction will proceed spontaneously under given conditions.
Why are pure solids and liquids excluded from K expressions?
Pure solids and liquids have constant concentrations (or more precisely, constant activities equal to 1) regardless of how much is present. Adding more solid CaCO₃ to a decomposition reaction does not change the equilibrium position because the concentration of a pure substance is determined by its density and molar mass, which are fixed properties. Only species whose concentrations can vary (gases and dissolved solutes) appear in the equilibrium expression.
How do you calculate K for a reversed or multiplied reaction?
When a reaction is reversed, the new K is the reciprocal: K' = 1/K. When a reaction is multiplied by a factor n, the new K is raised to that power: K' = K^n. When reactions are added together, their K values are multiplied: K_total = K₁ × K₂. These rules allow you to calculate equilibrium constants for complex reactions from simpler, tabulated values. For example, if K for A → B is 10 and K for B → C is 5, then K for A → C is 50.
What is Le Chatelier's Principle?
Le Chatelier's Principle states that when a system at equilibrium is subjected to a stress (change in concentration, pressure, or temperature), the system shifts to partially counteract that stress and establish a new equilibrium. Adding more reactant shifts equilibrium toward products; increasing pressure favors the side with fewer gas moles; increasing temperature favors the endothermic direction. Note that adding a catalyst does NOT shift equilibrium — it only helps the system reach equilibrium faster by equally accelerating both forward and reverse reactions.
Understanding Chemical Equilibrium
Chemical equilibrium is a dynamic state where the forward and reverse reactions occur simultaneously at equal rates, resulting in no net change in concentrations over time. This concept is fundamental to understanding chemical behavior in nature, industry, and biological systems. Unlike a static situation where nothing happens, equilibrium involves continuous molecular-level activity — molecules are constantly reacting in both directions, but the macroscopic properties remain unchanged.
In industrial chemistry, equilibrium considerations determine the feasibility and optimization of chemical processes. The Haber process for ammonia synthesis (N₂ + 3H₂ ⇌ 2NH₃, K = 6 × 10⁵ at 25°C) is thermodynamically favorable at room temperature but kinetically too slow. At the operating temperature of 400-500°C needed for reasonable rates, K drops to about 0.04, requiring high pressures (150-300 atm) to shift equilibrium toward products. This illustrates the fundamental tension between thermodynamic favorability and kinetic accessibility that chemical engineers must navigate.
Biological systems maintain homeostasis through carefully controlled equilibria. The oxygen-hemoglobin binding equilibrium (Hb + 4O₂ ⇌ Hb(O₂)₄) is modulated by pH, CO₂ concentration, and temperature through allosteric effects. In the lungs (high O₂, low CO₂), equilibrium favors oxygen binding; in tissues (low O₂, high CO₂), equilibrium favors oxygen release. This elegant system ensures efficient oxygen delivery throughout the body without requiring energy input.
Environmental chemistry relies heavily on equilibrium concepts. The dissolution of CO₂ in ocean water involves multiple equilibria: CO₂(g) ⇌ CO₂(aq) ⇌ H₂CO₃ ⇌ HCO₃⁻ + H⁺ ⇌ CO₃²⁻ + 2H⁺. As atmospheric CO₂ increases, these equilibria shift, producing more H⁺ ions and lowering ocean pH — a process known as ocean acidification. Understanding these coupled equilibria is essential for predicting the environmental impact of rising CO₂ levels on marine ecosystems and coral reef dissolution.
Solubility equilibria (described by Ksp) govern the dissolution and precipitation of ionic compounds. These equilibria are critical in water treatment, geological processes, and pharmaceutical formulation. The common ion effect, where adding an ion already present in solution decreases solubility, is a direct application of Le Chatelier's principle to solubility equilibria. Selective precipitation, used in qualitative analysis and wastewater treatment, exploits differences in Ksp values to separate ions from solution.